Probability And Statistics - Key Words Flashcards

Question

What formula should be used to work out the number of outcomes when order matters, and there is no replacement?

Answer 1

n!/(n-r)! , where r is the objects (quantity of numbers in the lottery) and n is the number of possible numbers to choose from.

Answer 2

n^r, where n is the range of numbers you can select, and r is the quantity of numbers needed to be picked.

Answer 3

n! / [r! (n-r)!], these are called combinations.

Answer 4

(n + r - 1)!/[r! (n-1)!]

Answer 5

This is when you have to find the probability of an event A happening, given event B has already occurred. Essentially, event B is the new sample space.

Answer 6

-52 cards -13 denominations (2 to 10, ace, Jack, King, Queen) -4 suits: Diamonds and hearts are red, Club and spades are black

Answer 7

P(AnB) = P(B) x P(A|B) = P(A) x P(B|A). Remember, the denominator cannot equal zero.

Answer 8

P(AuB) = P(A) + P(B) - P(AnB)

Answer 9

P(A) x P(B) = P(AnB)

Answer 10

Answer 11

A variable X will be a random variable (r.v) if it’s values can be thought to be determined by outcomes of a random experiment. This means that X will depend on the outcome we get is.

Answer 12

This is a random variable with values determined by a finite or a countably infinite number of events.

Answer 13

A continuous random variable is a random variable with values determined by an infinite number of events which are not countable, for example drawing a random real number between 0 and 1.

Answer 14

If X is a discrete random variable, and S is the set of values determined by outcomes of the random experiment, then for function f(x): -f(x) >= 0 for all real numbers x -the summing over all the values that have a positive probability of occurrence will equal 1. There may also be values of X that aren’t determined by the outcomes of a random experiment, and these will have a value of zero. When this occurs, we think of X=0 as implying the experiment has not taken place.

Answer 15

P(X=x) = f(x). This means the probability the random variable X takes is equal to pmf evaluated at x.

Answer 16

For a continuous random variable where f(x) >= 0 for all real values of x, and the definitive integral from -infinity to infinity is 1, you get a probability density function.

Answer 17

F(x) = P(X <= x) As it is cumulative, the probability X is the always either equal to or less than what p(x) will be, as it will be at least as high as the current probability.

Answer 18

E(X) = sum of x(i) x f(xi) I’n words it is the weighted average of its values, where weights are the corresponding values of the pmf, it is a measure of central tendency of probability around the random variable. The probability of the random variable could be zero, but the expected value is just how the probability is distributed around this point.

Answer 19

V(X) = sum of (xi- mu(xi))^2 x f(xi) In words, the variance of a random variable is the weighted average of the squared deviations of its values from the mean, where weights are corresponding to values of the pmf (their probability. The standard deviation is the positive square root of the variance.

Answer 20

V(x) = the finite integral of (x - mu(x))^2) x f(x).

Answer 21

The median of a random variable is the point that splits the distribution of X into two parts which overlap at the median, each part will have a probability of at least 0.5. If P(X = m) = 0, then both parts of the distribution are exactly equal to 0.5. This will always be the case for a continuous random variable, and may (but not necessarily) apply to a discrete random variable. -Unlike the mean, the median doesn’t necessarily have to be a unique value.

Answer 22

The mean is the constant that minimises E[(X - c)^2], and the median is the constant that minimises E(|x - c|). From this, we can see that the mean will be more affected by extremes, because it will square the prediction error, where as the median keeps a constant absolute value no matter how large the error is. This means when large anomalies are present, the median will be the more accurate measure.

Answer 23

If we let a random variable Y = g(X) be a function of X, where X has a pmf or pdf f(x), then Y is also a random variable. Therefore, E(Y): -If discrete = sum of g(x) multiplied by f(x). -If continuous = the integral of f(x) times g(x). We have however seen this before, as g(x) is equal to (x - mu(x))^2.

Answer 24

This is because the marks are dependent on the random factors of each student that impact their ability to do well on the exam (e.g sleep quality night before, cognitive ability, time revised for etc).

Answer 25

To do this, we can use random variable techniques, so we can see what would’ve happened if something else didn’t happen. To see if one variable is associated to another, we can use conditional probabilities, and see what the correlation between, say, grades in 2nd year maths would be, given you get a specific grade in 1st year maths.

Answer 26

1) We cannot hope to easily summarise the relationship between the two marks. 2) We cannot hope to derive features of their relationship between these two variables to compare it to other random variables.

Answer 27

This if the joint probability of two random variables both occurring at the same time, and this can be displayed by a table.

Answer 28

If you have two discrete random variables X and Y, then there will be a function f(X, Y) such that: 1) f(X,Y) >= 0 for all pairs of real numbers (x, y). 2) the double summation of all the pairs of values X and Y will equal 1. To see how to do double summation, look at lecture slides 4. The probability assignment P(X = x, Y = y) = f(X = x, Y = y)

Answer 29

For continuous random variables X and Y, a function f(x,y) such that: -f(x,y) is non-negative for all real numbers (x,y) -The double integral is equal to 1. Bare in mind, as this is a 2 variable function, the function will make a 3D graph. The probability will be represented by the volume under the p.d.f surface.

