Let $X_n$ be independent Poisson random variables with $E[X_i] = \mu_i$, and let $Y_n = X_1+...+X_n$. I want to show that if $\sum_n \mu_n = \infty $ then $Y_n/E[Y_n] \rightarrow 1$ almost surly.
What I do know is that if $X_1,...$ are independent, and $E\left[X^4\right] < \infty$, then $Y_n/n \rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-\mu_n$. The $C$'s are independent, $E[C_n^4]< \infty, E[C_n] = 0$, which means:
$\frac{C_1 +...+C_n}{n} \rightarrow 0$ a.s., or $\frac{\sum X_i - \sum \mu_i}{n} \rightarrow 0$ a.s. which, I think, completes the proof.

However I am not sure why the requirement $\sum_n \mu_n = \infty $ is necessary, so I suspect this proof is incorrect. Is it?

Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?

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    $\begingroup$ You cannot apply the result you recall since it requires the increments are identically distributed -- which is not true in your case (except if $\mu_n$ does not depend on $\mu$). $\endgroup$ – Did Dec 16 '18 at 13:04
  • $\begingroup$ To see why the $\sum_n\mu_n=\infty$ requirement is needed consider what happens if $\mu_1=1$ and $\mu_n=0$ for $n>1$. $\endgroup$ – kimchi lover Dec 16 '18 at 13:07
  • $\begingroup$ @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7 $\endgroup$ – SomeoneHAHA Dec 16 '18 at 13:15
  • $\begingroup$ Except you misquoted heavily. $\endgroup$ – Did Dec 16 '18 at 13:17
  • $\begingroup$ @kimchilover If $\mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem $\endgroup$ – SomeoneHAHA Dec 16 '18 at 13:17

If, in your argument, $M=\sum_n \mu_n<\infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.

And, the theorem you quote needs a uniform 4th moment bound $EX_k^4\le K$, which need not hold given your hypotheses.

A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=\sum_{t\le n} \mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/t\to 1$ almost surely.

Alternatively, you could use your argument with the added hypothesis that $\mu_n\le1$ for all $n$ and then use that to prove the general result by representing each $\mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.

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  • $\begingroup$ I see. So: the overall argument is correct, except the punch: $C_1+...+C_n /n \rightarrow 0$ implies the claim only if $M = \infty$, or should I change the entire argument? The comments in the original question made me completely not sure about is correctness. Thanks! $\endgroup$ – SomeoneHAHA Dec 16 '18 at 13:38
  • $\begingroup$ Re the edit: I see now that my argument is not true due to the uniform bounded requirement. Do you think that there is a proof which is not going through Poisson process? $\endgroup$ – SomeoneHAHA Dec 16 '18 at 13:49
  • $\begingroup$ I suppose there could be a non-Poisson process proof, but it would probably be ugly, like an apparatus designed by a lawyer to not infringe on a particular patent. $\endgroup$ – kimchi lover Dec 16 '18 at 13:58
  • $\begingroup$ Thanks. May you elaborate more on this way of proof? I am not familiar with Poisson processes, so I read about it but still have no idea how to use it here $\endgroup$ – SomeoneHAHA Dec 16 '18 at 14:55

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