ML Flashcards

Question

ML: What is the naive bayes assumption, both in words and mathematically?

Answer 1

In words, the naive bayes assumption is that there are *no interactions* between the features of n-dimensional feature vector X, or that the *context* of one of the features X_i, and how it interacts with other features X_j, is not relevant to prediction. (An example is assuming seeing "Donald Trump" is has the same predictive value as seeing those two words 10 words apart and in reverse order.) Mathematically, we have for D-dimensional X:

Answer 2

Generative

Answer 3

For each Y=y, we need to estimate P(Y=y), as well as P(X_i=x_i|Y=y) for every possible value of X_i. P(Y=y) is generally estimated through MLE. So, we just estimate it as the proportion of the training examples for which Y=y. For P(X_i=x_i|Y=y) it can vary. You could also do basic population MLE (The proportion of X_i=x_iamong training examples where Y=y), or you could learn a gaussian N(µ,σ²) as with *gaussian naive bayes*, or something else!

Answer 4

The disadvantage is that we cannot capture context, which is often very important. Hence this algorithm being "naive". The advantage is that it's simple, and is much easier to train: because we ignore interactions, we only need to learn a bunch of univariate distributions rather than one giant *joint* distribution, which is much harder (I believe exponentially harder).

Answer 5

We're finding n linear (or hyperplane) separators, one for each of the K possible Y=k which models P(Y=k) vs P(Y not k). As such, it looks like a bunch of k linear (or hyperplane) separators working together on the cartesian grid.

Answer 6

At each step, simply greedily add the split that most decreases classification error on the training set (or MSE in the case of regression).

Answer 7

Pruning. Once you have your full tree trained on training data, at each step, greedily remove the branch that causes the largest improvement on validation error. Continue until you can't find any more improvements to validation error.

Answer 8

Simply work your way a branch until you get to the bottom. For classification, return the most common label among the training examples at that branch. For regression, return the average training label.

Answer 9

They're very interpretable, especially to non-technical people

Answer 10

They don't predict very well. Variance is very high, as small changes in training data can cause big differences in how the splits occur. They also have a fair amount of bias, as they can only capture certain types of decision bounds (no "diagonal" decision bounds). They also handle categorical predictor variables poorly if there are too many predictors.

Answer 11

If the variable has N possible valuest each split, it must consider each of the 2^N-1 subsets of the values as a splitting rule, which is of course very computationally hard.

Answer 12

Bias decreases (as we increase complexity), and variance increases.

Answer 13

Several models are learned, and they vote on what the prediction is.

Answer 14

In boosting, you fit several (typically very basic) classifiers *iteratively*, with each new classifier trying to do well on the training points that the *previous classifiers did poorly on.*

Answer 15

Because it uses very simple predictors (for example, stumps of decision trees), each one individually has *very high bias*. But, by fitting a lot of them and correcting for previous mistakes, *bias is decreased*.

Answer 16

For each new point, you find the K points in the training set closest to it (by some measure such as euclidean distance); then for classification you'd return the most common label among the nearest neighbors, and for regression you'd return the average label of the nearest neighbors.

Answer 17

We start with each training example X_i having equal weights w_i. Then, repeat: 1. Fit a new classifier that minimizes weighted training error, based on the weights w_i. (Points on which earlier classifiers have performed poorly have higher weights, and errors on those points cost more) 2. Based on the weighted error of that classifier, find a weight alpha_j for the *classifier* for when it votes later on. The lower its weighted error, the higher its alpha_j, and thus the higher its voting power. 3. Update each of the point weights w_i based on this round's results.

Answer 18

At each step, we find the current gradient of the loss function. We then find our next classifier to "approximate this gradient vector". So we in some way optimally take into account the gradient of the loss function at each step as we choose our classifiers.

Answer 19

Because the 0/1 loss function is not smooth, as well as nonconvex, optimizing the solution is very hard, and you *also* can't approximate it with iterative techniques like gradient descent (which is generally the solution when finding the true global minimum is impossible). To combat this, you can instead iteratively minimize some *similar* loss function that puts an upper bound on 0/1 loss, such as hinge loss (pink), or exponential loss (yellow).

Answer 20

Resample several samples from your training data; say m bootstrapped samples of size n, with replacement. Train a (typically complex) model on each of the m bootstrapped samples. Have these models vote on the prediction.

Answer 21

The classifiers in bagging are complex; if they're trees, they're deep trees. As a result, the individual classifiers have low bias, but high variance. However, when we use several classifiers and have them vote, we can drive down this variance because we're averaging votes from classifiers trained on different bootstrapped samples. (As n goes infinity, these samples become independent.)

Answer 22

Boosting fits several models that are not complex, and thus have high bias. But by fitting several and correcting errors from previous iterations, bias is reduced. Bagging fits several *complex* models, so bias is lower but variance is high. But by averaging many predictions on slightly different datasets, we can lower variance.

Answer 23

Rather than just picking the class with the most votes from each of the classifiers, average the *probabilities* of each class from each of the classifiers, then predict the class with the highest probability. (If we get to the end of a branch in one of the classifiers, the "probability" of class j predicted by that classifier is the proportion of training examples that had class j). This improvement makes performance more consistent, decreasing variance a bit.

Answer 24

Random Forests work basically just by using bagging with deep decision trees: bootstrapping several resampled training sets, training a deep tree on each of them, and having them vote for a classification or regressed value. The key change from traditional bagging is this: if there are p features of a training example X, at each split in a decision tree, only m \< p features are considered for that split. (Typically m = sqrt(p))

Answer 25

By only considering m \< p features at each split of a decision tree, the trees become more independent, which helps the random forest algorithm decrease the variance in the overall prediction (which is the general goal of bagging).

Answer 26

The models lack interpretibility. Training and prediction both take a long time.

Answer 27

You use *out-of-bag* error. Because each classifier is trained on a bootstrapped training set, each point in the original training set was *not* used in training some of the classifiers (about 36.8%). So, get a prediction on this point from these classifiers, and find the error on all the points from that method. This is preferable to cross-validation because *you don't need to retrain the model for each of the folds*; training takes a long time for random forests, so we want to avoid it. But with enough trees, out-of-bag error tends to be very similar to CV error.

Answer 28

Variable-importance measures look at how important a variable is to the predictions of the model. They're commonly used for random forests, because random forests lack interpretability without them. And more recently in my experience, they're big for xgboost. Ensemble tree methods just lend themselves to variable importance metrics, because you can look at how often a tree chooses to split on a given variable, how the model performs with and without trees that use a certain variable, etc.

Answer 29

1. Find all of the splits which use that variable in all of the decision trees, and then find how much on average those splits improve the predictions, using some measure of prediction performance like Gini index. 2. Randomly permute each variable one at a time, and see how much model performance decreases for each. (I'm guessing this means to calculate out-of-bag error, and for each point, before getting the prediction, choose the value of the variable in question randomly.)

