Intro Probability & Stats Flashcards

Question

What is the notation for a Bernoulli distribution? What is the notation for a binomial distribution?

Answer 1

Bernoulli X ∼ B(p) | Binomial Y ∼ B(n, p)

Answer 2

The Binomial Distribution is the total number of successes from n repetitions of the same Bernoulli experiment. Binomial random variable takes values {0,1,2,…n}

Answer 3

f(x) = (n x) p^(x).(1-p)^(n-x) for x∈{0,1,2,...,n} and 0 elsewhere

Answer 4

n!/x!(n-x)!

Answer 5

n being high

Answer 6

1. There are a fixed number of trials, n, that is determined in advance and is not a random variable 2. There are two possible outcomes for each trial: success and failure 3. The outcomes are independent from one trial to the next 4. The probability of success and the probability of failure remains the same across all n.

Answer 7

Verbal: The expected value of a discrete random variable is the sum of all possible values the random variable X can take weighted by their probabilities. Check formula in notes as can't write on here.

Answer 8

``` E[X] = 1*P(X=1) + 0*P(X=0) E[X] = P(X=1) = p ```

Answer 9

By integrating x multiplied by the probability density function

Answer 10

1. If c is a constant and P(X = c) = 1, then E[X] = c 2. If c is a constant and P(X=c) = 1, then E[g(X)] = g(c) 3. If b and c are constants then E[b + cX] = b + cE[x]

Answer 11

E(aX + bY) = ∑(axP(X=x)+byP(Y=y)) E(aX + bY) = ∑(axP(X=x))+∑(byP(Y=y)) E(aX + bY) = a∑(xP(X=x))+b∑(yP(Y=y)) E(aX + bY) = aE[x] + bE[Y]

Answer 12

1. If X is a random variable and g(X) is a convex function, then E[g(x)] ≥ g(E[X]) 2. If X is a random variable and g(X) is a concave function, then g(E[X]) ≥ E[g(x)]

Answer 13

Variance is the expectation of the squared difference between the random variable and its expected value. Var(X) = E[(X-E[X])^2]

Answer 14

Variance is equal to standard deviation squared σ^2=Var(X)

Answer 15

For the variance of discrete random variables, for each possible value of the random variable, we find the variance and then we sum all the variances of the possible values, weighted by their probabilities. (check formula in notes)

Answer 16

((1-p)^2)p + ((0-p)^2)(1-p) | Var(X) = p(1-p)

Answer 17

1. Var(X) = E[(X-E[X])^2] | 2. Var(X) = E[X^2] – (E[X])^2

Answer 18

Both c^2Var(X)

Answer 19

By Jensen’s inequality, for a non-constant X, Var(X) > 0 since (X-E[X])^2 is a strictly convex function - By Jensen’s inequality, if g(x) is a strictly convex function, then E[g(X)] > g(E[X]) - Let the function g(x) be (X-E[X])^2 - E[(X-E[X])^2] > g(E[X]) - Var(X) > g(E[X]) - Var(X) > (E[X]-E[E[X]])^2 - Var(X) > (E[X] – E[X])^2 - Var(X) > 0

Answer 20

E[(X-E[X])^3] / Var(X)^(3/2)

Answer 21

The kurtosis of a distribution is a measure of how much mass is in its tails, and therefore is a measure of how much of the variance of X comes from extreme values. Kurt(X) = E[(X-E[X])^3] / Var(X)^2

Answer 22

Refers to situations where a large sample has been collected. Large sample = +30 observations

Answer 23

It is the probability that the random variables, X and Y, simultaneously take on certain values, x and y. Check notes for formula way of writing.

Answer 24

The marginal distribution of a random variable, X, is just another name for its probability distribution. The term is used to distinguish the X alone from the joint distribution of Y and another random variable The marginal distribution of X can be computed from the joint distribution of X and Y by adding up the probabilities of all possible outcomes for which X takes on a specific value So for marginal distribution of X sum all the possible realisations of Y where x = xi

Answer 25

For marginal distribution of x, you just integrate joint probability density function with respect to y For marginal distribution of y, you just integrate joint probability density function with respect to x

Answer 26

The distribution of a random variable Y conditional on another random variable X taking on a specific value.

Answer 27

P(A|B) = P(A∩B)/P(B) = P(B|A)P(A)/P(B)

Answer 28

