Joint Distribution of Two or More Random Variables

Transcription

1 Joint Distribution of Two or More Random Variables Sometimes more than one measurement in the form of random variable is taken on each member of the sample space. In cases like this there will be a few random variables defined on the same probability space and we would like to explore their joint distribution. Joint behavior of 2 random variable (continuous or discrete), X and Y determined by their joint cumulative distribution function n dimensional case ( x, y) = P( X x, Y ). F X, Y y ( x,..., x ) = P( X x1, K, X x ). F n X1,..., X 1 n 1 n n STA286 week 3 1

2 Discrete case Suppose X, Y are discrete random variables defined on the same probability space. The joint probability mass function of 2 discrete random variables X and Y is the function p X,Y (x,y) defined for all pairs of real numbers x and y by For a joint pmf p X,Y (x,y) we must have: p X,Y (x,y) 0 for all values of x,y and For any region A in the xy plane, ( x, y) = P( X = x and Y y) p X, Y = x y p ( x, y) 1 X, Y = [( X, Y ) A] p ( x y) P, = A XY STA286 week 3 2

3 Example for illustration Toss a coin 3 times. Define, X: number of heads on 1st toss, Y: total number of heads. The sample space is Ω ={TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}. We display the joint distribution of X and Y in the following table Can we recover the probability mass function for X and Y from the joint table? To find the probability mass function of X we sum the appropriate rows of the table of the joint probability function. Similarly, to find the mass function for Y we sum the appropriate columns. STA286 week 3 3

4 Marginal Probability Function The marginal probability mass function for X is ( x) p ( x y) p X = X, Y, The marginal probability mass function for Y is y ( y) p ( x y) p Y = X, Y, x Case of several discrete random variables is analogous. STA286 week 3 4

5 Example Roll a die twice. Let X: number of 1 s and Y: total of the 2 dice. There is no available form of the joint mass function for X, Y. We display the joint distribution of X and Y with the following table. The marginal probability mass function of X and Y are Find P(X 1 and Y 4) STA286 week 3 5

6 Conditional Probability Distribution Given the joint pmf of X and Y, we want to find and P ( X = x Y = y) ( = y X = x) P Y Back to the example on slide 5, ( = 2 X = 1) = 0 P Y ( = X = 1) P Y M = = P P 2 3 = 36 = ( = 12 X = 1) = 0 P Y ( X = x and Y = y) P( Y = y) ( X = x and Y = y) P( X = x) 1 5 These 11 probabilities give the conditional pmf of Y given X = 1. STA286 week 3 6

7 Definition For X, Y discrete random variables with joint pmf p X,Y (x,y) and marginal mass functions p X (x) and p Y (y). If x is a number such that p X (x) > 0, then the conditional pmf of Y given X = x is p Is this a valid pmf? ( y x) = p ( y X = x) Y X Y X = p X, Y p ( x, y) ( x) X Similarly, the conditional pmf of X given Y = y is p ( x y) = p ( x Y = y) X Y X Y = p X, Y p Y ( x, y) ( y) STA286 week 3 7

8 The Joint Distribution of two Continuous R.V s Definition Random variables X and Y are (jointly) continuous if there is a non-negative function f X,Y (x,y) such that (( X Y ) A) f ( x y) P,, X, Y A = for any reasonable 2-dimensional set A. dxdy f X,Y (x,y) is called a joint density function for (X, Y). In particular, if A = {(X, Y): X x, Y x}, the joint CDF of X,Y is X, Y ( x y) f ( u, v) F du, dv, = x y X, Y From Fundamental Theorem of Calculus we have f 2 x y 2 y x ( x y) = F ( x, y) = F ( x y) X. Y, X, Y X, Y, STA286 week 3 8

9 Properties of joint density function ( x, y) 0 f X, Y for all x, y R It s integral over R 2 is f X Y ( x, y), dxdy = 1 STA286 week 3 9

10 Example Consider the following bivariate density function f X, Y ( x, y) 2 ( x + xy) Check if it s a valid density function. = x 1, 0 otherwise y 1 Compute P(X > Y). STA286 week 3 10

11 Marginal Density Functions The marginal density of X is then f ' ( x) F ( x) = f ( x y) X X X, Y, = dy Similarly the marginal density of Y is f ( y) f ( x y) Y X, Y, = dx STA286 week 3 11

12 Example Consider the following bivariate density function f X, Y ( x, y) = 6xy x 1, 0 y 1 otherwise Check if it is a valid density function. Find the marginal densities of X and Y. STA286 week 3 12