Answer 30

Clustering is an unsupervised learning technique, so our training data are not labeled, and we try to divide the data into groups that are in some way "similar". So basically, we try to label each point such that similar points have the same label. Classification is similar because each point has a label, but classification is supervised: we are given a labeling scheme and want to learn how to predict it as best as possible. With clustering, we aren't given a labeling scheme, and want to find a sensible one.

Answer 31

After choosing your number of classes k, randomly initialize the locations of your k cluster centers. Then, repeat: 1. Assign all training points to the cluster center that is closest by euclidean distance 2. Set cluster center as the average of all points in the cluster

Answer 32

1. It's nondeterministic due to the random initializations, and is not guarenteed to reach the global optimum. 2. Cluster center isn't interpretable, as it's an average of a bunch of points

Answer 33

It will always converge, as each step will decrease the in-cluster variance (until it's at a local minimum)

Answer 34

Here the cluster centers are actual points in the dataset, rather than an average of a bunch of points. So we initialize k points randomly to be the initial cluster centers, then repeat: 1. Assign all training points to the cluster center (a real point) that is closest to it 2. For each cluster, choose the point in the cluster which minimizes mean squared distance from other points as the new cluster center. (In other words, minimize in-cluster variation.)

Answer 35

K-medoids provides more interpretable cluster centers, as they are actual points in the dataset rather than an average. However, this is at the cost of having slightly higher in-cluster variation for k-medoids, as you simply have "fewer choices" of where to put your cluster center.

Answer 36

Each training point starts as its own cluster. Then at each step, we merge the 2 clusters with the highest similarity into one. We use linkages to determine similarity.

Answer 37

1. Single linkage: Dissimilarity between 2 clusters is the distance between the two points *closest* to one another across groups 2. Complete linkage: Dissimilarity is the distance between the two points *farthest* to one another across groups 3. Average linkage: Dissimilarity is the average distance across all pairs of points between groups 4: Centroid linkage: Dissimilarity is the distance between the averages of the points in each group

Answer 38

1. It is deterministic, and thus has lower variance 2. You get clusterings for all values of k 3. You have hierarchical clusters! If your clusters naturally have a sub-cluster architecture, this is useful.

Answer 39

When looking at our training data, we assume they come from a probability distribution that is a *weighted sum* of several simple underlying distributions, such as several gaussians.

Answer 40

Rather than having each point discretely in one cluster, we compute the probability that each point is in each cluster. This is good for *overlapping clusters*, or for clusterings where some points have an unclear affiliation.

Answer 41

To produce a clustering from training data, we learn the parameters of the k underlying simple distributions (using a technique called expectation maximization) that we think may be forming the overall distribution. To "assign each point to a cluster", we compute the likelihood that each of the simple distributions would have produced this point. Here each simple distribution acts as a cluster, with a center at the simple distribution's mean, and we see probabilistically from which cluster the point was likely to come from. We thus don't have a "hard clustering"

Answer 42

1. Connect 2 points iff their distance is below a threshold epsilon (epsilon-nearest-neighbors) 2. Connect 2 points iff one is a KNN of the other, for some k 3. Form a weighted graph where points with close euclidean distances have high weights

Answer 43

A graph cut is simply putting the vertices into 2 (or more) subsets, but it can be visualized as "cutting" across the edges which go between your two subsets. We want our cuts to : 1. Have a low *total weight of cut edges*. So if we're using an unweighted graph, we want to cut as few edges as possible; if the graph is weighted, we want to cut as little weight as possible. This increases dissimilarity across clusters and similarity within clusters, as edges are drawn between similar points 2. We want cuts to be *balanced* (or approximately balanced), meaning that a similar number of points is in each subset. This avoids putting one point in a subset, for example.

Answer 44

It is very good at capturing *non-blob-like clusters*, or clusters with atypical shapes. This can be very useful

Answer 45

Form a graph from your training points, connecting points that are similar. Then, make graph cuts which separate dissimilar points to form your clusters.

Answer 46

Spectral embedding (it uses some fancy linear algebra that is related to graph-based dimensionality reduction, but not going to get into weeds here.)

Answer 47

The optimal regression function µ(x) = E[Y|X=x]; it is the expected value of the true function y=f(x) at every point x. This is the "optimal" regression function because it *minimizes the expected mean squared error*: E[(Y-µ(x))²]

Answer 48

In parametric bootstrapping, we sample new simulated data from a model (which is parameterized, hence the name), and that model is learned from the actual data we have. In non-parametric boostrapping, we don't learn a model, and instead just resample the points we already have.

Answer 49

Model-based bootstrapping, residuals resampling (which still involves a model, but uses it differently), and case resampling

Answer 50

Bootstrapping allows us to get an *approximate sampling distribution for a **statistic*** of our data. So typically we have a dataset of n values X_i, and from it we will calculate a variety of statistics: mean, standard deviation, regression coefficient for a model you're fitting to the data, difference in MSE between 2 models you're fitting, etc. But for each statistic, you get one value of the statistic per *dataset*, and to get an (approximate) *sampling distribution* for a statistic, you need to simulate many datasets, then use them to find the sampling distribution of the test statistic, which you calculate for every simulated dataset.

Answer 51

We use statistics to make estimations about the underlying data: maybe want the mean, or a coefficient in a model, or the difference in MSE between 2 models. By finding an approximate sampling distribution of such a statistic, we can approximately *quantify the uncertainty around our estimated value.* So if we have a sample mean estimating the expected value of the data µ, we can use bootstrapping to *find a* *confidence interval, or a standard error, or a variance, etc.* And quantifying our uncertainty regarding these estimates is always a responsible thing to do!

Answer 52

Residuals resampling: Here you have data in form of (x,y) pairs. You *learn a model* y~f\*(x) (so this is actually parametric), and use it to get the n residuals from your original dataset. Then, for a resampled dataset, the x’s stay the same, but the y’s are f\*(x) plus a resampled residual. So to reiterate, each "simulated dataset" has the *same* x values, which are not simulated. What's "simulated" is the y values: you find y from f(x), then add a resampled residual. It really strikes me personally as a sort of *hybrid between a parametric and non-parametric technique.* You're using a model, and generating a simulated value y from the model for each x. But then you get your residual, which is resampled in a quasi-non-parametric sort of way.

Answer 53

This one's pretty simple: each simulated dataset is n full, unaltered data points resampled from the original data, with replacement.

Answer 54

We learn the parameters of some model from the data; then, we use the model to sample entire datapoints. Example 1: Say we have 1-dimensional datapoints X_i. We might learn a normal distribution, then sample points from that distribution. Example 2: Say we have datapoints of (x,y) pairs. Here we would need to generate a full (x,y) pair from a learned model. Perhaps, for example, we learn a normal distribution for x, and then a regression model for y given x. (We didn't go into this much in class, and tended to use residuals or case resampling instead.)

