This preview shows page 1. Sign up to view the full content.
Unformatted text preview: nt, we have that E X Y = E X E Y . Since X is an
exponential r.v. with parameter = 1, its mean is E X = 1= = 1.
3
A uniform r.v. in a; b has mean a + b=2; hence for Y U 1; 3 , we have that E Y = 2.
3
Therefore, the correlation between X and Y is E X Y = 1 2 = 2
2
e Due to the independence between X and Y , we have that
Z31
E X=Y = E X E 1=Y = 1 y 1 dy = 1 lnyj3=1 = ln3=2 = 0.5493
y
2
2
1
5
Note: it is wrong to write E X=Y = E X =E Y = 1=2.
f First, we need to nd out f xjy. Since X and Y are independent, we have that
f xjy = fX x, 8y. Hence, to maximize f xjy over all x, we only need to maximize
fX x = e,x over x 0. Obviously, the maximum is achieved at x = 0 . Hence, the best
estimate of X is x = 0. Note that due to the independence, the best estimate of X is
una ected what value of Y is observed.
5
3...
View Full
Document
 Fall '08
 Staff

Click to edit the document details