Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\alpha \to +\infty$?

share|improve this question
    
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 '12 at 12:14
    
Hans: That condition is already implied, since $N(\alpha)$ is discrete and $Y$ is continuous. –  Mark Meckes Apr 19 '12 at 12:17
    
I do have that $E(N(\alpha)) \to +\infty$ as $\alpha \to +\infty$. –  Ale Zok Apr 19 '12 at 13:00
add comment

2 Answers

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.

share|improve this answer
add comment

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

share|improve this answer
    
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 '12 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 '12 at 15:34
    
Of course that should be $\frac {S_N}N \rightarros E(X_i)$ –  mike Apr 19 '12 at 15:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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