Probability and Statistical Inference NINTH EDITION

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1 GLOBAL EDITION Probability and Statistical Inference NINTH EDITION Robert V. Hogg Elliot A. Tanis Dale L. Zimmerman

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3 Section 4.5 The Bivariate Normal Distribution 167 We close this section by observing another important property of the correlation coefficient ρ if X and Y have a bivariate normal distribution. In Equation of the product h(y x)f X (x), let us consider the factor h(y x) ifρ = 0. We see that this product, which is the joint pdf of X and Y, equals f X (x)f Y (y) because when ρ = 0, h(y x) is a normal pdf with mean µ Y and variance σ 2 Y.Thatis,ifρ = 0, then the joint pdf factors into the product of the two marginal probability density functions, and hence X and Y are independent random variables. Of course, if X and Y are any independent random variables (not necessarily normal), then ρ, if it exists, is always equal to zero. Hence, we have proved the following theorem. Theorem IfX and Y have a bivariate normal distribution with correlation coefficient ρ, then X and Y are independent if and only if ρ = 0. Thus, in the bivariate normal case, ρ = 0 does imply independence of X and Y. Note that these characteristics of the bivariate normal distribution can be extended to the trivariate normal distribution or, more generally, the multivariate normal distribution. This is done in more advanced texts that assume some knowledge of matrices [e.g., Hogg, McKean, and Craig (2013)]. Exercises Let X and Y have a bivariate normal distribution with parameters µ X = 3, µ Y = 10, σ 2 = 25, σ 2 X Y = 9, and ρ = 3/5. Compute (a) P( 5 < X < 5). (b) P( 5 < X < 5 Y = 13). (c) P(7 < Y < 16). (d) P(7 < Y < 16 X = 2) Show that the expression in the exponent of Equation is equal to the function q(x, y) given in the text Let X and Y have a bivariate normal distribution with parameters µ X = 2.00, µ Y = 120, σx 2 = 0.25, σy 2 = 256, and ρ = 0.8. Compute (a) P(1.9 < X < 2.2). (b) P(1.9 < X < 2.2 Y = 109) Let X and Y have a bivariate normal distribution with µ X = 75, µ Y = 5, σx 2 = 144, σ Y 2 = 225, and ρ = 0.6. Find (a) E(Y X = 76). (b) Var(Y X = 76). (c) P(Y 90, X = 76) Let X denote the height in centimeters and Y the weight in kilograms of male college students. Assume that X and Y have a bivariate normal distribution with parameters µ X = 185, σ 2 = 100, µ X Y = 84, σ 2 Y = 64, and ρ = 3/5. (a) Determine the conditional distribution of Y, given that X = 190. (b) Find P(86.4 < Y < X = 190) For a freshman taking introductory statistics and majoring in psychology, let X equal the student s ACT mathematics score and Y the student s ACT verbal score. Assume that X and Y have a bivariate normal distribution with µ X = 22.7, σ 2 X = 17.64, µ Y = 22.7, σ 2 Y = 12.25, and ρ = (a) Find P(18.5 < Y < 25.5). (b) Find E(Y x). (c) Find Var(Y x). (d) Find P(18.5 < Y < 25.5 X = 23). (e) Find P(18.5 < Y < 25.5 X = 25). (f) For x = 21, 23, and 25, draw a graph of z = h(y x) similar to Figure For a pair of gallinules, let X equal the weight in grams of the male and Y the weight in grams of the female. Assume that X and Y have a bivariate normal distribution with µ X = 415, σ 2 X = 611, µ Y = 347, σ 2 Y = 689, and ρ = Find (a) P(309.2 < Y < 380.6). (b) E(Y x). (c) Var(Y x). (d) P(309.2 < Y < X = 385.1).