- Two random variables are independent if knowing the value of one of the variables provides no information about the other - 1. X and Y are independent if the conditional distribution of Y given X equals the marginal distribution of Y P(Y|X) = P(Y), P(X|Y) = P(X) 2. If X and Y are independent, then their joint distribution is the product of their marginal distributions P(X).P(Y) = P(X∩Y)

Answer 29

Verbal: The covariance of two random variables is the expectation of the product of the two variables' deviation from the mean. It is a measure of the extent to which random variables move together. Formula: Cov(Y,X) = E[(Y-E[Y])(X-E[X])] Cov(Y,X) = E[YX] – E[Y]E[X]

Answer 30

The covariance is the product of X and Y, deviated from their means, so its units are the units of X multiple by Y. But, the correlation between X and Y is the covariance between X and Y divided by their standard deviations. Dividing by standard deviations means the correlation coefficient is unitless

Answer 31

Corr(Y,X) = Cov(Y,X) / squareroot(Var(Y)Var(X))

Answer 32

a^2Var(Y) + b^2Var(X) +2abCov(Y,X)

Answer 33

Probability: we know the population, and we are asking what is in the sample. Statistics: we don’t know the distribution. We need to make inferences about the distribution from a sample

Answer 34

Simple random sampling -> n objects are selected at random from a population and each member of the population is equally likely to be included in the sample Assumptions: 1. Identically distributed. Because Yi…Yn are randomly drawn from the same population, the marginal distribution of Yi is the same for each i…n. 2. Independently distributed.

Answer 35

1/n ∑{n, i=1}[Xi] It is a random variable. Members of the sample are random variables so the sample mean is a random variable.

Answer 36

Estimate the population parameter using the corresponding feature of the sample. (e.g. estimate population mean using the sample mean)

Answer 37

Population mean μx

Answer 38

expected value stays at population mean. variance goes to 0

Answer 39

When the sample size is large, the sample mean will be close to the population mean with high probability. The sample mean converges in probability to the population mean if the limit, as n tends to infinity, of the probability that the sample mean differs from the population mean in absolute magnitude by an amount greater than or equal to some positive value ε is equal to 0. See notes for formal expression (couldn't write in here)

Answer 40

(most of the time) we can demonstrate mean square convergence. This is because mean square convergence is more stringent than convergence in probability.

Answer 41

lim(n→∞)⁡E[(¯Xn-μx)^2]=0 2 conditions for mean square convergence: 1. the limit, as n tends to infinity, of the expectation of the sample mean is the population mean 2. the limits n tends to infinity, of the variance of the sample mean must be 0 Yes. The sample mean is an unbiased estimator so satisfied 1. Variance of sample mean is (σx^2)/n so as n tends to infinity, this will tend to 0

Answer 42

check notes

Answer 43

Markov’s Inequality: P(X≥c) ≤ E[X] / c If sub in ¯Xn-μx as random variable and use satisfaction of mean square convergence, will get to weak law of large numbers.

Answer 44

* Strong -> convergence almost surely * Weak -> convergence in probability (1) X1, X2…Xn needs to be i.i.d (2) E(|X|) < ∞ (require expectation of random variable to exist, it cannot be positive or negative infinity)

Answer 45

Convergence in probability (between estimator and what is being estimated)

Answer 46

An estimator θ ̂ is biased if E[θ ̂] ≠ θ | An unbiased estimator is one that is not biased. So an unbiased estimator satisfies E[θ ̂] = θ

Answer 47

θ ̂1 is relatively efficient to θ ̂2 if Var(θ ̂1) < Var(θ ̂2) Given that this is true for all values of θ ̂ and that both θ ̂1 and θ ̂2 are unbiased (we only compare variance given estimators are unbiased)

Answer 48

Best Linear Unbiased Estimator -> ¯Xn is the most efficient estimator among all estimators that ae unbiased and are linear functions of Y1…Yn see notes

Answer 49

1. X1,…Xn are i.i.d 2. E[Xi] = μx where μx < ∞ 3. Var(Xi) = σx^2 where 0 < σx^2 < ∞ (2 and 3 say that both mean and variance are finite (they exist))

Intro Probability & Stats Flashcards

(79 cards)