Answer 30

F(X = x, Y = y) = P(X <= x, Y <= y). We will add up all the probabilities which apply to this constraint, for both variable by looking at the joint frequency table.

Answer 31

When we have joint distribution we can derive the probability mass functions of each of the random variables that are jointly distributed. From the joint probability mass function f(X,Y) we can derive the pmf of X and Y denoted f1(X) and f2(Y), as they are an ordered pair, with X first and Y second.

Answer 32

1) We need to find f(x,y) for all of x such that f(x,y) is strictly positive 2) Sum all the strictly positive pairs for all the values of x which fit the criteria in step 1. 3) once all are summed, this should equal 1, it may not if the decimals have been rounded early.

Answer 33

-To do this, all you have to do is integrate the variable which you aren’t taking in the marginal distribution of. -Remember, for each part of the function from minus infinity to infinity where the function changes, a new integral is needed. The total join c.p.f

Answer 34

If X and Y are discrete random variables, then the function f(X, Y) such that: -f(x,y) >= 0 for all pairs of real numbers (x, y). -the double summation of the joint probabilities should all add to one. To see how to do this, refer to lecture notes.

Answer 35

Conditional expectations are the expectations of certain random variables given a condition of the other variable in the joint function. For example, for the random variables X and Y, the conditional mean/expectation of X conditional on y is: E(X | Y = y) = mu(X|y).

Answer 36

We have seen how we can condition the probability of an event on the occurrence of another event, so the joint distribution functions allows us to condition distributions of one random variable on the values of another random variable, allowing us to calculate the probability of one random variable, conditional upon the value of another. Mathematically, this looks exactly the same as normal conditional probability.

Answer 37

The variance of conditional distribution is called the conditional variance. For a random variable X and Y, the conditional variance of X given y is: V(X|Y=y) = E[(X - mu(X|y)^2 | y] = (sigma)^2 X|y

Answer 38

The conditional expected function of, say random variables X and Y, where we want the expectation of X given Y is any y, just do the same as a conditional pdf of X, but without specifying the value of Y just leaving it as y in the equation.

Answer 39

If we let Z = g(X, Y) be a function of random variables X and Y, where X and Y have a joint pmf of pdf, the expected value of Z is given given by: E(Z) = sum of x(i)(sum of y(j))[g(x(i), y(j)) x f(x(i), y(j)) for discrete values, and: E(Z) = intergral g(x, y) x f(x, y) dx dy

Answer 40

g(X, Y) = (X - mu(X))(Y - mu(Y)). The expectation of this function made up of two different random variables is called the covariance, denoted Cov(X, Y). The formula for how to calculate this will vary depending on in the random variables are discrete if continuous. Both can be found in the lecture slides/notes. NOTE: The calculations of the covariance can be shortened using the formula Cov(X, Y) = E(XY) - mu(X)mu(Y)

Answer 41

If X takes its value above the mean, and Y also does, then covariance will be positive. But if when X takes a value above it’s mean, Y tends to take a value below its mean, the covariance will be negative. Also, the value of the covariance will depend on its units. If X and Y were measured in GDP, but then changed to pence, covariance will be multiplied by 10,000, without any relation of X and Y changing. Therefore, the covariance value itself has little meaning, but it is meaningful to compare the signs of covariance of random variables.

Answer 42

If we let X and Y be random variables with covariance Cov(X, Y) and variances V(X) and V(Y), then the correlation coefficient or correlation is given by: Corr(X, Y) = Cov(X, Y)/(sqrt(V(X)) x sqrt(V(Y))) The correlation will have the same sign that the covariance does. The correlation coefficient will take a value between -1 and 1 inclusive. -1 signifies a perfect negative correlations, and 1 a perfect positive correlation.

Answer 43

This requires that for two random variables X and Y, the probability of any event defined in terms of X doesn’t change when know any value of Y has occurred. If g1(x | y) = f1(x) , and g1(x | y) = f(x, y)/f2(y), then we can say that if: X and Y are jointly distributed random variables with joint pmf or pdf f(X,Y)bans marginal pmf or pdf’s f1(X) and f2(Y), then X and Y are statically independent if and only if: f(X, Y) = f1(X) x f2(Y). This must stand for ALL values of x and y.

Answer 44

They will have a covariance of zero (can be proven mathematically), but the reverse of this will not necessarily be true. also, if X and Y are independent, then U =f(X) and V = g(Y) are also independent.

Answer 45

For two random variables X and Y, the expressions for expectations of linear functions are as shown, if we let Z = a + bX + cY: E(Z) = a + bE(X) + cE(Y) Var(Z) = (b^2)Var(X) + (c^2)Var(Y) + 2bcCov(X,Y)

Answer 46

If we have Z1 = a1 + b1X + c1Y and Z2 = a2 + b2X + c2Y: Cov(Z1,Z2) = b1xb2xVar(X) + c1xc2xVar(Y) + (b1xc2+b2xc1)Cov(XY)

Answer 47

If the conditional expectations of Y doesn’t vary as X varies, then we can say that Y is mean independent of X, denoted as E(Y|X) = mu(y). This means the CEF of Y conditional on X will be constant.