4 168 Chapter 4 Bivariate Distributions Let X and Y have a bivariate normal distribution with parameters µ X = 12, µ Y = 15, σ 2 = 16, σ 2 X Y = 25, and ρ = 0. Find (a) P(14 < Y < 16). (b) E(Y x). (c) Var(Y x). (d) P(14 < Y < 16 X = 12) Let X and Y have a bivariate normal distribution. Find two different lines, a(x) and b(x), parallel to and equidistant from E(Y x), such that P[a(x) < Y < b(x) X = x] = for all real x. Plota(x), b(x), and E(Y x) whenµ X = 2, µ Y = 1, σ X = 3, σ Y = 5, and ρ = 3/ In a college health fitness program, let X denote the weight in kilograms of a male freshman at the beginning of the program and Y denote his weight change during a semester. Assume that X and Y have a bivariate normal distribution with µ X = 75.00, µ Y = 5.00, σx 2 = 100, σ Y 2 = 2.56, and ρ = 0.6. (The lighter students tend to gain weight, while the heavier students tend to lose weight.) Find (a) P(1.8 Y 8.2). (b) P(1.8 Y 8.2 X = 80) For a female freshman in a health fitness program, let X equal her percentage of body fat at the beginning of the program and Y equal the change in her percentage of body fat measured at the end of the program. Assume that X and Y have a bivariate normal distribution with µ X = 3.9, µ Y = 3.4, σx 2 = 0.5, σ Y 2 = 0.6, and ρ = 0.8. Find (a) P(3.1 Y 3.7). (b) µ Y x, the conditional mean of Y, given that X = x. (c) σy 2 x, the conditional variance of Y, given that X = x. (d) P(3.1 Y 3.7 X = 4) Let ( ) 1 [ ] f (x, y) = e (x2 +y 2 )/2 1 + xye (x2 +y 2 2)/2, 2π < x <, < y <. Show that f (x, y) is a joint pdf and the two marginal pdfs are each normal. Note that X and Y can each be normal, but their joint pdf is not bivariate normal An obstetrician does ultrasound examinations on her patients between their 16th and 25th weeks of pregnancy to check the growth of each fetus. Let X equal the widest diameter of the fetal head, and let Y equal the length of the femur, both measurements in mm. Assume that X and Y have a bivariate normal distribution with µ X = 60.6, σ X = 11.2, µ Y = 46.8, σ Y = 8.4, and ρ = (a) Find P(40.5 < Y < 48.9). (b) Find P(40.5 < Y < 48.9 X = 68.6). HISTORICAL COMMENTS Now that we have studied conditional distributions, it might be appropriate to point out that there is a group of statisticians called Bayesians who believe in the following approach (it is considered again in Chapter 6): They treat the parameter θ (such as µ, σ 2, α, and β in the various distributions) as a random variable with a pmf (or pdf), say, g(θ). Suppose another random variable X, given θ, has the pmf (or pdf) f (x θ). Say the prior probabilities can be described by g(θ) so that the marginal pmf (or pdf) of X is given by the sum (or integral) h(x) = θ g(θ)f (x θ). Thus, the conditional pmf (or pdf) of θ, given that X = x, is k(θ x) = g(θ)f (x θ) h(x) g(θ)f (x θ) = g(τ)f (x τ). With a little thought you can recognize this formula as Bayes theorem. Here the posterior probabilities, k(θ x), of θ change from the prior probabilities given by g(θ), after X is observed to be x. Repeating the experiment n independent times (see Chapter 5), we obtain n values of x say, x 1, x 2,..., x n. The Bayesians use the τ

5 Section 4.5 The Bivariate Normal Distribution 169 posterior distribution k(θ x 1, x 2,..., x n ) to make their inferences about the parameter θ because they then know the conditional probabilities of θ, given x 1, x 2,..., x n. It is interesting to note that the Reverend Thomas Bayes, a minister who started this method of thinking, never published a mathematical article in his lifetime, and his famous paper was published about two years after his death. Clearly, he was not working for tenure at some university! More will be noted about Bayesian methods in Chapter 6.

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7 Distributions of Functions of Random Variables Chapter Functions of One Random Variable 5.2 Transformations of Two Random Variables 5.3 Several Random Variables 5.4 The Moment-Generating Function Technique 5.5 Random Functions Associated with Normal Distributions 5.6 The Central Limit Theorem 5.7 Approximations for Discrete Distributions 5.8 Chebyshev s Inequality and Convergence in Probability 5.9 Limiting Moment-Generating Functions 5.1 FUNCTIONS OF ONE RANDOM VARIABLE Let X be a random variable of the continuous type. If we consider a function of X say, Y = u(x) then Y must also be a random variable that has its own distribution. If we can find its cdf, say, G(y) = P(Y y) = P[u(X) y], then its pdf is given by g(y) = G (y). We now illustrate the distribution function technique by two examples. Example Let X have a gamma distribution with pdf f (x) = 1 Ŵ(α)θ α xα 1 e x/θ, 0< x <, where α>0, θ>0. Let Y = e X, so that the support of Y is 1 < y <. For each y in the support, the cdf of Y is That is, G(y) = P(Y y) = P(e X y) = P(X ln y). G(y) = ln y 0 1 Ŵ(α)θ α xα 1 e x/θ dx, and thus the pdf g(y) = G (y) ofy is ( ) 1 g(y) = Ŵ(α)θ α (ln y)α 1 (ln y)/θ 1 e, 1< y <. y 171