Let $(a_i)_{i\ge1}$ be a bounded positive sequence and $X_i$ be iid random variables with mean $0$ and finite variance. Let $s_n=\frac{\sum_{i=1}^n a_i X_i}{\sqrt{\sum_{i=1}^n a_i}}$.

If $a_i=1$ for all $i$, then we have the central limit theorem for the limit. Is there anything known for the general case? Do we need to restrict $a_i$'s more to have a similar limit?

PS. I don't use the word converge because I don't want to guess what kind of convergence this may be.

  • $\begingroup$ Did you mean to say law of large numbers instead of central limit theorem above? $\endgroup$ – