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The real random variables $X$ and $Y$ are independent and both have a Poisson distribution with the parameter 1, i.e. Po(1).

Find:

$$\mathbb{E}\left[ \left( 2^{2X}+2^{Y} \right)^2|X+Y \right]$$

Answer: $$\left(\frac{9}{2}\right)^{X+Y} +2\cdot 3^{X+Y}+\left(\frac{5}{2} \right)^{X+Y}$$

My steps: $$\left( 2^{2X}+2^{Y} \right)^2=2^{4X}+2^{2Y}+2\cdot 2^{2X+Y}=2^{4X}+2^{2Y}+2^{X+1}2^{X+Y}$$ Hence we have $$\mathbb{E}\left[ \left( 2^{2X}+2^{Y} \right)^2|X+Y \right]=2^{X+Y}\mathbb{E}\left[ 2^{X+1}|X+Y\right]+\mathbb{E}\left[ 2^{4X}|X+Y\right]+\mathbb{E}\left[ 2^{2Y}|X+Y\right]$$

So, if we know how to calculate $\mathbb{E}\left[ 2^{cX+d}|X+Y\right]$ for $c,d\in \mathbb{R}$, we'll get the answer, but I don't know how to do it. Please help.

Vadim Omelchenko
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    The expected value of a random variable can't contain any random elements. The answer can't be $\left( \frac{9}{2} \right)^{X+Y}$ because that contains $X$ and $Y$. – Patrick Stevens Sep 11 '15 at 07:31
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    @PatrickStevens: The expectation is conditioned on a certain value of $X+Y$. It would be cleaner to write this as $\mathbb E[\cdot\mid X+Y=s]$ and then write $s$ where the question has $X+Y$, but conceptually it's correct that the sum can occur in the answer. – joriki Sep 11 '15 at 07:35
  • @PatrickStevens What you wrote is completely wrong since we are talking about conditional expectations.. The conditional expectation with respect to $X+Y$ is a random variable measurable with respect to the sigma-algebra generated by $X+Y$, and $\frac{9}{2}^{X+Y}$ obviously satisfies this. – air Sep 11 '15 at 07:37
  • @joriki Ah, I've never seen that abuse of notation before. Would it be appropriate for me to edit $\mid X + Y = s$ into the question? – Patrick Stevens Sep 11 '15 at 07:37
  • Patrick, it's a conditional expectation. It's a bit different notion. https://en.wikipedia.org/wiki/Conditional_expectation A conditional expectation $\mathbb{E}(X|Y)$ is non random e.g. if X and Y are independent. The unconditional expectation is always non random.

    However, if $X$ and $Y$ are not independent, then $E(X|Y)$ is usually a random variable.

     – Vadim Omelchenko Sep 11 '15 at 07:39
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    @PatrickStevens: I tend to be conservative about editing others' posts, unless it's for obvious mistakes. Since air seems to think that the notation is unobjectionable, I'd just suggest the edit to the OP. – joriki Sep 11 '15 at 07:39
  • @Vadim I recognise $\mathbb{E}(X \mid A)$ as a random variable, and am warily OK with $\mathbb{E}(A^3 \mid A) = A^3$ - for some reason I didn't recognise the manipulations to the corresponding $\mathbb{E}(X \mid f(A, B))$. I would personally always prefer to be conditioning on an event rather than on a variable, for concreteness, but of course this is more compact. – Patrick Stevens Sep 11 '15 at 07:43
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    @joriki Yes, it is not an abuse of notation. This is how Kolmogorov defined conditional expectations. Of course we could also condition on $X+Y=s$, which is closer to elementary intuition of conditional expectation, and there also exist factorization theorems which make the equivalence precise. But since I am rather uncomfortable with these, I definitely prefer the notation as is. – air Sep 11 '15 at 07:47
  • @PatrickStevens If you want to better acquaint with conditioning, you can take a look at my questions. There are the solutions to some very complicated problems on this topic. – Vadim Omelchenko Sep 11 '15 at 07:48
  • @PatrickStevens $\mathbb{E}(X|Y=s)$ is an intuitive representation of conditioning. $\mathbb{E}(X|Y)$ is a short hand.

    $\mathbb{E}(X|Y=s)=f(s)$ where $f$ is a function. And hence $f(Y)$ is a random variable. Hence $\mathbb{E}(X|Y)=f(Y)$ is a random variable.

     – Vadim Omelchenko Sep 11 '15 at 07:49
  • What about the solution to the problem? Any ideas? :) – Vadim Omelchenko Sep 11 '15 at 07:52
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    This might be helpful $X\sim Poin(\lambda_1)$ indep of $Y\sim Poi(\lambda_1)$ then $X|X+Y \sim Bin(X+Y,\frac{\lambda_1}{\lambda_1+\lambda_2})$. Here calculation will be simpler as $\lambda_1=\lambda_2=1$ i.e. $X|X+Y=x+y \sim Bin(x+y,1/2)$ and $Y|X+Y=x+y \sim Bin(x+y,1/2)$ i.e. $Y|X+Y=x+y$ and $X|X+Y=x+y$ are identically distributed as $Bin(x+y,1/2)$ – Saty Sep 11 '15 at 07:54

1 Answers1

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These exercises are really interesting!

As a first observation, note that we just need to calculate $\mathbb{E}\left[ 2^{cX}\mid X+Y\right]$ since $\mathbb{E}\left[ 2^d\mid X+Y\right]=2^d$.

Now we need a standard result on conditioning on sums of independent poisson random variables (which I have shown previously here), which in this case implies that:

$$ X \mid X+Y \sim \text{Binomial}\left(X+Y, \frac{1}{2}\right) $$

Therefore (with help of the binomial theorem), we get:

$$ \begin{aligned} \mathbb{E}\left[ 2^{cX} \mid X+Y\right] &= \sum_{k=0}^{X+Y}2^{ck}{X+Y \choose k}\left(\frac{1}{2}\right)^{X+Y-k}\left(\frac{1}{2}\right)^{k} \\ &=\left(\frac{1}{2}\right)^{X+Y}\;\sum_{k=0}^{X+Y}(2^c)^k1^{X+Y-k}{X+Y \choose k}\\ &=\left(\frac{1}{2}\right)^{X+Y}\left(2^c+1\right)^{X+Y} \end{aligned} $$

Community
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air
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    +1. Both the question and the answer would look nicer if you used \mid instead of | or \vert -- it provides the appropriate spacing. – joriki Sep 11 '15 at 08:17
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    Oh wow, thanks a lot, I did not know the \mid command! All the time I have been using some weird combinations of ; \vert ; or | or \vert and this makes things so much nicer.. – air Sep 11 '15 at 08:20