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In a problem I am asked to find $\Bbb P(X=1|\frac{X+Y}{2}=2)$, and $X$ and $Y$ are independent random variables. In a previous part of the problem I defined $X+Y$ to be $Z$. So I simplified the problem to look a little better saying $\Bbb P(X=1|Z=4)$ which by law of conditional probability would be $$\frac{\Bbb P(X=1\cap Z=4)}{\Bbb P(Z=4)}$$ Is it correct to say $$\frac{\Bbb P(X=1)\Bbb P(Z=4)}{\Bbb P(Z=4)}$$ In which can be simplified to $$\Bbb P(X=1)$$ Given that $X$ and $Y$ are independent and that $Z=X+Y$?

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How do you justify going from $P(X=1\cap Z=4)$ to $P(X=1)P(Z=4)$ without first proving that events $(X=1)$ and $(Z=4)$ are independent? – Dilip Sarwate Oct 28 '12 at 22:22
@Delip I'm wondering if the fact that $X$ and $Y$ are independent will mean that $X+Y$ which equals $Z$ is independent of X. I don't think I'm correct here which is why I'm asking. – Kyle Oct 28 '12 at 22:43

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Not quite, $X$ and $Z$ are not independent. To see this, note that $Y$ cannot be equally likely on all integers, so if $Y$ is more likely to be $a$ than $b$, then $\mathbb{P}(Z=0)$ is higher if $X=-a$ and lower if $X=-b$. So information about $X$ tells us about $Z$ and the two cannot be independent.

The term $\mathbb{P}(X=1 \cap Z=4)$ is equivalent to $\mathbb{P}(X=1 \cap Y=3)$ which are independent events. I don't think any further simplification can be done without knowing more about the relative distributions of $X$ and $Y$.

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Why can $Y$ not be equally likely on all integers? Did you mean to say that $Z$ cannot be equally likely on all integers? (which is a correct assertion) – Dilip Sarwate Oct 28 '12 at 22:41
sorry dismiss my last comment, I didn't mean to post that comment. Heres what I did mean to post: The distributions are Poisson distributed, but I don't think that will help any. So you're saying that hypothetically if were looking for $\Bbb P(X=a|Z=b)$ and we had the term $\Bbb P(X=a \cap Z=b)$, the latter would also be equivalent to $\Bbb P(X=a \cap Y=b-a)$ – Kyle Oct 28 '12 at 22:52
@DilipSarwate There is no uniform probability distribution on the integers, because the sum of all the probabilities has to be 1. The sum of a finite constant over all integers is either 0 or infinity. – Logan Oct 28 '12 at 23:03
@Logan I am fully aware that there is no uniform probability distribution on $\mathbb Z$; I thought you meant $Y$ was uniformly distributed on a finite set of integers. In any case, I still don't see what $P\{Y = a\} > P\{Y=b\}$ has to do with $P\{Z=0\mid X=-a\} > P\{Z=0\mid X=-b\}$. Could you state your the argument about this in more detail? – Dilip Sarwate Oct 29 '12 at 2:40
@DilipSarwate To show that $Z$ is not independent of $X$, given $Z=X+Y$ and $X$ and $Y$ are independent. Specifically, because $Y$ is not uniform there exist $a,b$ so $P(Y=a)>P(Y=b)$. Then $P(Z=0|X=-a)=P(X+Y=0|X=-a)=P(Y=a|X=-a)=P(Y=a)>P(Y=b)=P(Y=b|X=-b)=P(X+Y=0|X=-b)=P‌​(Z=0|X=-b)$. That's a long string, but each step follows from the definition of $Z$, the independence of $X$ and $Y$, and the already established inequality. But the fact that the first and last sentences are not equal proves that $Z$ is not independent from $X$. – Logan Oct 29 '12 at 3:14
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Suppose $X$ and $Y$ can each take each of the values $0,1,2,3$ independently with equal probabilities $\frac14$. Then $\Pr(X+Y=4)=\Pr(X=1,Y=3)+\Pr(X=2,Y=2)+\Pr(X=3,Y=1)=\frac{3}{16}.$

So $\Pr(X=1|Z=4) =\frac13$ but $\Pr(X=1)=\frac14$ is a counter-example to your assertions.

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Note that $\mbox{Cov}(X, X + Y) = \mbox{Var}X + \mbox{Cov}(X, Y) = \mbox{Var} X$ since $X$ and $Y$ are independent. Hence, a nessecary condition for $X$ and $Z$ to be independent is that $\mbox{Var}X = 0$, i.e. $X$ is equal to some constant (almost surely). If $X$ is equal to a constant, then it is straight forward to check to that $X$ and $X + Y$ are independent, so your proposed assertion holds if and only if $X$ equal to some constant almost surely. So, in all but the most trivial of situations, your assertion does not hold.

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