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 $\alpha \to +\infty$?

  • $\begingroup$ One can certainly say something if $N(\alpha)$ does not depend on $\alpha$. Do you also want to require that $E(N(\alpha)) \to \infty$? $\endgroup$ Apr 19, 2012 at 12:14
  • $\begingroup$ Hans: That condition is already implied, since $N(\alpha)$ is discrete and $Y$ is continuous. $\endgroup$ Apr 19, 2012 at 12:17
  • $\begingroup$ I do have that $E(N(\alpha)) \to +\infty$ as $\alpha \to +\infty$. $\endgroup$
    – alezok
    Apr 19, 2012 at 13:00
  • $\begingroup$ Do you want $E(N(\alpha))$ in the denominator ? $\endgroup$
    – mike
    Apr 19, 2012 at 15:01
  • $\begingroup$ 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. $\endgroup$
    – alezok
    Apr 19, 2012 at 15:09

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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