Does ${S_n\over \Bbb E (S_n)}$ converges in probability?

Question

Let $X_n\sim \text{Poi}(\lambda_n)$, with $X_i$'s independent and $\sum\limits_{n=1}^\infty\lambda_n=\infty.$ Define $S_n=\sum\limits_{i=1}^nX_i$ . Is it true that $${S_n\over \Bbb E (S_n)}$$ converges in probability?

I don't know what it converges to, but I guess it should be $1$.

Answer 1

Yes, your guess is correct. Since the sum of independent Poisson r.v.'s is again Poisson with parameter the sum of the parameters, you know that $$σ^2_n=\text{Var}(S_n)=\mathbb E[S_n]=\sum_{k=1}^nλ_n$$ and hence by Chebyshev's inequality \begin{align}P\left(\left|\frac{S_n}{\mathbb E[S_n]}-1\right|\ge\epsilon\right)&=P\left(\left|S_n-\mathbb E[S_n]\right|\ge σ_n^2\epsilon\right)\\[0.3cm]&=P\left(\left|S_n-\mathbb E[S_n]\right|\ge σ_n\cdot(σ_n\epsilon)\right)\le^{\text{Chebyshev's inequality}}\\[0.3cm]&\le\frac{1}{(σ_n\epsilon)^2}=\frac{1}{\epsilon^2\sum_{k=1}^nλ_k}\longrightarrow 0\end{align} as $n\to\infty$ for any $\epsilon >0$.

Answer 2

Even strongly, $\displaystyle \frac{S_n}{\mathbb{E}[S_n]}\to 1$ almost surely. To prove this, let us use the following steps.

1) First, notice that by Chebyshev's inequality, we have $$ \mathbb{P}\left(\left|\frac{S_n}{\mathbb{E}[S_n]}-1\right|>\epsilon\right)\leq \frac{\mathbb{VAR}\left(\frac{S_n}{\mathbb{E}[S_n]} \right)}{\epsilon^2}=\frac{1}{\epsilon^2}\frac{1}{\sum_{k=1}^n \lambda_k}. $$

2) Now, we will consider a subsequence $n_k$ determined as follows. Let $$ n_k \triangleq \inf\left\{n : \sum_{i=1}^n \lambda_i \geq k^2\right\}. $$ For this subsequence, $$ \sum_{k=1}^{\infty}\mathbb{P}\left(\left|\frac{S_{n_k}}{\mathbb{E}[S_{n_k}]}-1\right|>\epsilon\right)\leq\frac{1}{\epsilon^2} \sum_{k=1}^{\infty}\frac{1}{\sum_{i=1}^{n_k} \lambda_i}\leq \frac{1}{\epsilon^2}\sum_k \frac{1}{k^2}<\infty. $$

3) By Borel-Cantelli lemma, we have $$ \mathbb{P}\left(\left|\frac{S_{n_k}}{\mathbb{E}[S_{n_k}]}-1\right|>\epsilon, \text{ infinitely often}\right)=0, \forall \epsilon>0. $$ Hence, taking a countable sequence of $\epsilon$'s converging to $0$, we get that $$ \frac{S_{n_k}}{\mathbb{E}[S_{n_k}]} \rightarrow 1, \text{almost surely}. $$

4) Now to finish the argument, take $n_k \leq n \leq n_{k+1}$ (arbitrary). Notice that $S_{n_k}\leq S_n \leq S_{n_{k+1}}$, hence we have, $$ \frac{S_{n_k}}{\mathbb{E}[S_{n_k}]}\frac{\mathbb{E}[S_{n_k}]}{\mathbb{E}[S_{n_{k+1}}]}\leq \frac{S_n}{\mathbb{E}[S_n]}\leq \frac{S_{n_{k+1}}}{\mathbb{E}[S_{n_{k+1}}]}\frac{\mathbb{E}[S_{n_{k+1}}]}{\mathbb{E}[S_{n_k}]}. $$ Now, we will finish the proof by showing that $\displaystyle\frac{\mathbb{E}[S_{n_{k+1}}]}{\mathbb{E}[S_{n_k}]}\to 1$. To do so, notice from the definitions that, $$ k^2 \leq \mathbb{E}[S_{n_k}]\leq \mathbb{E}[S_{n_{k+1}}]\leq (k+1)^2 + \lambda_{n_{k+1}} \implies 1\leq \frac{\mathbb{E}[S_{n_{k+1}}]}{\mathbb{E}[S_{n_{k}}]} \leq \frac{(k+1)^2}{k^2} + \frac{\lambda_{n_{k+1}}}{k^2}. $$

Finally, we can assume all variances are bounded (by possibly representing the Poisson random variables with $\lambda_n$ as the sum of a few other independent Poisson's where we can control the rate and make sure it is bounded), and therefore, we are done.