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The joint probability function of the random variables $X$ and $Y$ is given by

$p(x,y)=\Biggl\{\frac 1{e^2y!(x-y)!}$ if $x=1,2...$ and $y=0,1,2,...,x$

and $p(x,y)=0$, $otherwise$.

Prove that $E[Y| X=x]=x/2$

I have doubts on how to attack this problem, I feel that it has to do something with the binomial variable, although I am not sure how to properly use the expected conditional value? Any contribution would be very helpful. Thanks, I'm new to the probabilities course!

Hendrik Matamoros
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1 Answers1

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Bayes Formula tells us $$\mathbb P(Y=y|X=x) = \frac{\mathbb P(X=x,Y=y)}{\mathbb P(X=x)}.$$

We already have the numerator in the hypothesis. $$\mathbb P(X=x)= \sum_{y=0}^xp(x,y) = \frac{1}{x!e^2}\sum_{y=0}^x {x \choose y} = \frac{2^x}{x!e^2}.$$

Now $$\mathbb E(Y|X=x) = \sum_{y=0}^x yP(Y=y|X=x) = \frac{1}{2^x}\sum_{y=0}^x y{x \choose y} = \frac{1}{2^x}\cdot x \cdot 2^{x-1} = \frac x 2.$$

Maffred
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  • Thank you very much, I understood, the main idea of the proof, but are there any additional details that you have used? Or did you only use the definition of conditional distribution? – Hendrik Matamoros Feb 15 '18 at 22:20
  • I used the definition of conditional distribution and 2 identities involving binomial sums. Which part is not clear? – Maffred Feb 15 '18 at 22:22
  • It would not be that we already have the denominator in the hypothesis? , what binomial identities did you use? because I see that that is the definition of combinatorial number in general. – Hendrik Matamoros Feb 15 '18 at 22:25
  • I don't understand the first question. I made algebra tricks to let appear the needed parts of the definition of binomials in the sums. Then I used this https://math.stackexchange.com/questions/734900/proof-by-induction-sum-of-binomial-coefficients-sum-k-0n-n-k-2n and this https://math.stackexchange.com/questions/683733/sum-of-k-n-choose-k-is-n2n-1. – Maffred Feb 15 '18 at 22:28
  • Thank you very much for attaching those links. In the first question, I meant that in your answer you wrote "We already have the numerator in the hypothesis". It would not be that we already have the denominator in the hypothesis?, sorry. – Hendrik Matamoros Feb 15 '18 at 22:34
  • We know $p(x,y)$, which is $\mathbb P(X=x, Y=y)$. – Maffred Feb 15 '18 at 22:42
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    Thank you very much for everything!, greetings from Venezuela. – Hendrik Matamoros Feb 15 '18 at 22:50