## Limit of a rescaled random sum of i.i.d. random variables

Consider a sequence of i.i.d. random variables $(X_i)_{i \in \mathbb N}$ and let $S_n=X_1+\dots+X_n$

For every $\alpha \in ]0,+\infty[$, let $N(\alpha)$ be a discrete random variable on $\mathbb N$, independent of $X_i$ for every $i \in \mathbb N$. Suppose that, as $\alpha \to +\infty$, $$\frac{N(\alpha)}{E(N(\alpha))} \stackrel{d}{\longrightarrow} Y,$$ where $Y$ is a non-degenerate continuous probability distribution on $[0,+\infty[$.

Is possible to say something about the limit in distribution of $\displaystyle \frac{S_{N(\alpha)}}{E(S_{N(\alpha)})}$, as $\alpha \to +\infty$?

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 One can certainly say something if $N(\alpha)$ does not depend on $\alpha$. Do you also want to require that $E(N(\alpha)) \to \infty$? – Hans Engler Apr 19 2012 at 12:14 Hans: That condition is already implied, since $N(\alpha)$ is discrete and $Y$ is continuous. – Mark Meckes Apr 19 2012 at 12:17 I do have that $E(N(\alpha)) \to +\infty$ as $\alpha \to +\infty$. – Ale Zok Apr 19 2012 at 13:00

A closely related problem was treated by H. Robbins, The asymptotic distribution of the sum of a random number of random variables, Bull. AMS 54(1948), 1151--1161, Math Reviews MR0027974.

In essence, under suitable nondegeneracy assumptions and assuming the existence of finite second moments, Robbins proves that the asymptotic (as $\alpha \to \infty$) distribution of $\frac{S_N - E(S_N)}{\sqrt{var(S_N)}}$ is related to the asymptotic distribution of a linear combination of $\frac{N - E(N)}{\sqrt{var(N)}}$ and another normal r.v. $Z$.

For the case where $var(N) = o(E(N)^2)$, the implication seems to be that $\frac{S_N}{E(S_N)} \stackrel{d}{\longrightarrow} 1$, a constant.

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Do you want $E(N(\alpha))$ in the denominator ?

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It is clear that $E(S_{N(\alpha)})=E(N(\alpha))E(X)$ and so, without loss of generality one may assume $E(X)=1$. In this case, I agree that the denominator becomes simply $E(N(\alpha))$. Did I get your question correctly? Anyway, yours seems more a comment than an answer. – Ale Zok Apr 19 2012 at 15:09
as long as $E(X_i) \ne 0$ I am ok with it. But if you are assuming $E(X_i) = 1$ as long as $P(Y=0) = 0$ that is going to converge in distribution to $Y$ by Slutsky's Thm , $\frac {S_N}{E(N)} = \frac {S_N}{N} \frac {N}{E(N)}$ and ${S_N}{N} \rightarrow E(X_i)$. I think it is also true in this case if $Y$ can be zero, but the proof would be a mess of details. – mike Apr 19 2012 at 15:34
Of course that should be $\frac {S_N}N \rightarros E(X_i)$ – mike Apr 19 2012 at 15:36