So p x 0 y y e 7 y also note that y is exponentially

Info iconThis preview shows pages 4–5. Sign up to view the full content.

View Full Document Right Arrow Icon
So P ( X = 0 | Y = y ) = e - 7 y . Also note that Y is exponentially distributed with rate 5, so f Y ( y ) = 5 e - 5 y . Therefore, P ( X = 0) = Z 0 P ( X = 0 | Y = y ) f Y ( y ) dy = Z 0 5 e - 12 y dy = ± - 5 12 e - 12 y ² 0 = 5 12 . (We saw another way to solve this problem in class. See Example 2 in Section 5.2 and note that X = 0 is just the event that the first woman arrives before the first man.) 5. [8 points] Let U 1 ,U 2 ,U 3 ,U 4 ,U 5 be independently and uniformly chosen from (0 , 1). Let Y be the largest and X be the third largest. (In the book’s notation, X = U (3) and Y = U (5) .) Find the joint density of ( X,Y ). Solution: For 0 < x < y < 1, P ( X dx and Y dy ) = 5 · 4 · P ( U 1 is the third largest and U 1 dx,U 2 is the largest and U 2 dy ) = 20 · 3 · P ( U 3 ,U 4 (0 ,x ) ,U 1 dx,U 5 ( x,y ) ,U 2 dy ) = 60[ x 2 ][1 · dx ][ y - x ][1 · dy ] = 60 x 2 ( y - x ) dx dy. Therefore, the joint density of X and Y is f ( x,y ) = 60 x 2 ( y - x ), for 0 < x < y < 1. Page 4
Background image of page 4

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Math 431: Final Exam 6. Let X be a random variable with exponential λ distribution. Given X = x , let Y be uniformly distributed on (0 ,x ). (a) [3 points] What is E ( Y | X )? Solution: The mean of a uniform distribution on an interval is the midpoint, so E ( Y | X = x ) = x/ 2. More compactly, E ( Y | X ) = X/ 2. (b) [3 points] What is E ( Y )? Solution: The expectation of X is 1 , so E ( Y ) = E [ E ( Y | X )] = E ± X 2 ² = 1 2 λ . (c) [6 points] What the joint density of X and Y ? Solution: f X ( x ) = λe - λx , for x > 0, f Y ( y | X = x ) = 1 x , for y (0 ,x ), so f X,Y ( x,y ) = f X ( x ) f Y ( y | X = x ) = λe - λx x , when 0 < y < x . 7. [8 points] Assume that X 1 ,...,X 5 are independent identically distributed random vari- ables with nonzero variance. Let Y = X 1 + ··· + X 5 . What is Corr ( Y,X 1 )? Solution: Cov ( Y,X 1 ) = Cov ( X 1 + ··· + X 5 ,X 1 ) = Cov ( X 1 ,X 1 ) + Cov ( X 2 ,X 1 ) + ··· + Cov ( X 5 ,X 1 ) = V ar ( X 1 ) , V ar ( Y ) = V ar ( X 1 ) + ··· + V ar ( X 5 ) = 5 V ar ( X 1 ), so Corr ( Y,X 1 ) = Cov ( Y,X 1 ) SD ( Y ) SD ( X 1 ) = V ar ( X 1 ) p 5 V ar ( X 1 ) p V ar ( X 1 ) = 1 5 . Page 5
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page4 / 5

So P X 0 Y y e 7 y Also note that Y is exponentially...

This preview shows document pages 4 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online