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My question may seem so simple but I have problem to prove it. Suppose that $X$ follows a Poisson distribution with parameter $\lambda$ which is defined per each period of time. How we can prove that the probability distribution of arrival over $t$ follows a Poisson with parameter $\lambda.t$? Thanks

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    $\begingroup$ You may have to make a further assumption that they are generated by a homogeneous Poisson process. Then the result is essentially immediate. $\endgroup$ – Henry Nov 5 at 20:06
  • $\begingroup$ It seems trivial but I like to know the proof. thanks $\endgroup$ – Katatonia Nov 5 at 21:28
  • $\begingroup$ Wikipedia has an explanation, though perhaps closer to a definition than a proof. $\endgroup$ – Henry Nov 5 at 21:39
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    $\begingroup$ My post at stats.stackexchange.com/a/215253/919 derives this result from basic principles and definitions, without any advanced math. $\endgroup$ – whuber Nov 5 at 22:03
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This is an attempt at a detailed demonstration. Whether it is a proof or not is up to you.

The characteristic function of a Poisson distribution with rate $\lambda$ is $\varphi(s)= \exp(\lambda(e^{is}-1))$

So the sum of $t$ i.i.d. such random variables is $\exp(\lambda(e^{is}-1))^t = \exp(\lambda t(e^{is}-1))$ which is the characteristic function of a Poisson random variable with parameter $\lambda t$. That shows your desired result for positive integer $t$

Now consider rational $t=\frac m n$. We now know that the characteristic function of a Poisson random variable with parameter $\lambda m$ is $\exp(\lambda m(e^{is}-1))$ and this is equivalent to the characteristic function of the sum of $n$ i.i.d. random variables each with characteristic function $\sqrt[n]{\exp(\lambda m(e^{is}-1))}=\exp(\lambda \frac m n(e^{is}-1))$, i.e. $\exp(\lambda t(e^{is}-1))$ which is the characteristic function of a Poisson random variable with parameter $\lambda t$. That shows your desired result for positive rational $t$

You can extend from the positive rationals to positive real $t$ by continuity and thus get your final result. You could use moment generating functions or probability generating functions instead of characteristic functions and still have essentially the same demonstration.

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