# Strong law of large numbers for Poisson rvs with different parameter

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?

• 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 $\mu$). – Did Dec 16 '18 at 13:04
• To see why the $\sum_n\mu_n=\infty$ requirement is needed consider what happens if $\mu_1=1$ and $\mu_n=0$ for $n>1$. – kimchi lover Dec 16 '18 at 13:07
• @Did - The result I cited does not require i.d. See "Probability with martingales" Chapter 7 – SomeoneHAHA Dec 16 '18 at 13:15
• Except you misquoted heavily. – Did Dec 16 '18 at 13:17
• @kimchilover If $\mu_n = 0$ it would mean that $X_n = 0$, so I am not sure where is the problem – 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.
• 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! – SomeoneHAHA Dec 16 '18 at 13:38