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The classic example... $X \sim Po\left (\lambda\right ), Y \sim Po\left (\mu\right)$, X and Y are independent. Show that the conditional distribution of X is binomially distributed. Or in other words, $P(X=k\mid X+Y = n) = P (\tilde{X} = k), \tilde{X} \sim B\left (n ,\frac{\lambda}{\lambda + \mu}\right )$.

I've so far managed to reach to this step, and have been stuck since. Just somehow gotta get a $\frac{1}{n!}$ in the denominator, that would then complete the proof..or at least I think..

$$P(X=k\mid X+Y=n) = \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{P(X+Y = n)}= \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{\sum_{i=1}^{n} \frac{\lambda^{i}\mu^{n-i}}{i!(n-i)!}}$$

Thanks for the help!

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  • $\begingroup$ The summation in the denominator must run from $i=0$ to $n$. $\endgroup$
    – user17762
    May 29, 2012 at 19:50
  • $\begingroup$ There is a typo, you wrote $B\left (\frac{\lambda}{p} ,\frac{\lambda}{n}\right )$. Presumably you mean something like $B\left(\frac{\lambda}{\lambda+\mu},n\right)$. $\endgroup$ May 29, 2012 at 20:18
  • $\begingroup$ @AndréNicolas did you mean the other way around? Isn't it represented in B(n,p)? And yep thanks for the correction! I thought the probability and n were just merely derived from the $\lambda$ without considering the $\mu$. Makes more sense now anyway! $\endgroup$
    – Jon Gan
    May 30, 2012 at 6:19
  • $\begingroup$ Sure. Binomial $n$ trials prob. of success each trial $\frac{\lambda}{\lambda+\mu}$. $\endgroup$ May 30, 2012 at 12:30
  • $\begingroup$ Where does the numerator in the first step come from? It doesn't look like the PDF of Poisson. $\endgroup$ Apr 14, 2022 at 1:11

3 Answers 3

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HINT

Remember that $$\sum_{i=0}^{n} \binom{n}{i} \lambda^i \mu^{n-i} = \left( \lambda + \mu\right)^n$$ The above gives us that $$\sum_{i=0}^{n} \dfrac{n!}{i! (n-i)!} \lambda^i \mu^{n-i} = \left( \lambda + \mu\right)^n$$ which inturn gives us that $$\sum_{i=0}^{n} \dfrac{ \lambda^i \mu^{n-i}}{i! (n-i)!} = \dfrac{\left( \lambda + \mu\right)^n}{n!}$$

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\begin{equation} \begin{aligned} P(X=k | X+Y = n) &= \dfrac{P(X+Y=n|X=k)P(X=k)}{P(X+Y=n)} \\&= \dfrac{P(Y=n-k)P(X=k)}{P(X+Y=n)} \\&= \dfrac{Poisson_{\mu}(n-k)Poisson_{\lambda}(k)}{Poisson_{\mu+\lambda}(n)} \\&= \dfrac{\frac{\mu^{n-k}e^{-\mu}}{(n-k)!}\frac{\lambda^{k}e^{-\lambda}}{k!}}{\frac{(\mu+\lambda)^{n}e^{-\mu - \lambda}}{n!}} \\&= {n \choose k} \left( \dfrac{\lambda}{\lambda+\mu}\right)^k \left( \dfrac{\mu}{\lambda+\mu}\right)^{n-k} \end{aligned} \end{equation}

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Instead of using the summation in the denominator to calculate $P(X+Y=n)$, we might use the property that the sum of two independent Poisson distributions is a Poisson distribution, whose parameter is the sum of the parameters of two independent Poisson (can be shown using mgf; see Theorem 3.2.1 in Introduction to Mathematical Statistics by Hogg et al.). Therefore $P(X+Y=n)=\frac{1}{n!}(\lambda+\mu)^ne^{-(\lambda+\mu)}$, which can be plugged into the denominator of last expression.

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