Let $X, Y$, and $Z$ be three independent Poisson Y3 random variables with parameters $\lambda_1, \lambda_2, \lambda_3$, respectively. For $y=0, 1, 2,\ldots, t$, calculate:

$$P(Y = y \mid X + Y + Z = t).$$

Unsure of how to approach this. Given formulas from my book, I get $$e^{-3}\cdot \frac{3^t}{t!}$$


\begin{align} & \Pr(Y=y \mid X+Y+Z = t) = \frac{\Pr(Y=y\ \&\ X+Y+Z = t)}{\Pr( X+Y+Z = t )} \\[10pt] = {} & \frac{\Pr(Y=y\ \&\ X+Z = t-y)}{\Pr( X+Y+Z = t )} = \frac{\Pr(Y=y)\cdot\Pr(X+Z = t-y)}{\Pr( X+Y+Z = t )} \tag 1 \end{align} Do you know that the sum of independent Poisson-distributed random variables also has a Poisson distribution? If so, you get $X+Y+Z\sim\mathrm{Poisson}(\lambda_1+\lambda_2+\lambda_3)$ and $X+Z\sim\mathrm{Poisson}(\lambda_1+\lambda_3)$ and $Y\sim\mathrm{Poisson}(\lambda_2)$, so plug those into the last expression in $(1)$ and simplify. Every occurence of the number $e$ will cancel out. You will find that the conditional distribution of $Y$ is a binomial distribution.

  • $\begingroup$ thank you for this explanation @Michael. So Pr(Y=y) would be e^-lambda2* lambda2^y/y! ? Not entirely certain about Pr(X+Z=t-y) $\endgroup$ – samp1920 Jul 14 '15 at 1:47
  • $\begingroup$ @samp1920 : Same sort of thing: $\displaystyle \Pr(X+Z=w) = \frac{(\lambda_1+\lambda_3)^w e^{-(\lambda_1+\lambda_3)}}{w!}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 14 '15 at 2:52

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