Statistical Analysis of Random Variables - Homework Assignment Solutions, Assignments of Probability and Statistics

Download Statistical Analysis of Random Variables - Homework Assignment Solutions and more Assignments Probability and Statistics in PDF only on Docsity! Statistics 341 Fall 2008 - Homework Assignment #8 Answers We will go over some of these problems during the last week of classes 1. Let Y1 and Y2 have the joint probability density function given by f(y1, y2) = { 6(1 − y2) 0 ≤ y1 ≤ y2 ≤ 1 0 elsewhere (a) Find P (Y1 ≤ 3/4, Y2 ≥ 1/2). The region where Y1 ≤ 0.75 and Y2 ≥ 0.5 within the support of Y1 and Y2 is shown in Figure 1. This region must be split into two parts. The two parts I chose are shown in the graph. You could choose to break this region up in a different manner. 0.0 0.2 0.4 0.6 0.8 1.0 0. 0 0. 2 0. 4 0. 6 0. 8 1. 0 First region ∫ 1 0.5 ∫ 0.5 0 6(1 − y2)dy1dy2 = ∫ 1 0.5 6(1 − y2) ( y1| 0.5 0 ) dy2 = ∫ 1 0.5 3(1 − y2)dy2 = ( 3y2 − 1.5y 2 2 ) |10.5 = 1.5 − 1.125 = 0.375 Second region ∫ 0.75 0.5 ∫ 1 y1 6(1 − y2)dy2dy1 = ∫ 0.75 0.5 ( 6y2 − 3y 2 2| 1 y1 ) dy1 = ∫ 0.75 0.5 3 − 6y1 + 3y 2 1dy1 = ( 3y1 − 3y 2 1 + y 3 1 ) |0.750.5 = 0.984375 − 0.875 = 0.109375 1 Adding the two regions together gives 0.375 + 0.109375 = 0.484375. (b) Find the marginal distributions for both Y1 and Y2. The marginal distribution of Y1 is f1(y1) = ∫ 1 y1 6(1 − y2)dy2 = ( 6y2 − 3y 2 2| 1 y1 ) = 3y21 − 6y1 + 3 = 3(1 − y1) 2 for 0 ≤ y1 ≤ 1 and 0 otherwise. The marginal distribution of Y2 is f2(y2) = ∫ y2 0 6(1 − y2)dy1 = (6(1 − y2)y1| y2 0 ) = 6y2(1 − y2) for 0 ≤ y2 ≤ 1 and 0 otherwise. (c) Find the conditional density function of Y1 when Y2 = y2. f(y1|Y2 = y2) = f(y1, y2) f2(y2) = 6(1 − y2) 6y2(1 − y2) = 1 y2 for 0 ≤ y1 ≤ y2 and 0 elsewhere. (d) Find the conditional density function of Y2 when Y1 = y1. f(y2|Y1 = y1) = f(y1, y2) f1(y1) = 6(1 − y2) 3(1 − y1)2 = 2(1 − y2) (1 − y1)2 for y1 ≤ y2 ≤ 1 and 0 elsewhere. (e) Find P (Y2 ≥ 3/4|Y1 = 1/2). P (Y2 ≥ 0.75|Y1 = 0.5) = ∫ 1 0.75 2(1 − y2) (1 − 0.5)2 dy2 = ∫ 1 0.75 8(1 − y2)dy2 = ( 8y2 − 4y 2 2| 1 0.75 ) = 4 − (8(3/4) − 4(3/4)2) = 0.25 (f) Are Y1 and Y2 independent random variables? Explain your answer. We have found the marginal distributions as f1(y1) = 3(1 − y1) 2 for 0 ≤ y1 ≤ 1 f2(y2) = 6y2(1 − y2) for 0 ≤ y2 ≤ 1 2 y1 y2 0.0 0.5 1.0 1.5 2.0 0. 