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.

  • $\begingroup$ Hint: express $Y$ as a linear combination of $X_1$ and $X_2$ with weights depending (linearly as well) on $Z$. $\endgroup$
    – Xi'an
    Oct 25, 2012 at 11:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Oct 25, 2012 at 14:20
  • $\begingroup$ 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. $\endgroup$ Oct 27, 2012 at 15:11

1 Answer 1


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 $\mathbb E(X_1\mid X_1)=1$, $$\mathbb E(Y\mid X_1)=\mathbb E((1-Z)X_1\mid X_1)+\mathbb E(ZX_2\mid X_1)=X_1\mathbb E(1-Z)+(\mathbb EZ)(\mathbb EX_2).$$


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