Joint Probability Distributions â Examples Example 1 A certain ...
Joint Probability Distributions â Examples Example 1 A certain ...
Joint Probability Distributions â Examples Example 1 A certain ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
d. Compute the correlation for X and Y .Solution The correlation for X and Y is computed as follows:V (X) = X x (x ¡ ¹ X) 2 ¢ p X (x) = 12:45V (Y ) = X y (y ¡ ¹ Y ) 2 ¢ p Y (y) = 19:15Corr(X; Y ) = ½ XY =¡3:2025q(12:45)(19:15) = ¡0:2074:<strong>Example</strong> 3 Two components of a minicomputer have the following joint pdf for their useful lifetimesX and Y :( )x exp [¡x(1 + y)] x ¸ 0 and y ¸ 0f(x; y) =:0 otherwisea. What is the probability that the lifetime X for the …rst component exceeds 3?Solution The desired probability is:P (X > 3) =Z 1 Z 130x exp [¡x(1 + y)] dy dx =Z 13exp (¡x) dx = exp(¡3) = 0:0499b. What are the marginal pdf’s of X and Y ? Are the two lifetimes independent? Explain.Solution The marginal pdf’s of X and Y are:f X (x) = R 10 x exp [¡x(1 + y)] dy = exp(¡x) for x ¸ 0f Y (y) = R 10 x exp [¡x(1 + y)] dx =1(1 + y) 2 for y ¸ 0:It is now clear that f(x; y) is not the product of the marginal pdf’s, so the two randomvariables are not independent.c. What is the probability that the lifetime of at least one component exceeds 3?Solution P (at least one component lifetime exceeds 3) = 1 ¡ P (X · 3 and Y · 3)= 1 ¡= 1 4Z 3 Z 300x exp [¡x(1 + y)] dy dx =+ exp(¡3) ¡exp(¡12)4= 0:2998Z 30¡ exp(¡4x) + exp(¡x) dx<strong>Example</strong> 4 Let X and Y have the joint probability density function given by:f(x; y) =(k(1 ¡ y) 0 · x · y · 10 otherwisea. Find the value of k that makes this a probability density function.):3