Answer 55

Our 3 options have varying levels of assumptions: model-based bootstrapping requires more assumptions about the underlying data than residuals resampling, and residuals resampling requires more assumptions than case resampling. (Case resampling basically assumes nothing; residuals resampling learns a model and assumes that the distribution of residuals is the same for any (x,y) pair; and model-based bootstrap assumes that it can make full (x,y) pairs from a model that the data actually fits). More assumptions means more efficient use of the data *if you are correct*: you are fitting a more specific model if you use more assumptions. This means that your *confidence intervals will be tighter*, which is good. So basically, you want to choose the model with the most assumptions *that you can still feel confident about*. We can rarely feel confident about model-based simulation, so we rarely pick it. If the residuals assumption appears true, we pick residuals resampling; otherwise, we just use case resampling and settle for wide confidence intervals. (See 6.8 in the text, a brief section, for a complete discussion if desired)

Answer 56

We "smooth out" each datapoint by putting a "kernel" at each datapoint to act as a "little pdf" for that point. The kernels will generally look like normal distributions; if f(x) is higher than 1-dimensional, then a multivariate normal. We then just sum all of the little pdfs together into an estimated real pdf (the sum is normalized so it's a real probability dist). The following picture is for a janky example where the width of the normal distributions was too small and where the data weren't to enough decimal places, but it illustrates the idea:

Answer 57

DAGs (directed acyclic graphs)

Answer 58

Each vertex is one of the features of your data being measured. For example, if X = [x₁, x₂,...], then a vertex would be x₁ or x₂. An edge between 2 attributes means that the two attributes have a nonzero covariance, or a nonzero linear correlation. The direction defines the causal influence, so A -\> B means A influences B.

Answer 59

Because we assume that there are no causal cycles. (for some reason; it seems logical at times. The reasoning was something about how it's time based, and with A -\> B, a change in A results in a change in B *later*, and we can't "go back in time" or something.)

Answer 60

We must control for a set of variables that satisfies the backdoor criterion. In programming terms, this means we include all of the variables in the set, along with A, as predictors of B, and then look at the coefficient of A in predicting B.

Answer 61

S must: 1. Block all backdoor paths from A to B (so any paths with an edge pointing out of A and into B, except for the "direct" path) 2. Contains no descendants of A

Answer 62

Fork: A \<- B -\> C Chain: A -\> B -\> C Collider: A -\> B \<-C

Answer 63

Forks and chains are open when B is not controlled for. Colliders are open when B *is* controlled for.

Answer 64

When it is open at every vertex in between, meaning that every vertex in the middle of a fork or chain is not controlled for, and every verted in the middle of a collider is controlled for.

Answer 65

1. Guess one (generally assisted by domain knowledge), and check if it's correct. 2. Use a discovery algorithm.

Answer 66

The PC algorithm. It of course assumes linearity, which can be wrong. It also sometimes can't figure out the direction of an arrow, in which case you may have to account for both possibilities. Lastly, each relationship is found with some confidence, say 95%, and with a lot of relationships, some are likely to be wrong; so taking the DAG as gospel, or any causal inference DAG for that matter, can be risky.

Answer 67

We can't really ever know for sure if a graph is correct. But, there are ways to check if it's incorrect, and if you go through all of those and find no issues with your graph, you can start to feel confident. You can check the assumptions: linearity for example. You can also *look at conditional independences* among lots of different subsets of the variables, which can potentially disprove the presence/absence of an edge, or the direction of an edge. (Conditional independence can be considered somewhat intuitively, or using the fork/collider/chain rules. Can go back in course materials for more info.)

Answer 68

Rather than writing the hyperplane as w^Tx + b = 0, you write it as w^Tx = 0, and implicitly assume that the first (zeroth) entry in w is your intercept b, and the first (zeroth) entry in your x vector is a 1.

Answer 69

Just draw the hyperplane by normally drawing a line, by converting into y = mx + b form. To figure out which side has which label and draw that arrow, do some easy-to-calculate sample point back in the original form of w^Tx = 0, and see whether w^Tx is above or below 0.

Answer 70

MAP allows you to include some prior distribution of the parameter before factoring in the data. So, if you have suspicions about the value of a parameter, perhaps because of domain knowledge, then MAP could be better as it will allow you to encorporate that knowledge.

Answer 71

We want to take our D-dimensional dataset, and represent it using fewer dimensions (or fewer features). And we want to preserve as much of the structure and information in the data as we can: we want similar points in our original dataset to also be similar in the dimension-reduced dataset, and want dissimilar points to also remain dissimilar (by whatever measure of similarity makes sense in context).

Answer 72

With D-dimensional data, PCA finds up to D new linear dimensions, or vectors, or "number lines" as I think of them, which are linear combinations of the original dimensions (e_1,e₂,...). Each of these dimensions is called a Principal Component, and the first is the vector v₁ maximizing the variance of Xv₁, the projections of the points in design matrix X onto the number line of v₁. The remaining principal components maximize the variance of the projection Xv_i while also making v_i orthogonal to all previous principal components. (All D principal components thus form a basis for R^D, as they're all orthogonal.) Using these principal components, you can choose k \< D of them to represent your data in the k dimensional subspace preserving as much of the variance in the data as possible (I bet it's actually an approximation of that, cuz it feels like it's a greedy algorithm, but that's the idea).

Answer 73

A principal component v_i is one of the new dimensions which you're projecting your data onto, and using to store information. A PC score Xv_i is the projection of all D-dimensional points in your design matrix X down on v_i. It is a nx1 vector, with each entry being a scalar showing the location of a point in X on the "number line" that is principal component v_i.

Answer 74

For some subset of the D principal components, say k of them, the proportion of variance explained is how much of the variance from the original dataset is preserved by this k-dimensional representation. (Or what is explained by each individual principal component; both definitions work.) Specifically, it is (I assume) the variance in the k-dimensional representation divided by the variance in the original data. We are often interested in proportion of variance explained *as a function of k*, so we can see when we have a k high enough to explain a reasonable proportion of the variance, such as maybe 90%.

Answer 75

It plots proportion of variance explained vs number of principal components used, as a means of visualizing how much information will be preserved by each additional principal component. This is the general idea, but there are many versions. Some plot number of PCs used vs eigenvalue associatied with its eigenvector, because these are related (as described in another flashcard). Another type, shown below, has the proportion of variance at each individual PC.

Answer 76

It's just the component of a that's in the direction of b. Think 2-d velocity calculations from physics: when we divide the velocity of an object into its x component and y component, that's the velocity vector's projection onto the vectors of the axes, (1,0) and (0,1), respectively.

Answer 77

An orthonormal matrix is a square DxD matrix where every column has a length of 1, and all columns are orthogonal to each other. The columns thus form a basis for R^D.

Answer 78

A basis is a set of vectors such that any vector in the vector space can be represented as a linear combination of the vectors in the set.

Answer 79

Mv = ÿv | (Mv = lambda\*v)

Answer 80

When it is real and symmetric.

Answer 81

M = UDU^T, where D is a diagonal matrix with M's eigenvalues on the diagonal, and U is an orthonormal matrix with the associated eigenvectores as columns. (It is assumed as convention that the values in D are reordered such that |d₁₁| \> |d₂₂| \> ...), and that the columns of U with the corresponding eigenvectors are reordered to match. This is an extremely powerful decomposition with lots of truly great uses, such as direct applications to PCA, MDS, and spectral clustering.

