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If $X,Y$ are independent Poisson random variables with parameter $\lambda_1, \lambda_2$, then $X+Y$ is Poisson random variable with parameter $\lambda_1+\lambda_2$. I am wondering whether the converse if true, given a poisson random variable on a probability space $(\Omega, \mathcal{F},P)$, can we always decompose it into independent poisson random variables with parameter $\lambda_1,\lambda_2$ such that there sum is the given random variable?

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  • $\begingroup$ I am almost sure that on given sufficiently 'small' probability space where $Z\sim Poiss(\lambda)$ is given with $\lambda =\lambda_1+\lambda_2$, it is impossible to construct $X\sim Poiss(\lambda_1)$, without any further requirements. $\endgroup$ – NCh Mar 11 '20 at 5:04
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    $\begingroup$ For sufficiently large sample space it is always possible: look at math.stackexchange.com/a/506891 $\endgroup$ – NCh Mar 11 '20 at 5:09
  • $\begingroup$ @NCh Excellent answer! Thank you $\endgroup$ – Zorualyh Mar 11 '20 at 5:16
  • $\begingroup$ @NCh : What's this about "sufficiently large"? See my answer below. There's no such hypothesis. $\endgroup$ – Michael Hardy Mar 11 '20 at 6:40
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    $\begingroup$ Sufficiently large - i.e. there exists all the variables you states without any proof of their existence. Please construct such $X$ on $\Omega=\{0,1,2,\ldots \}$ with $\mathcal F=2^\Omega$ and $P(\{k\})=\frac{\lambda^k}{k!}e^{-\lambda}$, for $W(\omega)=\omega$. $\endgroup$ – NCh Mar 11 '20 at 10:02
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Suppose $W\sim\operatorname{Poisson}(\lambda).$

Suppose $0<\lambda_1 <\lambda,$ and let $\lambda_2 = \lambda - \lambda_1.$

Let $p = \dfrac{\lambda_1}\lambda = \dfrac{\lambda_1}{\lambda_1+\lambda_2}.$

Let $X\mid W \sim\operatorname{Binomial}(W,p),$ i.e. this is the number of successes in $W$ independent trials with probability $p$ of success on each trial. Let $Y= W-X.$ Then $Y\mid W \sim\operatorname{Binomial}(W,1-p).$

Given all of this, one can conclude that

  • $W=X+Y.$
  • $X\sim\operatorname{Poisson}(\lambda_1).$
  • $Y \sim\operatorname{Poisson}(\lambda_2).$
  • $X,Y$ are independent.

Proving this is a standard exercise. How to do it is a question that has been posted here a number of times.

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  • $\begingroup$ Let $\Omega=\{0,1,2,\ldots \}$ with $\mathcal F=2^\Omega$ and $P(\{k\})=\frac{\lambda^k}{k!}e^{-\lambda}$ and $W(\omega)=\omega$. Does there exist $X$ as you claim on this probability space? $\endgroup$ – NCh Mar 11 '20 at 10:04
  • $\begingroup$ @NCh : I take your comment as a criticism of the practice of identifying the theory of probability with what Kolmogorov said it is. $\endgroup$ – Michael Hardy Mar 11 '20 at 16:28
  • $\begingroup$ I don’t really understand how the request to construct a random variable on a given probability space can be perceived as a criticism of the Kolmogorov approach to the axiomatization of probability theory. Perhaps, due to poor language skills, I do not catch any nuances. But in the question it is emphasized that the random variable has already been set somewhere, and it is on this space that it is required to construct the summands. You offer to expand it by introducing extra r.v.'s, without noticing it yourself. This is the only thing I object to. By the way, this is not my downvote. $\endgroup$ – NCh Mar 11 '20 at 17:16
  • $\begingroup$ Despite this, many thanks for the comment! I realize that the excessive craving for constructions underlying the foundations of probability prevents me from properly appreciating true probabilistic facts. $\endgroup$ – NCh Mar 11 '20 at 17:34
  • $\begingroup$ @NCh : ok, The question can be construed as wanting the decomposition to be on the same probability space, and in some contexts it might be the most reasonable way to construe it, and possibly that's what was intended. $\endgroup$ – Michael Hardy Mar 11 '20 at 20:11

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