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Let $X$ and $Y$ be independent Poisson random variables with parameters $1$ and $2$ respectively. Then, what is the following probability?

$$\mathbb P(X=1 \mid (X+Y)/2=2)$$

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    It is not clear what your formula is. Do you mean $\mathbb{P}(X=1|X+Y=4)$? That is, the probability that $X=1$ given that you know the two variables sum to 4. – 2017-01-27
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    Probability X=1 given that (X+Y)/2 is 2. The average of X and Y is 2 – 2017-01-27
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    See, this is why the condition separator is a vertical pipe | rather than a slash which is easily confused with division. – 2017-01-27
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    Yeah sorry, I edited it now. – 2017-01-27
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    So, have you tried applying the definition of conditional probability ? – 2017-01-27
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    Yes. I am having problem with, what will be the distribution of (X+Y)/2. The sum of two Poisson variates is a Poisson but I don't know about the average of two Poisson variates. – 2017-01-27
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    You are correct that $W=(X+Y)/2$ is not Poisson; it takes non-integer values. Also $E(W) = 3/2$ and $Var(W)=3.4.$ But you don't need the distribution of $W$ to solve the problem. – 2017-01-27

3 Answers 3

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You don't need the distribution of the average: Use $X + Y \sim Pois(3).$

$$P(X = 1|(X+Y)/2 = 2) = \frac{P(X = 1,X+Y=4)}{P(X+Y=4)} = \frac{P(X = 1,Y = 3)}{P(X+Y=4)} = ??,$$

where you can finish the computation, using independence and the formulas for $X \sim Pois(1)$ and $Y \sim Pois(2).$

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    To check your computation: numerical answer is close to 0.4. – 2017-01-27
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Just apply the definition of conditional probability and the Law of Total Probability, using the fact of independence.

$\mathsf P(X=1\mid X+Y=4) = \dfrac{\mathsf P(X=1, Y=3)}{\mathsf P(X+Y=4)}$

Argue that the sum of two independent Poisson random variables is a Poisson random variable whose rate is the sum of their rates, and thus that $(X+Y)\sim\mathcal{Pois}(3)$.

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I think you can solve your question knowing that if $X$ and $Y$ are two Poisson-distributed r.v.s with $\lambda_{x}$ and $\lambda_{y}$ respectively, then the distribution of $X$ conditional on $X+Y=k$ is binomial with $n=k$ and $p=\frac{\lambda_{x}}{\lambda_{x}+\lambda_{y}}$. See the discussion here.