13 Example Consider the joint density f X, Y ( x, y) = 2 λ e 0 λ y 0 x y otherwise where λ is a positive parameter. Check if it is a valid density. Find the marginal densities of X and Y. STA286 week 3 13

14 Conditional densities If X, Y jointly distributed continuous random variables, the conditional density function of Y X is defined to be ( y x) if f X (x) > 0 and 0 otherwise. f Y X = f X, Y f X ( x, y) ( x) Similarly, the conditional density of X Y is given by ( x y) if f Y (y) > 0 and 0 otherwise. f X Y = f X, Y f Y ( x, y) ( y) STA286 week 3 14

15 Example Consider the joint density f X, Y ( x, y) 2 λ e = 0 λy Find the conditional density of X given Y and the conditional density of Y given X. 0 x y otherwise STA286 week 3 15

16 Independence of Random Variables Definition Random variables X and Y are independent if the events ( X A) and ( Y B) are independent. Theorem: Two discrete random variables X and Y with joint pmf p X,Y (x,y) and marginal mass function p X (x) and p Y (y), are independent if and only if p ( x, y) p ( x) p ( y) X, Y = X Y Note: If X, Y are independent random variables then P X Y (x y) = P X (x). Question: Back to the rolling die 2 times example, are X and Y independent? STA286 week 3 16

17 Theorem Suppose X and Y are jointly continuous random variables. X and Y are independent if and only if given any two densities for X and Y their product is the joint density for the pair (X,Y) i.e. f ( x, y) f ( x) f ( y) X, Y = X Y If X, Y are independent then f ( y x) f ( y ) Y X = Y STA286 week 3 17

18 Example Suppose X and Y are discrete random variables whose values are the nonnegative integers and their joint probability function is p X 1 x y ( λ+ μ ), Y ( x, y) = λ μ e x, y = x! y! 0,1,2... Are X and Y independent? What are their marginal distributions? Factorization is enough for independence, but we need to be careful of constant terms for factors to be marginal probability functions. STA286 week 3 18

19 Example and Important Comment The joint density for X, Y is given by f X, Y ( x, y) 4 = 2 ( x + y ) 0 x, y > 0, x + otherwise y 1 Are X, Y independent? Independence requires that the set of points where the joint density is positive must be the Cartesian product of the set of points where the marginal densities are positive i.e. the set of points where f X,Y (x,y) >0 must be (possibly infinite) rectangles. STA286 week 3 19

20 Expectation In the long run, rolling a die repeatedly what average result do you expact? In 6,000,000 rolls expect about 1,000,000 1 s, 1,000,000 2 s etc. Average is: 1,000,000 1 For a random variable X, the Expectation (or expected value or mean) of X is the expected average value of X in the long run. It is also called the mean of the probability distribution of X. Symbols: μ, μ X, E(X) and EX. ( ) + 1,000,000( 2) + L + 1,000,000( 6) 6,000,000 = 3.5 STA286 week 3 20

21 Expectation of Random Variable For a discrete random variable X with pmf p X (x) E ( X ) = x p( x) whenever the sum converge absolutely (i.e. x p( x) < ) For a continuous random variable X with density f X (x) E whenever this integral converge absolutely. x ( X ) x f ( x) = dx X x STA286 week 3 21

22 Examples 1) Roll a die. Let X = outcome on 1 roll. Then E(X) = ) Bernoulli trials P(X = 1) = p and P(X = 0) = 1- p. Then 3) X ~ Uniform(a, b). Then ( X ) = 1 p + 0 ( 1 p) p E = 4) X is a random variable with density 2 x for x > 1 f X ( x) = 0 otherwize (i) Check if this is a valid density. (ii) Find E(X). STA286 week 3 22

23 Theorem For g: R R If X is a discrete random variable then [ g( X )] g( x) p X ( x) = If X is a continuous random variable E E x [ g( X )] g( x) f ( x) = dx X Note, this theorem can be generalized to bivariate distributions. STA286 week 3 23

24 Examples 2 1. Suppose X ~ Uniform(0, 1). Let Y = X then, 2. Suppose X has the following probability mass function X Let, then Y = e p X x λ e x! λ ( x) = x = 0,1, 2,... STA286 week 3 24

25 Properties of Expectation For X, Y random variables and a, b R constants, E(aX + b) = ae(x) + b Proof: Continuous case If X is a non-negative random variable, then E(X) = 0 if and only if X = 0 with probability 1. If X is a non-negative random variable, then E(X) 0 E(a) = a E[g(X) ± h(x)] = E[g(X)] ± E[h(X)] STA286 week 3 25

