I have $n$ Poisson random variables $X_i$ each sampled from an independent Poisson distribution parametrized by $\lambda_i$. I want to compute $P(X_1+X_2+...+X_n=z|X_1+X_3=y)$. The distribution of the unconditional sum would be a Poisson, but I am not sure how to condition it.
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4$\begingroup$ It is equivalent to computing $P(X+Y=z|Y=y)$ when $Y$ is Poisson with parameter $2\lambda_n$ and $X$ is Poisson with parameter $(n-2)\lambda_n$ $\endgroup$– Xi'anCommented Jul 22, 2021 at 17:40
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$\begingroup$ I don't think so cause if the sum and one of the variables are known, the answer is easy to calculate and deterministic. If there were two free variables like $P(X+Y+Z=a|Y=y)$, then it would be similar to this llc.stat.purdue.edu/2014/41600/notes/prob1805.pdf, and the answer would be $\frac{e^{-\lambda}\lambda^{a-y}}{(a-y)!}$. However, when I have more free variables, the summations get complicated, and I am unsure about how to reduce them to a similar Poisson variable. $\endgroup$– ItsKalvikCommented Jul 22, 2021 at 19:00
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2$\begingroup$ One proof of the statement by @Xi'an begins by observing that $Y=X_1+X_3$ and $X=X_2+X_4+\cdots+X_n$ have Poisson distributions with parameters $\lambda_1+\lambda_3$ and $\lambda_2+\lambda_4+\cdots+\lambda_n,$ respectively, and are independent. You can go further and rewrite the question as asking for $\Pr(X=z-y\mid Y=y),$ which permits you to write the answer immediately. $\endgroup$– whuber ♦Commented Jul 22, 2021 at 19:44
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The sum of independent Poisson random variables is a Poisson random variable as shown here. So for $Z = X+Y$ where all are Poisson r.v., $P(Z=z)=\frac{\exp^{-\lambda}\lambda^z}{z!}$. Here, $\lambda=\lambda_x+\lambda_y$. If $P(Z)$ was conditioned with $X$, $P(Z|X)$ the solution would be deterministic as the sum of two variables is known and one of the variables is known. As per @whuber and @Xi'an suggested, $P(X_1+X_2+X_3+X_4+...+X_n=z|X_1+X_3=y)=P(X_2+X_4+X_5...+X_n=z-y)$.
So, the conditional can be reduced to an unconditional that is just a sum of r.v. which in turn is just an extension of the attached derivation. Thanks @Xi'an and @whuber, appreciate the help.
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2$\begingroup$ That's right: but take some care to specify that when $z\lt y,$ the chance is zero! $\endgroup$– whuber ♦Commented Jul 22, 2021 at 20:49