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For $n\in \mathbb{N}$ let $X_n$ be geometric with parameter $p_n \in (0,1)$, that means $\mathbf{P}[X_n = k] = p_n(1-p_n)^k$, $k\in\mathbb{N}_0$.

How must the sequence $(p_n)_{n\in\mathbb{N}}$ look like so that $X_n/n\overset{\mathcal{D}}{\longrightarrow} \mathrm{Exp}(\alpha)$ with $\alpha>0$?

Note: "$\overset{\mathcal{D}}{\longrightarrow}$" denotes convergence in distribution.

I'm pretty sure that the condition for the sequence is $p_n = \alpha/n$.

Let $A \in \mathcal{B}([0, \infty))$. We have to prove that the associated image measures \begin{equation*} \mathbf{P}_{X_n/n}[A] = p_n(1-p_n)^k \, \delta_{k/n}(A) = \frac{\alpha}{n} \Bigl(1-\frac{\alpha}{n}\Bigr)^k \delta_{k/n}(A) \end{equation*} converge weakly against the image measure \begin{equation*} \mathbf{P}_{\mathrm{Exp}(\alpha)}[A] = \int_A \alpha e^{-\alpha x} \, dx\,. \end{equation*}

Let $f \in C_b([0, \infty))$, i.e. an arbitrary real valued function on $[0, \infty)$ which is bounded and continuous, then \begin{align*} \int f \, d \mathbf{P}_{X_n/n} = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty} f\biggl(\frac{k}{n}\biggr) \, \frac{\alpha}{n}\Bigl(1-\frac{\alpha}{n}\Bigr)^{k} = \lim_{n\rightarrow\infty} \sum_{k=0}^{\infty} f\biggl(\frac{k}{n}\biggr) \, \frac{\alpha}{n}\biggl(\Bigl(1-\frac{\alpha}{n}\Bigr)^{n}\biggr)^{\frac{k}{n}} \, . \end{align*} Also using dominated convergence with the majorant $2 \|f\|_\infty \alpha e^{-\alpha x}$ we get: \begin{align*} \int f \, d\mathbf{P}_{\mathrm{Exp}(\alpha)} &= \int f(x)\, \alpha e^{-\alpha x} \, dx \\ &= \int \lim_{m\rightarrow\infty} \sum_{k=0}^{m^2} f\biggl(\frac{k}{m}\biggr) \, \mathbf{1}_{\bigl[\frac{k}{m}, \frac{k+1}{m}\bigr)}(x) \, \alpha e^{-\alpha k/m} \, dx \\ &= \lim_{m\rightarrow\infty} \int \sum_{k=0}^{m^2} f\biggl(\frac{k}{m}\biggr) \, \mathbf{1}_{\bigl[\frac{k}{m}, \frac{k+1}{m}\bigr)}(x) \, \alpha e^{-\alpha k/m} \, dx \\ &= \lim_{m\rightarrow\infty} \sum_{k=0}^{m^2} f\biggl(\frac{k}{m}\biggr) \, \frac{\alpha}{m} e^{-\alpha k/m}\, . \end{align*}

So, obviously it seems that we're very near the solution. But we have to partially take some limits and I don't know how to do that. For example, I can't use the M-test to draw the limit back into the infinite sum. Also, though it seems that I can substitute the $m^2$ by $\infty$ in the last equation, I'm not so sure. And even if possible, I don't know what to do about the limit there, too.

If anybody could help me, please?

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What about a simpler approach? Let $p_n = \alpha/n$. Then for any $t>0$:

\begin{align} Pr[X_n/n > t] &= Pr[X_n > nt] \\ &= Pr[X_n > \lfloor nt \rfloor] \\ &= (1-p_n)^{\lfloor nt \rfloor} \\ &= \left(1 - \frac{\alpha}{n}\right)^{\lfloor nt \rfloor}\\ &= \left(1-\frac{\alpha}{n}\right)^{nt} \left(1-\frac{\alpha}{n}\right)^{-\beta(n,t)} \end{align} where $\beta(n,t) \equiv nt - \lfloor nt \rfloor$ and satisfies $0 \leq \beta(n,t) \leq 1$ for all $n,t$. Taking a limit as $n\rightarrow\infty$ makes the right-hand-side converge to $e^{-\alpha t}$.


Indeed, recall that: $$ \lim_{n\rightarrow\infty} \left(1 - \frac{\alpha}{n}\right)^n = e^{-\alpha} $$

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  • $\begingroup$ Thank you. That was easy money, but I'm glad somebody finally answered. An easy exercise, actually, but I had a pretty bad misunderstanding about weak convergence / convergence in distribution that led to this problem. $\endgroup$ – Qyburn May 9 '15 at 10:10
  • $\begingroup$ @Michael, what is the theorem to show what you have written underneath "recall that"? $\endgroup$ – jaja Jan 6 at 19:19
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    $\begingroup$ @jaja : This is a well known limit, see here: math.stackexchange.com/questions/882741/… $\endgroup$ – Michael Jan 7 at 1:44
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    $\begingroup$ Basically just define $y_n = \log[(1-\alpha/n)^n]$, note that $y_n = \frac{\log(1-\alpha/n)}{1/n}$, and use L'Hopital's rule for 0/0 to prove $y_n \rightarrow -\alpha$. $\endgroup$ – Michael Jan 7 at 1:50

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