# how to prove that ${S_N\over E[S_N]}$ converges to an exponential distribution

Suppose that $\{X_1,X_2,\ldots\}$ is a sequence of iid $L^1$-random variables such that $E[X_1]\neq 0$. Define for every $n$, $$S_n=X_1+\cdots+X_n.$$ Let $N$ be a geometric random variable such that $$P(N=k) = q^{k-1}p,\quad k=1,2,\ldots,$$ where $q=1-p$ and $p\in(0,1)$. Assume that $N$ and $\{X_1,X_2,\ldots\}$ are all independent. Show that as $p\to0$, $${S_N\over E[S_N]}$$ converges in distribution to an exponential distribution with some rate $\lambda$, and identify $\lambda$.

• I have no ideas of addressing this kind of problem. Using characteristic function or in any other ways? – Ulton Prinsphield Dec 16 '15 at 9:46
• I think the starting point will be to use Wald's lemma which states that $\mathsf{E}(S_N)=\mathsf{E}(X_1)\mathsf{E}(N)=\frac{\mathsf{E}(X_1)}{p}$ (since $X_1,\ldots, X_n$ are $IID$. Hence \begin{align} \frac{S_N}{\mathsf{E}(S_N)}=\frac{p}{\mathsf{E}(X_1)}S_N \end{align} – Shahid M Shah Dec 16 '15 at 9:52
• @ShahidMShah This can also be shown directly with condition expectations (nested). I am not sure how to continue after that, but this seems a good start (conditional expectations should help you "decouple" the $N$ from the $X_k$'s in the expectations). – Clement C. Dec 16 '15 at 10:04
• suppose the moment generating function of X is $\psi$, then the mgf of left hand side is $[\psi({pt\over m_1})]^N$, where $m_1=E(X_1)$. So we need to show that $[\psi({pt\over m_1})]^N\to {\lambda\over \lambda-t}$ as $p\to0$. – Ulton Prinsphield Dec 16 '15 at 11:44
• The exponent $N$ is a geometric random variable related to $p$, and the mgf function also contains $p$. So the problem remains to be: how could we get the limit of the above formula in this kind of form. I tried to expand the mgf, however I really have no idea how to handle the random variable $N$. – Ulton Prinsphield Dec 16 '15 at 11:49

You are already on the right track and I complete the latter part for you. Let $\mu = E[X_1]$ be the common mean and $\varphi(t) = E[e^{itX_1}]$ be the common characteristic function

As stated above, $$E[S_N] = E[E[S_N|N]] = E[N\mu] = \frac {\mu} {p}$$

Now consider the characteristic function of $\displaystyle Z = \frac {S_N} {E[S_N]} = \frac {pS_N} {\mu}$:

\begin{align*} \varphi_Z(t) & = E\left[e^{it\frac {pS_N} {\mu}}\right] \\ & = E\left[E\left[e^{it\frac {pS_N} {\mu}}|N\right]\right] \\ & = E\left[E\left[e^{i\frac {tp} {\mu} X_1}\right]^N\right] \\ & = E\left[\varphi\left(\frac {tp} {\mu} \right)^N\right] \\ & = \sum_{k=1}^{+\infty} \varphi\left(\frac {tp} {\mu} \right)^k (1 - p)^{k-1}p \\ & = p\varphi\left(\frac {tp} {\mu} \right) \sum_{k=1}^{+\infty} \left[\varphi\left(\frac {tp} {\mu} \right) (1 - p)\right]^{k-1} \\ & = \frac p\varphi\left(\frac {tp} {\mu}\right)} 1 - \varphi\left(\frac {tp} {\mu} \right) (1 - p) } \end{align*

Note the infinite geometric series converges as the characteristic function is bounded by 1. Since $X_1 \in \mathcal{L}^1$, $\varphi$ is differentiable and thus we can evaluate the limit via L'Hopital Rule:

\begin{align*} \lim_{p\to 0} \varphi_Z(t) &= \lim_{p\to 0} \frac p\varphi\left(\frac {tp} {\mu}\right)} 1 - \varphi\left(\frac {tp} {\mu} \right) (1 - p) } \\ & = \lim_{p\to 0} \frac \varphi\left(\frac {tp} {\mu}\right) + p\varphi'\left(\frac {tp} {\mu}\right)\frac {t} {\mu}} \varphi\left(\frac {tp} {\mu} \right) - (1 - p)\varphi'\left(\frac {tp} {\mu}\right)\frac {t} {\mu} } \\ & = \frac {\varphi(0) + 0} \varphi(0) - \varphi'(0)\frac {t} {\mu}} \\ & = \frac {1} 1 - i\mu \frac {t} {\mu}} \\ & = \frac {1} {1 - it} \end{align*

which is the characteristic function of $\text{Exp}(\lambda = 1)$

• impressive! Thank you very much! – Ulton Prinsphield Dec 16 '15 at 16:02