Answer 48

Mean independence is weaker than independence, and it doesn’t imply independence, but independence does imply mean independence. -If Y is mean independent of X, the covariance between X and Y is zero, but the reverse isn’t necessarily true. -Just because Y is mean independent of X, X won’t always be mean independent of Y.

Answer 49

The same as we would do with two r.v functions, we could obtain the CEF of Y conditional on X, Y conditional on Z, or Y conditional on X and Z. If we wanted E(Y|X), we would first have to get the joint marginal probability of X and Y, to Mel image the 3rd random variable Z. To do E(Y|XZ), we would first need the marginal probability function of XZ, so we can then find Y, conditional on the combinations of X and Z. Finding these can allow us to do more in depth analysis and calculations about data.

Answer 50

It means that the CEF will be far easier to interpret, but it is important to note we mean linear in their parameters, not their variables. E(Y|X,Z) = b0 + b1X + b2Z+ b3XZ, for the example in the slides, see lecture notes for a more thorough explanation. When the conditioning variables are binary (can only take 0 or 1), the CEF can always be expressed as an equation linear to the parameters.

Answer 51

If we have n random variables, then we just have to find the summations of all the different random variable combinations such that the outcome of the function is > 0 (as we can’t have a negative probability). From this, we can then do probability assignment, the same way in which we would’ve done before. If continuous, you need to do the integral for as many random variables as you have to get the probability density functions. Marginal probabilities can also be extended in a similar way, but note not all binary relationships will be transitive, just because X is independent of Y, and Y is independent of Z, this doesn’t mean X will be independent of Z.

Answer 52

If you have f(x,y,z) and x and z are independent, but x and y aren’t the joint pmf can be written: f(x,y,z) = f3(z) x f12(x,y) It also applies that the expectation of their product will be product of their expectations if independent.

Answer 53

A parametric family is when probability functions have the exact same mathematical formulations except for one or more constants are grouped together in parametric families. The same formulation apart from one constant is a one-parameter family of distribution, and so on.

Answer 54

The normal distribution of a random variable is a probability functions with two parameters, the standard deviation and the mean (expected value). When the formula is applied to this random variable, it is known a normal random variable.

Answer 55

-They’re symmetrical about the mean. -Linear functions of normal random variables are also normal random variables. If X is a normal r.v and Y = a + bX, then Y~N(a + b(mu), b^2(s.d)^2)

Answer 56

If we have X as a normal random variable, and Y = a + bX: If a = -mu/s.d and b = 1/s.d the normal distribution of Y will be the standard normal r.v, Y~N(0,1).

Answer 57

It will give the cdf of that random variable, so if you want to find the probability if it being between the value n and the mean, you need to minus 0.5, as the negatives side of the mean will also be included in the cumulative probability.

Answer 58

Statistical inference allows us to use a sample from an unknown distribution (also known as a population), to make inferences about unknown distributions. We do this by first learning how to deal with samples from known populations, and then using this knowledge to see how we can extract information from samples in order to infer feature of an underlying unknown population.

Answer 59

A random sample of random variables X are independent drawings from the distribution (or population). The ordered set (from X1 to Xn) is called the random sample, and has a size on n on the random variable X. Each value of Xi is a random variable as their values are determined by an experiment. Xi is independently and identically distributed from other Xi’s.

Answer 60

This is the total summation of all the different random samples from the sample mean squares, divided by (n - 1).

Answer 61

Gn(X) = the product individual marginal probability of each random variable in the sample, as they are independent and the laws of independent random variables apply.

Answer 62

Let T = h(X1, X2, X3, X4…… Xn) = h(x). So, the values of T are decided by a function of h() of the random sample. This makes T a sample statistic.

Answer 63

This is the arithmetic average of the random sample. The sample variance then follows this, also using the sample mean to get the sample variance.

Answer 64

The distribution of a sample statistic is called a sampling distribution, it depends on 3 things in general: 1) The function that determines the values of the sample statistic 2) The distribution from which we choose the sample (the pmf or pdf) 3) The size n of the random sample.

Answer 65

Given a random sample size of n, from a population with E(X) = mu and V(X) = population variance: -Expected value of the sample mean will equal the population mean -Variance of the sample mean will equal population mean/n.

Answer 66

When sampling the r.v X, with sample size n, X ~ N(mu, variance): Sample Mean ~ N(mu, variance/n). This follows directly from the sample mean theorem.

Answer 67

In random sampling size of n on any r.v with E(X) = mu and V(X) = variance, as the size of the random sample increases, the sampling distribution of the sample mean approaches a normal distribution with mean mu and variance: variance/n. The standardised sample mean Z = (sample mean - mu)/(population s.d/sqrt(n)) approaches the standard normal distribution N(0,1).