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I'm studying almost-surely convergence and convergence in probability of $S_{n}=X_{1}+\cdots+X_{n},$ where $X_{n}$ is distributed $\mathrm{ Poisson}\left(1/n\right)$ with $n\in\mathbb{N}$ and the sequence of $X_{n}$ are independent.

I'm stuck with this because I cannot applied Chebyshev inequality because of divergence of $\displaystyle\sum_{i=1}^{\infty}\frac{1}{n}$, neither Borel-Cantelli Lemma because I don't find a useful bound to use it.

I tried use the convergence of $\displaystyle\sum_{n=1}^{\infty}E\left(X_n\right)<\infty$ implies $\displaystyle\sum_{n=1}^{\infty}X_n$ convergence a.s., but I don't get it.

Any kind of help is very thanked.

Suiz96
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  • The Poisson r.v. are non negative; this implies that $S_n$ converges a.s.. However the limit is infinity. Indeed, as the Poisson are integer variables, the set $S_n$ converges to a finite number is equal to $\liminf X_n=0$. However, this has probability zero as it's complement, $\limsup X_n>0$, has probability one by second Borel Cantelli. – Kolmo Nov 09 '16 at 13:47

2 Answers2

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Let $a_n:=\Pr\left(S_{2n}-S_n\geqslant 1\right)$. Using independence, we can see that $S_{2n}-S_n$ is Poisson distributed with parameter $\sum_{i=n+1}^{2n}1/i$. As a consequence, we have $$a_n\geqslant 1-\exp\left(-\sum_{i=n+1}^{2n}1/i\right) \geqslant 1-\exp\left(-\frac 12\right)\gt 0.$$

Therefore, the sequence $\left(S_n\right)_{n\geqslant 1}$ cannot be convergent in probability.

Community
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Davide Giraudo
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The (finite) sum of independent Poisson distributed random variables is Poisson distributed with a parameter which is the sum of the individual parameters.

So $S_n$ is Poisson distributed with harmonic number parameter (and mean and variance) $H(n) = \displaystyle\sum_{m=1}^{n}\frac{1}{m} \approx \log_e{n}+\gamma+\frac{1}{2n}- \cdots$.

The only real convergence here is that $\dfrac{S_n}{H(n)} \to 1$ in probability and almost surely in a law of large numbers sort of sense and $\dfrac{S_n}{\sqrt{H(n)}} -\sqrt{H(n)} \xrightarrow{d} N(0,1)$ in a central limit theorem sort of sense.

The infinite series is almost surely greater than any given number.

Henry
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  • How do you prove $\dfrac{S_n}{H(n)} \to 1$ almost surely ? – Gabriel Romon Sep 13 '17 at 08:35
  • @GabrielRomon $\dfrac{S_n}{H(n)}$ has characteristic function $\exp\left(H(n)(e^{it/H(n)}-1)\right) \to \exp(it)$ as $n\to \infty$, which is the characteristic function of a degenerate distribution with a point probability at $1$. – Henry Sep 13 '17 at 09:45
  • I think I'm missing something... Wouldn't that only imply convergence in distribution ? – Gabriel Romon Sep 13 '17 at 10:18
  • It is convergence to a constant rather than to a wider distribution. You can also show the variance of $\frac{S_n}{H(n)} -1$ is $\frac1{H(n)}$, the other moments are finite and indeed you know the full distribution. But you raise a good point which needs further thought – Henry Sep 13 '17 at 11:37
  • @GabrielRomon - with further thought, although I still think $\frac{S_n}{H(n)} \to 1$ in probability, I seem to have persuaded myself that $\mathbb P \left( \frac{S_n}{H(n)} \gt 2 \right) \gt \frac{0.08}{\sqrt{n}}$ so can be expected to happen infinitely often, meaning that it does not converge to $1$ almost surely. I have edited my response. Thank you – Henry Sep 13 '17 at 23:16
  • your worries are founded. It actually does converge almost surely, see for example my answer here https://math.stackexchange.com/a/2427732/66096 – Gabriel Romon Sep 14 '17 at 05:24
  • @GabrielRomon So should I worry about my belief that $\mathbb P \left( \frac{S_n}{H(n)} \gt 2 \right) \gt \frac{0.08}{\sqrt{n}}$ or that this may mean that I could expect to see $\frac{S_n}{E[S_n]} \gt 2$ infinitely often? – Henry Nov 02 '17 at 18:00