Welcome to MathOverflow

Question

1

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 $\alpha \to +\infty$?

flag	asked Apr 19 2012 at 11:40 Ale Zok 103●1●3

	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

Hans Engler · Answer 1 · 2012-04-19 19:53:06Z UTC

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.

mike · Answer 2 · 2012-04-19 15:01:01Z UTC

0

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

answered Apr 19 2012 at 15:01

mike
100●1●3

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

1

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

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

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.