Answer 82

Its entry at position i,j is the sample covariance of feature i and feature j (based on the observations/data points in design matrix X).

Answer 83

X_new = XV_k, where the columns of V_k are the k principal components v_i. In other words, you find all k PC scores, Xv_i, which is just the projection of each point x onto the number line of direction v_i, and then you combine all of those projections into your new representation X_new.

Answer 84

Given data matrix X, you can: 1. Find the eigendecomposition VDV^Tof its covariance matrix (Sigma-hat) 2. Find the singular value decomposition of X = UD\*V^T. 3. Take the inner product matrix XX^T and find XX^T = U(D\*)²U^T, where U and D\* the same as in option 2.

Answer 85

The columns of V are the principal components, and the corresponding values in diagonal matrix D\* are the *amount* of variance explained. (So in order to get the proportion of variance explained for each PC, you'd take its entry in D\* divided by the sum of all the entries in D\*.)

Answer 86

If covariance matrix Sigma-hat = VDV^T, then the principal components are the columns of V, and the corresponding values in diagonal matrix D are the *amount* of variance explained. (Specifically, the absolute value of the entries of D.) (So in order to get the proportion of variance explained for each PC, you'd take the absolute value of its entry in D divided by the sum of all the absolute values of the entries in D.) This means that for any dataset X, the PCs are just the eigenvectors of the sample covariance matrix, in order by the absolute value of their eigenvalues, and the eigenvalues define the proportion of variance explained.

Answer 87

UD gives the PC scores. We can thus do PCA dimension reduction using only the inner product matrix.

Answer 88

Kernel PCA allows you to find *nonlinear embeddings* using PCA, which typically only finds optimal linear embeddings. You'll recall that you can do normal PCA by taking the inner product matrix as XX^T = U(D\*)²U^T, where XX^T is the inner product matrix with (XX^T)_i,j = x_i^Tx_j. Well instead, use the kernel trick and make a different similarity matrix, a kernel matrix K, where the entries K_i,j = k(x_i,x_j) for some nonlinear kernel. Then do the same decomposition into K = U(D\*)²U^T for some other matrices U and D, and find the PC scores in the new nonlinear space as UD just as you would when using XX^T. This way, you can find the PCA dimension reduction of X, but after transforming the points of X into any new subspace you want, likely a non-linear one, using kernels.

Answer 89

If you have n points, and you have a matrix of *distances between points*, then you can find a similarity matrix M, similar to XX^T, on which you can perform PCA to get a dimension-reduced version of the data. (Each entry M_i,j is the distance between points i and j.) But, you can use *any distance measure*, including *non-euclidean distance measures*, which will help you find nonlinear dimension reductions which can preserve nonlinear structure in the data.

Answer 90

You'll recall that you can use PCA to find a linear dimension reduction of data using only similarity matrix XX^T. Additionally, in MDS, you feed PCA a similarity matrix that instead uses some non-euclidean distance measure. Well, ISOMAP is simply an instance of MDS where the distance matrix is calculated using graphs. You make a graph where the vertices are your data points, and every edge from point i to j has a weight of their euclidian distance apart. Then the graph distance is the shortest path in the weighted graph from i to j; feed these non-euclidean distances into MDS, and you'll get a nonlinear embedding which preserves structures like swirls or curves.

Answer 91

Overfitting is when you allow your model to learn the patterns in the training data *too* well, so that it follows the noise in the training data. This results in validation error that is much higher than training error. This typically happens when your model has too much complexity (so that it can learn all the idiosyncracies of the training data), and/or when you don't regularize.

Answer 92

Regularization is when you limit the values your models parameters can take, or (more commonly) penalize your model for taking certain values. Typically, this looks like penalizing your model for having parameter values/coefficients that are large nonzero. So your objective function now needs to balance model fit and model complexity. Regularization is generally used to *combat overfitting*.

Answer 93

Regularization increases bias: by penalizing a model for taking certain forms or parameter values, it becomes less capable of taking the form of the true model in certain situations; its ability to fit the data decreases. Conversely, variance decreases, meaning that the model becomes less impacted by noise and fluctuations in the data. This is really the whole point of regularization: to combat overfitting to noise.

Answer 94

***Almost always***. This is a good thing to remember; it was driven home in 462, but often falls by the wayside.

Answer 95

If you have p parameters, it takes the value below for some balancing value lambda. (Note you don't regularize the intercept B₀)

Answer 96

If you have p parameters, it takes the value below for some balancing value lambda. (Note you don't regularize the intercept B₀)

Answer 97

A coefficient basically says that there is a substantive relationship between one variable and your output variable. By shrinking the coefficients, or penalizing large coefficients, you limit the number and magnitude of substantive predictive relationships that your model says exist. This in theory means, if tuned right, it can only capture the important or "real" trends, but can't capture the minutia in the noise of the training data. Hence, it combats overfitting.

Answer 98

Lasso regression *particularly wants coefficients to be zero* (rather than them being very small, but not zero). This means it will result in more zero coefficients, so it makes models that are more interpretable and that show which variables "actually matter". However, lasso thus zero's out small relationships, which means it has more bias. Ridge doesn't do this: the difference between 0 and .001 isn't as big for ridge, which results in basically no zero coefficients. Thus, ridge would make more sense if you're more concerned with accuracy than with interpretability.

Answer 99

As lambda goes to zero, we move towards no regularization, so the solution moves towards being the *same as if we hadn't regularized*. As lambda goes to infinity, our model will just make *all coefficients go to zero*, as this minimizes the objective function.

Answer 100

A high lambda means that the shrinking of coefficients has a lot of weight in the objective function, and so combatting overfitting will be very important to this model. Conversely, low lambda means more liberty to have high coefficients, and we are more able to fit training data.

Answer 101

Increased lambda means more regularization. As such, training error will monotonically increase as lambda increases. Test error will decrease, then increase. So there will be an optimal lambda value where test error is minimized.

Answer 102

The one where things go quickly to zero, or the right one, is lasso, because lasso zero's out coefficients. The one where coefficients asymptote towards zero is ridge, as it decreases coefficients but doesn't actually zero them out.

Answer 103

If the scales of your variables are very different, then a small parameter might mean a large change, or a large parameter might be a smart change. Because regularization tries to decrease parameter values without considering which variables they correspond to or what those variables' scales are, the incentives for decreasing the influence of certain variables will be accidentally too high or too low based on scale.

Answer 104

The goal of regularization is to shrink coefficients of predictor variables, making it less likely our model will say it has predictive value. Take Y = aX + b for example. By shrinking our learned value for a, we limit the amount of influence our model can percieve that X has on Y. But b *doesn't describe how much impact X has on Y.* It merely offsets the relationship and says what Y is likely to be when X is zero, but it doesn't impact the model's perception of whether X is related to Y. So if we were to regularize b, and penalize our model for having large values of b, we would be limiting our model's ability to find the correct relationship (increasing bias) *without* helping its ability to combat overfitting, or finding substantive relationships between variables which don't actually exist. Hence, don't regularize b.

