Expected value and variance of a conditional pdf (I think I have it)

[phiiota]

Homework Statement

Suppose the distribution of X2 conditional on X1=x1 is N(x1,x1²), and that the marginal distribution of X1 is U(0,1). Find the mean and variance of X2.

Homework Equations

Theorem: E(X_{2})=E_{1}(E_{2|1}(X_{2}|X_{1}))
Var(X_{2})=V_{1}(E_{2|1}(X_{2}|X_{1}))+E_{1}(V_{2|1}(X_{2}|X_{1}))

The Attempt at a Solution

If I understand the above right, which I'm not sure I do, in words, the first part says that the expected value of X2 is going to be the expected value of the conditional function. Here, the expected value of X2|X1 is X1, itself a random variable, and the expected value of that is 1/2 (just the expected value of the marginal uniform distribution.
The second part, a little trickier for me, is the same idea, I'm finding the variance of the random variable X1 (which is the expected value of the conditional distribution X2, and that is 1/12. Then I add that to the expected value of the variance of the conditional distribution. That variable is X1^2, which is equal to the variance of X1 plus the mean of X1 squared.
Am I understanding this right?

Long story short, E(X2)=1/2, Var(X2)= 5/12 (this part is 1/12 + 1/12 + (1/2)^2

Sorry, I hope this is readable.b]

 

[haruspex]

5/12 looks right to me.

 

[phiiota]

Awesome, thanks. This concept took me a little while to get, but I think I got it. It's not very hard, but my book is kind of vague.

 

[Ray Vickson]

Homework Statement

Suppose the distribution of X2 conditional on X1=x1 is N(x1,x1²), and that the marginal distribution of X1 is U(0,1). Find the mean and variance of X2.

Homework Equations

Theorem: E(X_{2})=E_{1}(E_{2|1}(X_{2}|X_{1}))
Var(X_{2})=V_{1}(E_{2|1}(X_{2}|X_{1}))+E_{1}(V_{2|1}(X_{2}|X_{1}))

The Attempt at a Solution

If I understand the above right, which I'm not sure I do, in words, the first part says that the expected value of X2 is going to be the expected value of the conditional function. Here, the expected value of X2|X1 is X1, itself a random variable, and the expected value of that is 1/2 (just the expected value of the marginal uniform distribution.
The second part, a little trickier for me, is the same idea, I'm finding the variance of the random variable X1 (which is the expected value of the conditional distribution X2, and that is 1/12. Then I add that to the expected value of the variance of the conditional distribution. That variable is X1^2, which is equal to the variance of X1 plus the mean of X1 squared.
Am I understanding this right?

Long story short, E(X2)=1/2, Var(X2)= 5/12 (this part is 1/12 + 1/12 + (1/2)^2

Sorry, I hope this is readable.b]

To understand the above, just think of X1 as discrete (and X2 may be discrete, continuous or mixed). Then
E\,X_2 = \sum_{x_1} P\{ X_1 = x_1 \} \, E(X_2 |X_1 = x_1), \\<br /> E\,X_2^2 = \sum_{x_1} P\{ X_1 = x_1 \} \, E(X_2^2 |X_1 = x_1),\\<br /> \text{Var}(X_2) = E\,X_2^2 - (E\, X_2)^2.
When X1 is continuous, just replace a sum by an integral, etc.

In your case X2|X1=x1 ~ N(x1,x1^2) and X1~U(0,1), so
E\,X_2 = \int_0^1 x_1 \, dx_1 = 1/2, as you said, and
E(X_2^2 |X_1 = x_1) = \text{Var}(X_2|X_1=x_1) + (E(X_2|X_1=x_1))^2 = <br /> 2 x_1^2,\\<br /> \Longrightarrow \text{Var}(X_2) = \int_0^1 2x_1^2 \, dx_1 - (1/2)^2 = 5/12,
as you also stated. Personally, I prefer this last way of getting the variance, as opposed to using the formulas you gave. However, either way works if you are careful.

RGV