1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

$\newcommand\la\lambda$ $\newcommand\nt{\lfloor nt\rfloor}$ 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 $\la$.

$\endgroup$
3
  • $\begingroup$ How do you know that convergence relationship, i.e. $P(S^n_{\lfloor nt\rfloor}<i)\to P(S_{\lambda t} <i)$? $\endgroup$ – Math is like Friday Jun 7 '20 at 21:08
  • $\begingroup$ @MathislikeFriday : I have added this detail. $\endgroup$ – Iosif Pinelis Jun 7 '20 at 21:28
  • $\begingroup$ Thank you sir :) $\endgroup$ – Math is like Friday Jun 7 '20 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.