# Conditional Expectation

Let $$X_1$$ and $$X_2$$ and $$Z$$ denote independent, real valued random variables and $$Pr(Z=0) = 1-Pr(Z=1) = \alpha$$ for some $$0<\alpha<1$$ Define

$$Y$$ = $$X_1$$ if $$Z=0$$

= $$X_2$$ if $$Z=1$$

(a) Suppose that $$E(X_1)$$ and $$E(X_2)$$ exist, does it follow that E(Y) exists? (b) Assume that $$E(|X_1|) < \infty$$ and $$E(|X_2|) < \infty$$. Find $$E(Y|X_1)$$

This is NOT homework. This is from selfstudy. I would appreciate any kind of help. Intuitively, I feel that the answer for (a) is $$E(X_1).\alpha+(1-\alpha).E(X_2)$$ but I don't have any rigorous solution. For (b) I am totally confused not knowing even how to start. I don't even understand why the "$$< \infty$$" is necessary.

I appreciate it.

• Hint: express $Y$ as a linear combination of $X_1$ and $X_2$ with weights depending (linearly as well) on $Z$. – Xi'an Oct 25 '12 at 11:23
• The homework tag Wiki states this is "a routine question from a textbook, course, or test used for a class or self-study." Your question fits the bill perfectly. – whuber Oct 25 '12 at 14:20
• Some people say that "$E[X]$ exists" only if $E[|X|] < \infty$, others that "$E[X]$ is defined but unbounded" if exactly one of the series or integral expressions for $E[X_+]$ and $E[X_-]$ is divergent and the other is convergent. Both groups agree that if both $E[X_+]$ and $E[X_-]$ are divergent (as happens for Cauchy random variables, for example), then $E[X]$ is undefined. – Dilip Sarwate Oct 27 '12 at 15:11

The key is to notice that $$Y=(1-Z)X_1+ZX_2.$$ Then for the first question, we have $|Y|\leqslant |X_1|+|X_2|$.
For the second one, we have using independence and $\mathbb E(X_1\mid X_1)=1$, $$\mathbb E(Y\mid X_1)=\mathbb E((1-Z)X_1\mid X_1)+\mathbb E(ZX_2\mid X_1)=X_1\mathbb E(1-Z)+(\mathbb EZ)(\mathbb EX_2).$$