Answer 105

Linear regression and logistic regression (because they fit the formulas and ideas we've learned so well).

Answer 106

Our classifier becomes biased towards predicting the label that is more common in our training set, hurting its predictive ability.

Answer 107

Downsampling: Only train using a subset of the points with the common label Upsampling: Duplicate the training points with the less common label and use them multiple times in the dataset. Weights: Changes weights of your problem in a variety of ways. For example, penalize the model more for errors when the correct answer was the uncommon label.

Answer 108

Change the *amount of regularization* (more regularization is bias up, variance down) Or change the *model complexity* (more complexity means bias down, variance up)

Answer 109

* Model flexibility, or ability to fit a variety of true underlying models * Ability to capture nonlinearity * Interpretability * Ability to handle irrelevant features well (Don't necessarily need to get all of these or get it exactly right, but these are good things to keep in mind)

Answer 110

They both handle irrelevant features very well, which means you can throw a lot of features at them in hopes of finding a couple good ones, even if you suspect lots of them have little or no predictive power.

Answer 111

1. Split off some of your data, say 10 or 20%, to act as test data. **Do this first.** 2. Split off more data to act as your validation data; leave the rest as training data. 3. Try a bunch of models by training them with the training data, then finding their approximate out-of-sample error by predicting on the validation data. 4. Choose a final model. 5. Approximate your final model's true out-of-sample error by training it on both the training and validation data and finding its error on the test set. 6. Train the model on all of your data, and deploy it to do whatever it needs to do in the real world.

Answer 112

With normal validation, once you've set aside your test data, you split your remaining data into train and validation data. You train all your models on the train data, then validate them with the validation data. You pick a model and test it on the test data. With cross-validation (say with k folds, such as 5 or 10), after you set aside test data, you instead divide it into k groups, or k "folds". For every model you try, you train it on k-1 of the folds and find its validation error on the remaining fold, and you do this *for each of the k folds*. Finally, you get your validation error by averaging all k of your error measures on each of the folds. (So you train your model k times rather than once when you are validating.)

Answer 113

Cross-validation wastes less training data during the train/validation part of the model fitting process. It also decreases the variance that appears when you split train and test: what if an outlier shows up in the validation set, or what if it doesn't? For these reasons, cross-validation is much more common in practice than normal validation. However, cross-validation can take a long time: the more folds you have, the longer it's going to take to do all that training, whereas with normal validation you only train once.

Answer 114

Generally more folds is better, and you want as many folds as possible. The "ideal" cross validation is leave-one-out cross validation or LOOCV, where each datapoint is its own fold, with the intuition being that the more data you use to train for each fold, the better you simulate how the model will perform on actual held-out data. But of course, more folds means more time, and thus LOOCV is often not realistic. 5-fold and 10-fold CV are common and tend to do the trick.

Answer 115

Any choice that you make in model selection when working with training and validation data has the risk of *overfitting to the training and validation data*. The characteristics of your training and validation data that cause you to make certain decisions might be noise. Additionally, if you try enough models, one of them is likely to get a low validation error simply due to random chance. For these reasons, you need data to test your model on *after you choose your model* to accurately estimate your out-of-sample error. You need the data that you use to assess out-of-sample error to *not be involved in selecting a model in any way*, otherwise it is possible that you are underestimating your error for these reasons.

Answer 116

**All decisions and all actions.** Or as many as possible. You shouldn't do feature engineering, or consider candidate algorithms, or decide how to define an outlier, or even curiously look at a histogram of your data before you set aside your test data. This is because all of these decisions are a part of your model selection process, and you need a test set which *wasn't used for model selection* in order to get an accurate, unbiased estimate of your final model's out-of-sample error.

Answer 117

If you have enough data (or as your sample size n tends to infinity), your odds of seeing unexpected new things and/or things that cause bugs should be low, as your training and validation data should be representitive of all available data.

Answer 118

It's very important: finding the right features to feed into your model can easily be as important or more important as what algorithm you choose to use, and deciding/brainstorming these features will often take up a large majority of your time when trying and selecting models.

Answer 119

A model can really be thought of as *any aspect of the algorithm that outputs your predictions.* This of course includes the algorithm you choose (random forests, linear regression, etc) and the values of the model's parameters, but it also includes *the features that your model takes as inputs*, or how your model defines and handles outliers. That features idea is very important: **the presence or absence of features is a part of your model; feature engineering is a subset of model selection**. This is a good way to think because it emphasizes the importance of not making any decisions or looking at anything before you set aside your test set. It also encourages you to think deeply about (and remember to cross-validate) aspects of model selection like feature engineering that you might think less important than, say, choosing your algorithm.

Answer 120

You want to cross-validate *as many decisions as possible*. This includes the normal stuff like the value of your hyperparameters, but also which algorithm you use, which features you choose to include, etc. To cross-validate these things, you just find a cross-validation error for each combination of those decisions. So you try your algorithms with and without each feature you're considering, for each of your candidate algorithms, and each of their possible hyperparameter values. The issue with this strategy is it takes a lot of time, as you have a lot of combos. You address it by taking shortcuts that seem reasonable to you, and that your intuition says would still lead to a good final model. Maybe you don't try every combination of features, or every value of a hyperparameter. Maybe you choose an algorithm with only a couple possible sets of features and a couple sets of hyperparameters, then hone in on that algorithm and try more combos of features, more candidate features in general, and spend more time on more possible hyperparameter values.

Answer 121

* Training data* is used to learn the values of your model's *parameters*. For example, when learning a neural network, your training data would teach the model its weights and bias terms * Validation data* is used to learn the values of the *hyperparameters*. For example, in KNN, you might use validation data to decide k, the number of neighbors, which is a hyperparameter. In the neural network example, you might use validation data for defining number of layers, nodes per layer, which activation function, etc.

Answer 122

We use gradient descent to minimize (or maximize) differentiable objective functions, or to find the value of x sucht that f(x) is minimized. f(x) needs to be differentiable because we need to take its derivative in order to get the gradient.

Answer 123

Solving it with calculus is often extremely time-consuming or impossibly time-consuming, especially when you need to minimize an objective function with respect to, say, 1000 parameters as with NNs.

Answer 124

Depending on where your starting point is, you may descent to a *local minimum* rather than a global minimum, but your algorithm will still stop: it can't take a small step in a direction and improve its situation. This won't happen with convex functions, or where the gradient is always "negative" or zero (this is how to think of convexity in say 2 or 3 dimensions; the terminology might differ at higher dimensions) To combat this issue for non-convex functions, you can try several randomly chosen starts and see which returns the minimum; it still doesn't guarentee a global min though.

