# Convergence in distribution of sum of Bernoulli distributed random variables.

Let $$X_i^{(n)} \sim \operatorname{Ber}(p_{i,n})$$ for all $$n\in \mathbb N$$ and $$i\in \{1,\dots n\}$$ Bernoulli random variables on a probability space $$(\Omega, F, \mathbb P)$$, such that $$X_1^{(n)},\dots X_n^{(n)}$$ are independent for all $$n$$. And let $$\alpha >0$$ exist such that $$\lim_{n\to\infty}\max_{i\in \{1,\dots,n\}}\mid np_{i,n}-\alpha \mid =0$$+ Calculate the characteristic function of $$S_n:= \sum_{i=1}^n X_i^{(n)}$$ for all $$n$$. Then show that the distribution of $$S_n$$ converges weakly for $$n\to\infty$$ and identificate its limit.

My approach: Since $$X_i^{(n)}$$ are independent, for their characteristic functions holds $$\phi_{X_1^{(n)}+\dots +X_n^{(n)}}=\phi_{X_1^{(n)}}\dots\phi_{X_n^{(n)}}$$.

We have $$\phi_{X_j^{(n)}}=1-p_j+p_je^{it}$$ and therefore we get $$\phi_{S_n}=\prod_{j=1}^n(1-p_j+p_je^{it})$$

Now I want to show that this function converges pointwise for $$n\to\infty$$ to some function $$f$$ which is continuous in $$0$$ and then by Levy's Theorem the distribution of $$S_n$$ will converge to the distribution corresponding to $$f$$.

But here I am struggling. I tried to consider $$\lim_{n\to\infty} \exp(\log(\phi_{S_n}))= \exp(\sum_{j=1}^{\infty} \log(1-p_j + p_je^{it}))$$, but I am unable to continue.

I think that this is somehow related to Poisson approximation Theorem.

How does one find its limit here? And why did the author write $$X_i^{(n)}$$ instead of simply $$X_i$$, i.e. what is this index good for?

• It seems to me that $X_i^{(n)}\sim\text{Ber}(p_{i,n})$. As for a hint, consider first the case where $p_{i,n}p_n$ depends only on $n$ and guess what the limit would be. Sep 25, 2018 at 17:46
• if it only dependends on n it has to binomial distributed
– user563311
Sep 25, 2018 at 18:04
• That is no longer the case for the limit. Have you heard of this phenomenon? Sep 25, 2018 at 18:06
• yes I know this Phenomen. But the problem here is that $S_n$ is not binomial distributed or?
– user563311
Sep 25, 2018 at 18:12
• Although they are not the same, you have a uniform control over their convergence, so you can expect that the same phenomena will also happen in your case. Like CLT, Poisson convergence is a robust phenomena. Sep 25, 2018 at 18:16

As you noted, if $$p_{i,n}$$ only depends on $$n$$, then the $$S_n$$ are binomial, and the Poisson approximation theorem tells you the limiting distribution is Poisson with mean $$\lim_{n \to \infty} n p_n$$. Sangchul Lee's hint is that under the slightly more general conditions in your post, the limiting distribution is still Poisson.
The Poisson distribution with mean $$\alpha$$ has characteristic function $$\exp(\alpha (e^{it} - 1))$$. So you need to show $$\sum_{i=1}^n \log(1 - p_{i,n} + p_{i,n} e^{it}) \to \alpha (e^{it} - 1).$$
Using the Taylor approximation $$\log(1+x) = x + O(x^2)$$ with $$x = p_{i,n}(e^{it} - 1)$$ we have $$\sum_{i=1}^n \log(1 - p_{i,n} + p_{i,n} e^{it}) = (e^{it}-1)\sum_{i=1}^n p_{i,n} + O(n p_{i,n}^2) (e^{it}-1).$$ You can show the second term vanishes and that $$\sum_{i=1}^n p_{i,n} \to \alpha$$.