STATISTICAL LABORATORY, April 30th, 2010 BIVARIATE PROBABILITY DISTRIBUTIONS

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1 STATISTICAL LABORATORY, April 3th, 21 BIVARIATE PROBABILITY DISTRIBUTIONS Mario Romanazzi 1 MULTINOMIAL DISTRIBUTION Ex1 Three players play 1 independent rounds of a game, and each player has probability 1/3 of winning each round. 1) Find the joint distribution of the numbers of games won by each of the three players. 2) What are the probabilities of the following events: X 1 = X 2 = 5, X 1 = X 2 = 3? (Rice, 3.3) 1. Denote with X i, i = 1, 2, 3 the number of games won by the i-th player. The joint distribution of (X 1, X 2, X 3 ) is multinomial with parameters n = 1 (number of independent trials) and success probabilities p 1 = p 2 = p 3 = 1/3. The probability function is P (X 1 = x 1, X 2 = x 2, X 3 = x 3 ) = where x i satisfies the constraints x i 1, i = 1, We use R to answer the questions. 1! x 1!x 2!(1 x 1 x 2 )! (1/3)x 1 (1/3) x 2 (1/3) 1 x 1 x 2, > dmultinom(x = c(5, 5, ), size = 1, prob = rep(1/3, 3)) [1] > dmultinom(x = c(3, 3, 4), size = 1, prob = rep(1/3, 3)) [1] Ex2 Three cards are drawn at random and with replacement from the box containing 2 cards, each card with the name of a different italian region. Recall that there are 8 northern regions (N), 4 central regions (C) and 8 southern regions (S). Let (X N, X C, X S ) denote the joint distribution of the number of regions of the three areas. 1) What is the probability of no southern regions? one region from each area? 2) Describe the probability distribution X C X N = 1. 1

2 2 GENERAL BIVARIATE CONTINUOUS DISTRIBUTIONS 2 1. and P (X S = ) = P (X N = X C = X S = 1) = 3! = This is a Binomial distribution: X C X N = 1 Bi(2, 1/3), whose determinations are, 1, 2 with probabilities 4/9, 4/9, 1/9. 2 GENERAL BIVARIATE CONTINUOUS DISTRI- BUTIONS Ex1 A bivariate density function is defined as follows { 4x(1 y), x 1 y 1, f X,Y (x, y) =, elsewhere. 1) What are the marginal distributions? Are they uniform? 2) Are X and Y stochastically independent? 3) Compute the joint cdf values at the points (2, 1/2), ( 1/2, 1/2), (1/2, 1/2). 1. Marginal densities are obtained by integrating out the other variable. f X (x) = f Y (y) = f X,Y (x, y)dy = 4x f X,Y (x, y)dx = 4(1 y) (1 y)dy = 2x, x 1 and elsewhere, xdx = 2(1 y), y 1 and elsewhere. The marginal distributions are not uniform, because neither density is constant. 2. X and Y are stochastically independent because the joint density is identically equal to the product of the marginal densities: for all pairs of real numbers f X,Y (x, y) = f X (x)f Y (y). 3. Note that stochastic independence implies F X,Y (x, y) = F X (x)f Y (y), where F X and F Y are the marginal cdf s. Therefore F X,Y (2, 1/2) = F X (2)F Y (1/2) = F Y (1/2) = 2 F X,Y ( 1/2, 1/2) = F X ( 1/2)F Y (1/2) =, /2 F X,Y (1/2, 1/2) = F X (1/2)F Y (1/2) = (3/4)F X (1/2) = (3/2) 2(1 y)dy = 3/4, /2 xdx = 3/16.