Answer 125

The gradient of the function with respect to theta is just a vector of the partial derivatives of the function with respect to each of the p parameters in theta. Intuitively (especially when visualizing say 3 dimensions), it is the direction of the n-dimensional slope, or the direction in which a small step would most productively move you towards optimizing the objective function. (Technically, you might move in the opposite direction, but same idea.)

Answer 126

If you choose the point randomly, the *expected value* of the gradient with respect to a single point equals the gradient with respect to all n points.

Answer 127

A single pass throught the entire training set.

Answer 128

Finding the gradient with respect to the entire dataset takes a long time. (Specifically, I believe it takes about equal time to find the gradient with respect to each of the individual points, as I believe it's basically just a sum of those). So getting one update, which is albeit definitely in the direction of the gradient, takes a long time. Conversely, SGD may make some updates in the "wrong" direction, but you get to make n updates in the time normal gradient descent needs to make only one. In practice, this typically leads to faster convergence to the same result.

Answer 129

In normal GD, you make updates with respect to the entire datapoint; so, your update is always in the right direction, but updates take a long time. In SGD, you make updates with respect to a single datapoint; so, your update may be in the wrong direction, but the expected value of the direction is correct, and you can make many updates quickly. In mini-batch gradient descent, you combine these ideas by updating with respect to *small subsets of the training data* each time. This can sometimes lead to a happy medium between making updates quickly, and often being in an approximately correct direction.

Answer 130

The distance from the hyperplane to the *closest point*

Answer 131

If the data aren't linearly separable, no hard-margin solution exists, because the data can't be perfectly classified with a hyperplane classifier. Soft-margin SVM fixes this by allowing some points to cross the hyperplane without having the solution violate the constraints.

Answer 132

A support vector is a point that has an influence on the final solution. For hard-margin, it is one of the points on the margin of the hyperplane. For soft-margin, it is one of the points that is on the margin *or within it*, including points that are misclassified.

Answer 133

A higher C allows more points to have positive slack variables, and thus to be inside the boundary. The margin M is thus *increased* to accomodate the higher amount of points inside the margin. Increasing C also thus increases the number of points that influence the boundary, as the boundary is only influenced by points on or inside the margin. Because more points influence the boundary, variance goes down (we are less susceptible to outliers). Bias would I assume increase as a result, but I'm not sure why.

Answer 134

Because the sum of the slack variables is capped at C, and a slack variable above 1 means a point is misclassified, C is the maximum number of misclassified training points.

Answer 135

Similar to multiclass logistic regression, we just fit k binary-classification SVMs: one for each of the k labels, predicting whether it's that specific label vs *any of the others* (we treat all other labels as a single label for each binary classifier). We then predict via:

Answer 136

It maps a vector to a non-negative scalar in a way that measures its "length"

Answer 137

The *l₂* norm, or the euclidean norm, is simply the distance formula: it's the length of the vector in euclidean space. If x = [1,5,2], ||x||₂ = sqrt(1² + 5² + 2²).

Answer 138

For ||x||₁, you just sum the absolute values of the entries in x. ||x||₁ = |x₁| + |x₂| + |x₃|

Answer 139

Suppose we are trying to find a vector x that minimizes a norm, either the *l₁* or *l₂* norm. So in both cases, we're trying to find a vector x whose entries are near zero. The *l₂* norm doesn't care much whether an entry in x is exactly zero, or very nearly zero; as such, when we try to minimize the *l₂* norm, we get a bunch of values that are very nearly zero. Conversely, the *l₁* norm *does* care a lot whether an entry in x is exactly zero, or very nearly zero; as such, when we try to minimize the *l₁* norm, we get a bunch of values that are exactly zero, whereas with the *l₂* norm they weren't quite zeroed out. This pops up in regularization: when we do ridge regression, we're minimizing the parameter vector over the *l₂* norm, and so we get a bunch of values that are almost zero, but none that are really zeroed out. When we do lasso regression, we use the *l₁* norm and thus get a bunch of parameters that are exactly 0. There are tradeoffs here: lasso regression yields more interpretable results, but ridge regression is better at capturing very small relationships that lasso tends to just zero out.

Answer 140

Most classifiers find the probability that a point has each possible label, and then predicts the label with the highest probability. So, we can evaluate either the predicted *labels*, or the predicted *probabilities.* (Note: The classifiers don't exactly make probabilities, they make values between 0 and 1. But these values can often be interpreted as probabilities, especially if we choose to transform them into more of a probability distribution.)

Answer 141

Oftentimes it's important to know how likely the labels actually are, not just what the maximum-likelihood label: if a potential client has a 99% chance of signing a deal, we'll prioritize them differently than if they have a 51% chance.

Answer 142

Recall that a classifier generally outputs the probability that a point has each label. A classifier is well-calibrated if *the labels happen about as often as they're predicted to happen*. So if, in every case where the classifier says a point has a 70% chance of being label A, about 70% of them are actually label A while the other 30% are something else, we have a well-calibrated classifier (at least for that predicted percentage for that label). For binary classifiers, we can evaluate calibration using a *bin plot*. We group together all of the points where we predicted the odds of y=1 were between 0% and 10%, and then we see how many of those points were actually y=1. Similarly for 10% to 20%, and all the other bins.

Answer 143

For a classifier with k labels, it's a k-by-k grid that plots the frequency of every predicted label vs every actual label, so we can see what types of errors the classifier tends to make. For a binary classifier, it simply shows the number of true positives, false positives, true negatives and false negatives the classifier had, in the form of a 2x2 grid.

Answer 144

It is the proportion of true positive's, or of y=1's, that we get right. So it's : (true positives/[true positives + false negatives])

Answer 145

It is the proportion of true negativess, or of y=0's, that we get right. So it's : (true negatives/[true negatives + false positives])

Answer 146

Both of these cases occur when one error is much more costly than the other. Say we're trying to decide whether to administer a treatment that has few side effects. If we get a false negative, and incorrectly classify someone as "healthy", then we don't give them treatment, which is very costly. But, if we incorrectly say they're "sick" and treat them when they don't need it, it's not as bad. Here we heavily prioritize avoiding false negatives, or we *want sensitivity to be high*. Conversely, consider a spam filter. We're okay with accidentally sending spam to the inbox (false negative), but don't want to send important emails to spam (false positive). Now, we care a lot about the *specificity of our classifier being high*, as we want to avoid false positives.

Answer 147

Classifiers output a probability that y=1 on a given point. Typically we label a point with a 1 if the probability is above 50%, but we can *change the threshold that the probability must reach to label a point as y=1.* If we only label it y=1 if the probability of y=1 is above 75%, then we are okay with false negatives, but really don't want false positives; we're *prioritizing specificity.* Conversely, if we label it y=1 whenever the probability of y=1 is above 25%, then we are okay with false positives, but really don't want false negatives; we're *prioritizing sensitivity.* Another option would be to *initially train* the classifier so a false-negative causes higher loss in the loss function than a false positive, or vice versa. So we would have different loss amounts *L₀* and *L₁* for the two different errors.

