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The real random variables $X$ and $Y$ are independent and both have a uniform distribution $U([0,1])$. Find $$\mathbb{E}\left[e^{X+Y}|\quad |X-Y| \right]$$ Since $f(\cdot)=|\cdot|$ is not a monotone function, we can't remove it. I.e. if we had $\mathbb{E}\left[e^{X+Y}|\quad (|X-Y|)^2 \right]$ or $\mathbb{E}\left[e^{X+Y}|\quad e^{X-Y} \right]$ we'd be able to use the fact that the exponent and quadratic functions are monotone and simply rewrite the problem in the following form: $$\mathbb{E}\left[e^{X+Y}|\quad X-Y \right]$$ However, here it's impossible.

Is it a correct approach?

$$\mathbb{E}\left[ e^{X+Y}|\quad |X-Y|=s \right]=\mathbb{E}\left[ e^{X+Y}|\quad X-Y=s \right]+\mathbb{E}\left[ e^{X+Y}|\quad X-Y=-s \right]$$ and hence $$\mathbb{E}\left[ e^{X+Y}|\quad |X-Y| \right]=\mathbb{E}\left[ e^{X+Y}|\quad X-Y \right]+\mathbb{E}\left[ e^{X+Y}|\quad Y-X \right]$$

Vadim Omelchenko
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2 Answers2

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No, that's not correct. It would work for probabilities, but not for expected values. (You can check this easily by noting that adding a constant to the function whose expected value is being taken adds the constant once on the left but twice on the right.)

But since the problem is symmetric with respect to interchange of $X$ and $Y$, the two cases $X-Y=s$ and $X-Y=-s$ are equiprobable, so you can get the expected value as the average of these two cases; i.e. you're merely missing a factor $\frac12$ in the penultimate line.

joriki
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No, what you are doing is not correct. What would be correct is the following:

$$ \begin{aligned} \mathbb{E}\left[ e^{X+Y} \mid |X-Y|=s \right]=&\mathbb{E}\left[ e^{X+Y}\mid X-Y=s \right]\Pr\left[ X-Y \geq 0 \mid |X-Y|=s \right]\\ +&\mathbb{E}\left[ e^{X+Y}\mid X-Y=-s \right]\Pr\left[ X-Y < 0 \mid |X-Y|=s \right] \end{aligned} $$

As @joriki notes, in this case, out of symmetry reason, the probability factors are just $\frac{1}{2}$.

Nevertheless, because of the same symmetry, you can simplify your problem even more, to actually get what you originally wanted, namely:

$$\mathbb{E}\left[e^{X+Y} \mid X-Y \right] = \mathbb{E}\left[e^{X+Y} \mid |X-Y |\right]$$

This follows from the above result, since $$\mathbb{E}\left[e^{X+Y} \mid X-Y \right] = \mathbb{E}\left[e^{X+Y} \mid Y-X \right]$$

This in turn follows as in your previous question, simply by relabeling $X$ and $Y$.

Community
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air
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    I've removed the last paragraph of my answer in light of your answer (which is much better than mine). Thanks, I've learned to think properly about conditional expectations from your answers today :-) – joriki Sep 11 '15 at 09:28
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    Thank you for your kind words!! Conditional expectations are really annoying to learn (I am still quite uncomfortable with them).. There is a really big gap between Kolmogorov's measure theoretic definition and the elementary definition and I have never found a basic resource which tries to bridge that.. But maybe the paper "Conditioning as disintegration" by Chang and Pollard can do that job; at least the abstract resonates with me, though I have not read the whole paper yet. – air Sep 11 '15 at 09:47