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Conditional Expectation Function

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P(x1,x2)= (x1 choose x2) (1/2)^x1 (x1/15), x2= 0,1,2, ...., x1
x1= 1, 2, 3, 4 ,5

0 elsewhere
is the joint pmf of X1 and X2.

Determine E(X2|x1)

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Solution Preview

The answer is x1/2

(See Attachment)

p(x1,x2)= (x1 choose x2) (1/2)^x1 (x1/15), x2= 0,1,2, ...., x1

Solution Summary

The conditional expectation functions are determined.

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Statistics: Joint, marginal and conditional densities & expectation

** Please see the attached file for a Word formatted copy of the problem **

1. Consider the joint pmf p(x,y)=cxy, 1<x<y<3.
a) Find the normalizing constant c.
b) Are X and Y independent? Prove your claim.
c) Find the expectations of X, Y, XY.

2. Conceptual
Suppose (X, Y) have the joint pmf p(x, x+1) = 1/(n+1), x= 0,1,2,.....,n.
a) Are X, Y independent?
b) What is E( YlX = x)?
c) What is Var( YlX=x)?
d) What is Var(X)?
e) What is Var(Y-X)?

3. X, Y are jointly distributed Uniformly within the Unit circle. What is, the joint PDF is
f(x,y) = c, if X2+Y2 < 1
=0, otherwise.
a) Are X,Y independent?
b) Find the marginal PDF of X.
c) Find E(X).
d) Find E(Y).
e) Find E(X-Y).
f) Find Var(X-Y).

4. (expectation of a Quotient) suppose X and Y are independent and X~Be(2,2), Y~Be(3,3). Find E(X2/Y2).

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