Answer 148

Sensitivity describes the probability that we label a point y=1 when it is in fact y=1; it's the probability of avoiding a false-negative. Specificity describes the probability that we label a point y=0 when it is in fact y=0; it's the probability of avoiding a false-positive. So, we want high sensitivity when false-negatives are bad, and high specificity when false-positives are bad.

Answer 149

Recall that our trained binary classifier outputs a probability *p* that a point x has label y=1, and we can tradeoff the sensitivity and specificity of our classifier by moving the threshold of this probability where we predict y=1. So for some value *t*, we predict y=1 if *p* \> *t.* It is high, we are avoiding false positives and thus have high specificity; if t is low, we are avoiding false negatives and thus have high sensitivity. A curve on an ROC plot shows a classifier's sensitivity and its specificity *at every possible value of threshold t*, which can be anywhere in [0,1]. It plots (1 - specificity) on the x axis, and sensitivity on the y axis. What we want is classifiers whose curves are bowed out to the left, because then for each value of *t* they do a good job with both sensitivity and specificity. In this case, the classificer has a high AUC, or area-under-the-curve. *We want high AUC*, as it basically means our classifier is accurate. The diagonal curve on the ROC corresponds to a *random classifier*, where our probability value *p* is randomly chosen from the interval [0,1], and then we threshold that probability for some threshold value *t.* The AUC of this curve is 0.5. So a *curve* on the ROC plot is a *classifier*, and *point* on the ROC plot is a *classifier-threshold pair*. The top-left corner is a classifier-threshold pair always predicts the label correctly, and the bottom-right always predicts incorrectly. The bottom-left always predicts 0, and the top-right always predicts 1.

Answer 150

It's very common It's very easy to understand The math is simple and well-known (I'm not going to memorize it, but it follows from the statement of what it's trying to minimize.)

Answer 151

*None.* The least squares formula will always form the optimal linear predictor for that dataset. However, if the assuptions are wrong, *that linear predictor won't match the actual underlying model that produced the data*. In other words, without the assumptions, the linear regression won't be "correct," or useful in modeling these data or other data from the same source.

Answer 152

i.i.d. data, linearity, and heteroscedasticity. (pretty sure on heteroscedasticity, could check that) **Not gaussianity.** That's not needed.

Answer 153

Whenever we're making confidence intervals or prediction intervals, be them for our coefficients, point predictions, etc.

Answer 154

In the context of a point prediction, so predicting Y from X, a *confidence interval* is a range for E[Y|X], and a *prediction interval* is a range for the actual value Y itself. Confidence intervals thus discuss the distribution from which Y came, whereas prediction intervals discuss the actual, currently-existant value of Y. (I think. This is weird.)

Answer 155

*Plot the residuals* vs both the X values, or the predicted Y values. If there are any patterns in the residuals, linearity is not met. If the residuals look basically random, linearity may be correct. You can also plot the squared residuals vs these values.

Answer 156

A residual for a prediction Y\*=µ(x) is Y - Y\*, the difference between the real and predicted values.

Answer 157

One option is to plot your points vs some axis like Time, and assure that there is no pattern among points close in time. Basically look at ways the points might be related to one another, and see that they're not, and that there is no pattern. (This one was less well-defined in class)

Answer 158

Make a Q.Q. plot (or quartile-quartile plot) of the residuals. You're seeing where the quantiles (or the "lines" demarking percentiles) are in the observed distribution of residuals, vs what these quantiles would be if the residuals followed a normal distribution. If it forms a line with a slope of 1, the normality assumption is probably correct; if not, then probably not.

Answer 159

An additive model

Answer 160

When given a new X, we predict the Y by taking a weighted sum of all the Y's in the training set. The weighting w(x_i,x) is meant to be high when the points are "similar", and low otherwise.

Answer 161

In knn regression, we predict Y for a new X by just averaging the Y's of the k X's in the training set that are nearest to our new X. It is one option for nonlinear regression. Recall that a linear smoother predicts Y for a new point X by taking a weighted sum of all the Y_i's from the training set, for some weight function w(x_i,X) meant to express similarity between the two x's. KNN regression is just this, where the weight is 1/k if x_i is one of the k nearest neighbors of X, and 0 otherwise.

Answer 162

The (x,y) pairs farther from the mean have a larger weight on the prediction than closer ones. So basically, the farther out a point is, or the more outlier-ish it is, the more impact it has on predictions. Upon inspection, this makes sense if the underlying model is actually linear, but is very weird otherwise.

Answer 163

We have a dataset of (x,y) pairs, and we predict Y for a new X by taking a sum of the y's from the dataset, weighted by how close each y's x value is to the X being analyzed. *Kernel regression weights points closer to X more heavily, and points farther from x less heavily*. The weight is found using a kernel k shown below. Often k is a gaussian. * The bandwidth h determines how quickly the weight changes as we move away from our X value.* * We tune h as a hyperparameter via cross-validation to get the best predictions.*

Answer 164

The banwidth h is in the denominator of the term that determines how differently we weight close points and far points. When bandwidth is high, the differences in weights will be small, so we will encorporate all points more equally into the prediction. As h approaches infinity, we will thus get to a prediction of just the mean Y value in the dataset. So, as h rises, variance decreases, but bias increases. Conversely, when bandwidth is low, nearby points have higher relative weight and matter more. As h approaches 0, it turns into 1-nearest-neighbor regression, where we just predict the label of the nearest neighbor.

Answer 165

As n goes to infinity, the learned model must converge to the actual underlying model.

Answer 166

Kernel and knn regression are universally consistent. Linear regression and additive models are not. This means that given enough data, kernel and knn regression are guarenteed to converge to the true model, while the other two are not. This is because linear regression and additive models both impose some kind of special structure on the underlying model: linear regression assumes it's linear, additive assumes it's additive. And if this special structure is wrong, it'll never converge. Kernel and knn regression just use the actual data, and they can take the form of any underlying model, so it makes sense that given tons of data, they will converge.

Answer 167

Additive models converges faster than kernel regression. When we make assumptions about what type of underlying model we have, we can use data more efficiently, meaning we need less data to converge. (Of course, we will only converge if our assumptions about the model are correct, but if they are, we converge more quickly.) This is a general fact about all models, not just these regression models. An intuitive explanation as to why is, when we already know something about the model, we don't need to learn it from the data, and we can use all of the data just learning the parameters within that model structure, rather than both figuring out the model structure and learning its parameters. Kernel regression converges very slowly, in part because it makes no assumptions about the underlying model's structure. Additive models assume that the model is additive; this significantly decreases the number of possible underlying models, and as such it converges significantly more quickly.

Answer 168

Additive and linear

Answer 169

Additive models are an easy step beyond linear models, with a lot of advantages. A linear model is just a special case of additive models, and an additive model can fit a *far* wider array of underlying models without sacrificing that much efficiency or rate-of-convergence. So if the underlying model is nonlinear, additive might fit it while linear won’t. And if the underlying model is linear, an additive model can still fit it, and an additive model doesn’t need that much more data to converge to the true model than a linear regression does.