3 2 GENERAL BIVARIATE CONTINUOUS DISTRIBUTIONS 3 Ex2 A bivariate density function is defined as follows { x + y, x 1 y 1, f X,Y (x, y) =, elsewhere. 1) Describe the contours of the bivariate density. 2) Derive the marginal distributions. Are X and Y stochastically independent? 3) Compute the joint cdf value at the point (1/2, 1/2). 4) Obtain the conditional densities of Y X = x, X Y = y. 1. Observe that the joint pdf varies between (at the point, ) and 2 (at the point 1, 1). The contours are the subsets of the unit square Q with a constant value c 2 of the density, that is (x, y) Q : x + y = c. Therefore, the contours are parallel segments, more precisely, they are the intersections of the parallel lines x + y = c with Q. The figure below shows the plots of the contours and of the bivariate density. The corresponding R code is > f <- function(x, y) x + y > x <- seq(, 1, length = 5) > y <- seq(, 1, length = 5) > z <- outer(x, y, f) > contour(x, y, z, col = "black", lty = "solid", asp = 1, lwd = 2, + xlab = "X", ylab = "Y", main = "Contours of f(x,y) = x+y") > persp(x, y, z, theta = 3, phi = 3, expand =.5, col = "lightblue", + ltheta = 12, shade =.75, ticktype = "detailed", xlab = "X", + ylab = "Y", zlab = "Density", main = "Plot of f(x,y) = x+y") 2. Marginal pdf s are f X (x) = f Y (y) = f X,Y (x, y)dy = f X,Y (x, y)dx = Here, independence test fails because clearly 3. The joint cdf is (x + y)dy = x + 1/2, x 1 and elsewhere, (x + y)dx = y + 1/2, y 1 and elsewhere. f X,Y (x, y) = x + y (x + 1/2)(y + 1/2) = f X (x)f Y (y). F X,Y (x, y) = Hence, F X,Y (1/2, 1/2) = 1/8., x ory, xy(x + y)/2, x 1 y 1, x(x + 1)/2, x 1 y 1, y(y + 1)/2, y 1 x 1, 1, x 1 y 1.

4 3 BIVARIATE NORMAL DISTRIBUTION 4 Contours of f(x,y) = x+y Plot of f(x,y) = x+y Y Density X Y X 4. We use the definition of conditional density. For any fixed x 1, f Y X=x (y) = f X,Y (x, y) f X (x ) Similarly, for any fixed y 1, f X Y =y (y) = f X,Y (x, y ) f Y (y ) = x + y, y 1 and elsewhere. x + 1/2 = x + y, x 1 and elsewhere. y + 1/2 3 BIVARIATE NORMAL DISTRIBUTION Ex1 X and Y have a bivariate normal distribution with parameters µ X = 5, µ Y = 1, σ X = 1, σ Y = 5 and ρ >. It is also known that What is the value of ρ? P (4 < Y < 16 X = 5) =.954. What is required is the conditional distribution Y X = 5. From the general properties of the bivariate normal distribution, Y X = x is a univariate normal distribution with mean function (regression function) µ Y X (x) = µ Y + ρ σ Y σ X (x µ X ) and standard deviation (not dependent on x) σ Y X = σ Y (1 ρ 2 ) 1/2.

5 3 BIVARIATE NORMAL DISTRIBUTION 5 In the present case, replacing known parameters and x = 5, gives Now, µ Y X (5) = µ Y = 1, σ Y X = 5(1 ρ 2 ) 1/2. P (4 < Y < 16 X = 5) = F Y X=5 (16) F Y X=5 (4) = F XST (/(1 ρ 2 ) 1/2 ) F XST ( /(1 ρ 2 ) 1/2 ) = 2F XST (/(1 ρ 2 ) 1 where F XST is the cdf of the standard normal distribution. The previous equation holds iff F XST (/(1 ρ 2 ) 1/2 ) =.977, and the solution is the quantile of the standard normal distribution of the order p =.977, that is ) = (1 ρ 2 x(st ) 1/ We use R to obtain a very precise value of x (ST ).977. > qnorm(.977, mean =, sd = 1) [1] The value of ρ is the solution of the equation that is, ρ.799. = , (1 ρ 2 ) 1/2 Ex2 X and Y have a bivariate normal distribution with parameters µ X = 2, µ Y = 4, σ X = 3, σ Y = 2 and ρ =.6. What is shortest interval for Y X = 22 containing 9% of the probability? Y X = 22 has a normal distribution with parameters µ Y X (22) = 4.8, σ Y X = 1.6. As we know, the endpoints of the shortest interval with 9% probability are the quantiles of the orders.5 and.95, respectively. > qnorm(c(.5,.95), mean = 4.8, sd = 1.6) [1]