26 Variance The expected value of a random variable E(X) is a measure of the center of a distribution. The variance is a measure of how closely concentrated to center (µ) the probability is. It is also called 2nd central moment. Definition The variance of a random variable X is Var 2 2 [ ] = E[ ( μ) ] ( X ) E ( X E( X )) = X Claim: Proof: Var ( X ) = E( X ) ( E( X )) = E( X ) μ We can use the above formula for convenience of calculation. The standard deviation of a random variable X is denoted by σ X ; it is the square root of the variance i.e.. σ X = Var ( X ) STA286 week 3 26

27 Properties of Variance For X, Y random variables and are constants, then Var(aX + b) = a 2 Var(X) Proof: Var(aX + by) = a 2 Var(X) + b 2 Var(Y) + 2abE[(X E(X ))(Y E(Y ))] Proof: Var(X) 0 Var(X) = 0 if and only if X = E(X) with probability 1 Var(a) = 0 STA286 week 3 27

28 Examples Suppose X ~ Uniform(0, 1), then ( X ) = and E X = therefore Var( X ) = = E ( ) ( ) p 2. Suppose X ~ Bernoulli(p), then X and therefore, 2 Var X = p p = p 1 p ( ) ( ) E = E ( X ) = 1 p + 0 q = p 3. Suppose X has the following probability mass funcation p X x 1 ( x) = 0.2( 0.8) x = 1, 2,,... find the mean and variance of X. STA286 week 3 28

29 Properties of Expectations Involving Joint Distributions For random variables X, Y and constants a, b R Proof: E(aX + by) = ae(x) + be(y) For independent random variables X, Y E(XY) = E(X)E(Y) whenever these expectations exist. Proof: STA286 week 3 29

30 Covariance Recall: Var(X+Y) = Var(X) + Var(Y) +2 E[(X-E(X))(Y-E(Y))] Definition For random variables X, Y with E(X), E(Y) <, the covariance of X and Y is Covariance measures whether or not X-E(X) and Y-E(Y) have the same sign. Claim: Proof: Cov Cov ( X, Y ) = E[ ( X E( X ))( Y E( Y ))] ( X, Y ) = E( XY ) E( X ) E( Y ) Note: If X, Y independent then E(XY) =E(X)E(Y), and Cov(X,Y) = 0. STA286 week 3 30

31 Example Suppose X, Y are discrete random variables with probability function given by y x p X (x) -1 1/8 1/8 1/8 0 1/8 0 1/8 1 1/8 1/8 1/8 p Y (y) Find Cov(X,Y). Are X,Y independent? STA286 week 3 31

32 Important Facts Independence of X, Y implies Cov(X,Y) = 0 but NOT vice versa. If X, Y independent then Var(X+Y) = Var(X) + Var(Y). If X, Y are NOT independent then Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y). Cov(X,X) = Var(X). STA286 week 3 32

33 Properties of Covariance For random variables X, Y, Z and constants a, b, c, d R Cov(aX+b, cy+d) = accov(x,y) Cov(X+Y, Z) = Cov(X,Z) + Cov(Y,Z) Cov(X,Y) = Cov(Y, X) STA286 week 3 33

34 Correlation Definition For X, Y random variables the correlation of X and Y is ( ) Cov X, Y ρ( X, Y ) = V ( X ) V ( Y ) whenever V(X), V(Y) 0 and all these quantities exists. Claim: Proof: ρ(ax+b,cy+d) = ρ(x,y) This claim means that the correlation is scale invariant. STA286 week 3 34

35 Theorem For X, Y random variables, whenever the correlation ρ(x,y) exists it must satisfy -1 ρ(x,y) 1 STA286 week 3 35

36 Interpretation of Correlation ρ ρ(x,y) is a measure of the strength and direction of the linear relationship between X, Y. [ ] If X, Y have non-zero variance, then ρ 1,1. If X, Y independent, then ρ(x,y) = 0. Note, it is not the only time when ρ(x,y) = 0!!! Y is a linearly increasing function of X if and only if ρ(x,y) = 1. Y is a linearly decreasing function of X if and only if ρ(x,y) = -1. STA286 week 3 36

37 Example Find Var(X-Y) and ρ(x,y) if X, Y have the following joint density f X, Y ( x, y) = 3x 0 0 y x 1 otherwise STA286 week 3 37

38 Markov s Inequality If X is a non-negative random variable with E(X) < and a >0 then, Proof: P ( X a) E( X ) a STA286 week 3 38

39 Chebyshev s Inequality For a random variable X with E(X) < and V(X) <, for any a >0 Proof: P ( X E( X ) a) V a ( X ) 2 STA286 week 3 39