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Are there any random variables so that E[X] and E[Y] exist but E[XY] doesn't?

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You can probably look at some variant of the Cauchy Distribution. –  PEV Apr 9 '11 at 17:18
    
what do you mean by $E(XY)$ does not exist? –  picakhu Apr 9 '11 at 17:19
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would you be satisfied with probability space $[0,1]$, $X=Y=x^{-1/2}$? –  user8268 Apr 9 '11 at 17:20
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@Rasmus That's a different question. –  Douglas Zare Apr 9 '11 at 19:54
    
@Douglas: Oh, right, thanks you. –  Rasmus Apr 10 '11 at 9:00
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2 Answers

Here is an example which is similar to user8268's comment.

Flip a sequence of independent fair coins. Let $Z$ be the index of the first head, so $P(Z=1)=1/2, P(Z=2)=1/4, ... P(Z=n)=1/2^n,...$.

Let $c$ be a real number. The random variable $c^Z$ has an expected value $c/(2-c)$ if $|c| \lt 2$, and does not have an expected value if $|c| \ge 2$. So, if $X=Y=\sqrt{3}^Z$ then $E[X]$ and $E[Y]$ exist, but $E[XY] = E[3^Z]$ does not exist.

See also this question where there is the added assumption of independence.

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Having recently discussed here the difference between "$X$ has expectation" (in the wide sense) and "$X$ is integrable", let us give an example where $X$ and $Y$ are integrable (that is, have finite expectation) but $XY$ does not admit an expectation (that is, ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$).

Let $Z$ be any nonnegative random variable satisfying ${\rm E}(Z)<\infty$ and ${\rm E}(Z^2) = \infty$ (for example, $Z=1/\sqrt{U}$, where $U \sim {\rm uniform}(0,1)$; cf. user8268's comment above). Let $R$ be independent of $Z$ with ${\rm P}(R=1) = {\rm P}(R=-1) = 1/2$. Define the random variables $X$ and $Y$ by $X=ZR$ and $Y=Z$. Then, $X$ and $Y$ have finite expectation. Next note that $XY = Z^2$ if $R=1$ and $XY = -Z^2$ if $R=-1$. Hence, $$ (XY)^ + := \max \{ XY,0\} = Z^2 {\mathbf 1}(R = 1) $$ and $$ (XY)^ - := -\min \{ XY,0\} = Z^2 {\mathbf 1}(R = -1), $$ where ${\mathbf 1}$ denotes the indicator function. Since $Z$ and $R$ are independent (and, by assumption, ${\rm E}(Z^2) = \infty$), we get ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$, as desired.

In the example (mentioned above) where $Z=1/\sqrt{U}$, $U \sim {\rm uniform}(0,1)$, one can find that $XY$ has density function $f_{XY}$ given by $f_{XY} = 1/(2x^2)$, $|x|>1$. Thus, $\int_0^\infty {xf_{XY} (x)\,{\rm d}x} = \int_1^\infty {(2x)^{ - 1} \,{\rm d}x} = \infty $ and $\int_{ - \infty }^0 {|x|f_{XY} (x)\,{\rm d}x} = \int_{ - \infty }^{ - 1} {(2|x|)^{ - 1} \,{\rm d}x} = \infty $, corresponding to ${\rm E}(XY)^+ = \infty$ and ${\rm E}(XY)^- = \infty$, respectively.

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