0 0. 2 0. 4 0. 6 0. 8 1. 0 = ∫ 2/3 0 (2 − 3y2)dy2 = ( 2y2 − 3 2 y22)| 2/3 0 ) = 2/3 (b) Find the marginal distributions for both Y1 and Y2. The marginal distribution for Y1 is f1(y1) = ∫ 0.5y1 0 dy2 = 0.5y1 for 0 ≤ y1 ≤ 2 The marginal distribution for Y2 is f2(y2) = ∫ 2 2y2 dy1 = 2(1 − y2) for 0 ≤ y2 ≤ 1 (c) Find the conditional density function of Y1 when Y2 = y2. f(y1|Y2 = y2) = f(y1, y2) f2(y2) = 1 2(1 − y2) for 2y2 ≤ y1 ≤ 2 and 0 elsewhere. (d) Find the conditional density function of Y2 when Y1 = y1. f(y2|Y1 = y1) = f(y1, y2) f1(y1) = 1 0.5y1 = 2 y1 for 0 ≤ y2 ≤ 0.5y1 and 0 elsewhere. (e) Find the probability that the amount of pollutant in a sample taken above the stack without the cleaner (Y1) is 1.5 or greater given the amount of pollutant in a sample taken above the stack with the cleaner (Y2) is 0.5. We need to use f(y1|Y2 = 0.5) = 1 2(1−y2) = 12(1−0.5) = 1. P (Y1 ≥ 1.5|Y2 = 0.5) = ∫ 2 1.5 dy1 = 2 − 1.5 = 0.5 5 (f) Are Y1 and Y2 independent random variables? Explain your answer. We have found the marginal distributions as f1(y1) = 0.5y1 f2(y2) = 2(1 − y2) The joint distribution is not equal to the product of the marginal distributions and so the two variables are dependent. (g) Find the covariance between Y1 and Y2. Using the two marginal distributions, we have E(Y1) = ∫ 2 0 y1 ∗ 0.5y1dy1 = ∫ 2 0 0.5y21dy1 = 1 6 y31| 2 0 = 1 6 (8 − 0) = 4 3 E(Y2) = ∫ 1 0 y2 ∗ 2(1 − y2)dy2 = ∫ 1 0 2y2 − 2y 2 2dy2 = y22 − 2 3 y32| 1 0 = 1 − 2 3 = 1 3 E(Y1Y2) = ∫ 2 0 ∫ 0.5y1 0 y1y2dy2dy1 = ∫ 2 0 y1 1 2 y22| 0.5y1 0 = ∫ 2 0 y1 1 8 y21 = ∫ 2 0 1 8 y31 = 1 32 y41| 2 0 = 1 32 (16 − 0) = 0.5 Cov(Y1, Y2) = E(Y1Y2) − E(Y1)E(Y2) 6 = 0.5 − ( 4 3 )( 1 3 ) = 1 18 (h) Find the correlation between Y1 and Y2. For the correlation, we need V (Y1) and V (Y2). E(Y 21 ) = ∫ 2 0 y21 ∗ 0.5y1dy1 = ∫ 2 0 0.5y31dy1 = 1 8 y41| 2 0 = 1 8 (16 − 0) = 2 V (Y1) = E(Y 2 1 ) − (E(Y1)) 2 = 2 − ( 4 3 )2 = 2 9 E(Y 22 ) = ∫ 1 0 y22 ∗ 2(1 − y2)dy2 = ∫ 1 0 2y22 − 2y 3 2dy2 = 2 3 y32 − 1 2 y42| 1 0 = 2 3 − 1 2 = 1 6 V (Y2) = E(Y 2 2 ) − (E(Y2)) 2 = 1 6 − ( 1 3 )2 = 1 18 Corr(Y1, Y2) = Cov(Y1, Y2) √ V (Y1)V (Y2) = 1 18 √ ( 2 9 ) ( 1 18 ) = 0.5 7