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. $X_1^n \overset{d}{=}\operatorname{Be}(\lambda/n)$ 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?


1 Answer 1


$\newcommand\la\lambda$ $\newcommand\nt{\lfloor nt\rfloor}$ 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}<i)\to P(S_{\la t} <i) \\ &=\frac{\la^i}{\Gamma(i)}\int_t^\infty u^{i-1} e^{-\la u}\,du \end{align} (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 $\la$.

  • $\begingroup$ How do you know that convergence relationship, i.e. $P(S^n_{\lfloor nt\rfloor}<i)\to P(S_{\lambda t} <i)$? $\endgroup$ Commented Jun 7, 2020 at 21:08
  • $\begingroup$ @MathislikeFriday : I have added this detail. $\endgroup$ Commented Jun 7, 2020 at 21:28
  • $\begingroup$ Thank you sir :) $\endgroup$ Commented Jun 7, 2020 at 21:43

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