# Does sum of i.i.d. Bernoulli random variables with parameter $\lambda/n$ asymptotically converges to Gamma distribution?

Actually the question has more details than what it says in the title. Sorry about that I may described the question wrongly.

Let $$X_1^n, X_2^n,\dots$$ be i.i.d. Bernoulli random variables with parameter $$\lambda/n$$, i.e. $$X_1^n \overset{d}{=}$$Be$$(\lambda/n)$$ with fixed $$\lambda > 0$$. Consider $$T_i^n := \inf\{k : X_1^n + \cdots + X_k^n = i\}.$$ And I want to show that $$\frac{T_i^n}{n}\xrightarrow[n\to\infty]{d}\text{Gamma}(i,\lambda).$$

This confuse me since we know that the sum of Bernoulli random variables asymptotically converges to Poisson distribution and I don't see any relationship between Poisson and Gamma distribution.

Can anyone help me out?

$$\newcommand\la\lambda$$ $$\newcommand\nt{\lfloor nt\rfloor}$$ For any natural $$i,n,k$$, let $$S^n_k:=X^n_1+\dots+X^n_k$$, with $$S^n_0:=0$$. Then for any real $$t>0$$ \begin{align} P(T^n_i/n>t)&=P(T^n_i>\nt) \\ &=P(S^n_{\nt} (as $$n\to\infty$$), where $$S_{\la t}\sim Poisson(\la t)$$; the convergence holds by the Poisson limit theorem; the last displayed equality can be obtained by integrating by parts $$i-1$$ times.
Thus indeed, the distribution of $$T^n_i/n$$ converges to the gamma distribution with the shape paratemer $$i$$ and the rate $$\la$$.
• How do you know that convergence relationship, i.e. $P(S^n_{\lfloor nt\rfloor}<i)\to P(S_{\lambda t} <i)$? Jun 7, 2020 at 21:08