# Math Help - Bivariate RV Probability

1. ## Bivariate RV Probability

Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

3. Compute E(Z).

How do I do the questions?

2. Originally Posted by maibs89
Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

3. Compute E(Z).

How do I do the questions?
1. Is R(0, 1) a uniform pdf?
2. Do I understand you correctly ....... n = 1 in the pmf for Y, that is, there's only 1 trial?

3. Yes, that is precisely right.

I have no idea where to get the answer.

Help!

4. Originally Posted by maibs89
Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

3. Compute E(Z).

How do I do the questions?
Note that Y = 0 or 1 only. So perhaps it's this simple:

1. When Y = 0, Z = 1 and E(Z) = 1. So n(0) = 1.
When Y = 1, Z = X and E(Z) = E(X) = 1/2.

2. When Y = 0, Z = 1 and Var(Z) = 0. So S(0) = 0.
When Y = 1, Z = X and Var(Z) = Var(X) = 1/12. So S(1) = 1/12.

3. I think it will be

$(1) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = .........$

Someone might jump in to confirm or deny this ......

By the way, there's an interesting question if X = R(0,1) and Y = R(0,1) ......

5. What are the rv's of E(Z|Y) and var(Z|Y)?

And could you please explain part 3?

6. Originally Posted by maibs89
What are the rv's of E(Z|Y) and var(Z|Y)?

And could you please explain part 3?
I have already done parts 1. and 2.

Part 3: Clearly either Z = Z^0 = 1 or Z = X^1 = X. Each obviously occurs with probability 1/2. Obviously E(1) = 1. When Z = X, E(Z) = E(X) = 1/2, as stated in my answer to part 1. Remember that X ~ R(0, 1).

It should be clear from all this where my answer has come from.

You started off with "I have no idea where to get the answer". I've given you a lot of food for thought. It's time now for you to step up to the plate.