Answer 170

An additive model is similar to a linear model in that all the terms are being linearly summed (along with an intercept and a noise term) in order to get the prediction. They differ because, in a linear model, a term is a variable with a coefficient. In an additive model, a term is that same variable x_i but fed into some partial response function, which can be nonlinear. This allows the modeling of nonlinear relationships, while still maintaining some of the speedup benefits of an additive structure.

Answer 171

It basically just *maximizes likelihood*, so it finds the model with the highest likelihood of producing that data. However, it does so while *regularizing, or penalizing models that are too "wiggly,"* in order to avoid overfitting data. So we are finding the model that optimizes the objective function shown below, which balances maximizing likelihood and minimizing model complexity, using tuning parameter lambda. *It also chooses a value for lambda, so it finds the optimal balance between data fit and regularization, presumably using something like cross-validation.* The model is a combination of smoothing splines , plus linear terms if desired or necessary. A smoothing spline, or partial response function, for a particular predictor is essentially the smoothed predicted outcome variable based on just that predictor. A smoothing spline is calculated in gam by "summing a bunch of small functions in the data called basis functions." I'm a little unclear on how this works, but basically *gam() does not explicitly try a bunch of different functions like log(x_i), sin(x_i), x_i², etc,* which is what we would expect based on the original definition of additive models. Rather, it *uses* *fits a spline using a flexible, computational smoothing method that is _capable of taking a form very similar to each of these possible underlying partial response function forms._* It fits smoothers by iteratively improving each (improving them all, improving them all again...) until they all reach a local minimum and stop improving.

Answer 172

Basically in generalized additive models, we get to specify the type of probability distribution from which our outcome variable is drawn. If we think our outcome has a normal distribution, or a poisson distribution, we can specify that, which (I assume) increases efficiency in using the data if it's correct. We can also specify if it's categorical data and we want to do classification rather than regression, or if it's geospatial data, or other more niche things, but I'm less familiar with these applications

Answer 173

If I recall correctly, we can assess variable importance by looking at the partial response functions. If there appears to be a relationship in the partial responses, the variable is probably important. If there doesn't appear to be a relationship, or if "the confidence interval for the partial response contains zero for many values of the predictor variable" or something, then the variable probably isn't that important. This is referring to variable importance *in context*, so a variable might have predictive power but be predicting things similar to other variables in the model, for example. **But I'm remembering this somewhat hazily, would want to look back at notes and homeworks where we did this to get specific, if needed.** Looking on piazza regarding HW12, this observing the partial response function seems to be used for seeing if the outcome and a predictor are independent or dependent, so yeah that assesses variable importance and predictive power in a sense. **But this is described on piazza as an informal independence test; it is useful, but not completely rigorous.** **We could also assess variable importance by seeing the drop in predictive ability caused by removing a variable!** This was something we did often: fit the model with and without that variable, and see how much predictions got worse. We could use bootstrapping to get a confidence interval for the difference in MSEs of the two models and see if the confidence interval contains zero. To actually perform variable selection, we can include select=TRUE in our gam() call, which penalizes the inclusion of variables in prediction, similar to lasso. We didn't do this in class, but it exists and could definitely be useful.

Answer 174

The nature of additive models is to not include them by default, but if we suspect they're there, we can add interaction terms to the model and smooth them using smoothing functions like te() and ti(), which take multiple variables while s() takes 1.

Answer 175

They aren't as interpretable as say, linear models: they lack interpretable coefficients for each variable explaining the expected difference in the outcome variable as we vary the predictor. But we do have the partial response functions, which provide some interpretability of how different predictors impact outcome variables. Additionally, they aren't as robust as nonparametric models, as they require an additive form, but are still far more robust than linear models.

Answer 176

Below is the formula for the "true" value of R² for a model: if we knew the true value of the coefficients of the optimal linear model, as well as the true values of the variance of X and noise epsilon, this would be the value of R². (Recall that even if the underlying model isn't linear, there still exists an optimal linear model.) Here, we can see that R can be arbitrarily close to 1 even if model fit is bad: just make Var[X] high or σ² low. Doing the opposite, we can make R arbitrarily close to 0 even if the model fit is perfect. It also says nothing about your prediction error. **R² is typically described as the proportion of variance in the output variable that our regression model explains.** In other words, it's the variance of our predictions divided by the variance of the actual values. *Because it's just a matter of comparing two variances, it doesn't really say anything about prediction error*. We could have a really bad model with similar variance in the outputs that has a "great" R². Adjusted R² is very similar, it's only difference is a constant to make σ²-hat an unbiased estimator of σ². It doesn't solve any of these issues. * (See section 3.2 here for more:)* http: //www.stat.cmu.edu/~cshalizi/mreg/15/lectures/10/lecture-10.pdf

Answer 177

**Correct:** B_i is the expected difference in the outcome variable Y between two points whose difference in X_i is 1, assuming the values of all other predictors are held equal. **Incorrect:** B_i is the expected difference in Y caused by increasing X_i by 1. The incorrect one is wrong because *it assumes some sort of causal relationship*. Our linear model by itself says nothing about what will happen if we *change* the value of X_i, it just says that on average, points with higher X_i have higher (or lower, depending on sign of Bi) values of Y based proportionally on B_i, which is what the correct interpretation is saying.

Answer 178

We are going to do a hypothesis test. The null hypothesis is that the underlying data are linear. The alternative hypothesis is that they aren't *difference in MSEs between the linear model and the more complex additive model*. If the null hypothesis is true, the difference in MSEs should be about zero, as the linear model can fit the data as well as the additive model; if the alternative hypothesis is true, the difference in MSEs will be nonzero, since the additive model will do a better job of fitting the data. We start by finding the difference in MSEs on our actual dataset. To perform our hypothesis test, we need to find the p-value of our observation: *we need to find the probability of seeing our observed difference in MSEs **under the null hypothesis.*** If our observed difference in MSEs would occur with very low probability under the null hypothesis, we reject; otherwise, we fail o reject. So, we need to simulate datasets under the null hypothesis that the data are linear. We will find the difference in MSEs on each of these datasets, to get an approximate distribution for the difference in MSEs. *Each of our simulated datasets will have the same set of X values as the original dataset, and y values equal to the predicted y value under the linear model plus an error term*. In the example in lecture 12, this error term is found by assuming gaussianity of the residuals, and sampling from a normal distribution with variance equal to the sample variance of the residuals. This is very similar to residuals resampling; I expect doing normal residuals resampling is also a viable option. So, we get all these datasets, get our approximate distribution in the difference in MSEs between the two models under the null hypothesis, and look at the p-value of the difference in MSEs on the actual dataset to make a conclusion about the linearity assumption.

Answer 179

**Cross-validation**, to approximate the MSEs of each model on out-of-sample data.

Answer 180

Fit the model with all the variables, as well as the model missing each of the variables. Find the difference in MSE when you remove each of the variables. It would be good practice to use bootstrapping to find a confidence interval for the difference in MSE cause by removing each variable.