# probability -one more question about expectation

Yesterday I asked you whether there's a situation where $E[X]$, $E[Y]$ defined, but $E[X\cdot Y]$ ($E[Z]$) is $\infty$ or even Diverging?

You claimed that the answer is NO and it is not elementary to explain that without using theorems which I don't know yet. Here it is

I would like now to ask you if I can assume also that it is not possible that there is $E[X\cdot Y]$ which is well-defined but one of the expectationm $E[X]$ or $E[Y]$ is not defined ($\infty$ or diverging)?

Thank you.

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Do you assume that $X$ and $Y$ are independent? –  Shai Covo Apr 7 '11 at 14:53
@shai: yes, I do –  henry11 Apr 7 '11 at 15:10

One can find (uninteresting) counterexamples. For instance, for $n=1$, 2, 3, and so on, let $X$ take on the value $2^n$ with probability $1/2^n$. Let $Y$ take on the value $0$ for all $n$. Then the expectation of $X$ is infinite, but $X\cdot Y$ is $0$ everywhere.

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And one cannot find interesting counterexamples (assuming that $X$ and $Y$ are independent). –  Shai Covo Apr 7 '11 at 14:38
If $X$ and $Y$ are not assumed to be independent, then for a (relatively) nontrivial counterexample take $X=U$ and $Y=1/U$, where $U$ is a uniform$(0,1)$ random variable. Then, $E[XY]=1$ but $E[Y]=\infty$. –  Shai Covo Apr 7 '11 at 15:16
Here are slightly less degenerate examples of positive random variables $X$ and $Y$ such that $E(XY)$ exists but $E(X)$ and $E(Y)$ do not.
Let $U$ and $V$ be independent random variable such that $U$ is Bernoulli, for example $P(U=1)=P(U=0)=1/2$, and $V$ is not integrable, for example $V\ge1$ almost surely and $P(V\ge v)=1/v$ for every $v\ge1$. Let $X=0$ and $Y=V$ if $U=0$ and $X=V$ and $Y=0$ if $U=1$. Then $XY=0$ hence $XY$ is integrable but neither $X$ nor $Y$ is.
Or: let $Z$ be an almost surely positive random variable such that neither $Z$ not $1/Z$ is integrable, for example $Z$ has density $\frac12\min\{1,1/z^2\}$ for $z$ in $(0,+\infty)$. Let $X=Z$ and $Y=1/Z$. Then $XY=1$ hence $XY$ is integrable but neither $X$ nor $Y$ is. Note that $Z$ can be realized through $U$ defined previously and an independent random variable $W$ uniform on $(0,1)$, as $Z=W$ if $U=1$ and $Z=1/W$ if $U=0$.
Edit While I was writing the above, the OP mentioned that $X$ and $Y$ should be independent. Recall that if $X$ and $Y$ are nonnegative independent random variables, the relation $E(XY)=E(X)E(Y)$ always hold in $[0,+\infty]$ (with the obvious multiplication rules plus the fact that $0\times(+\infty)=+\infty\times0=0$). In particular, if $X$ and $Y$ are independent, $X$ is not integrable and $Y$ is not almost surely zero, then $XY$ is not integrable.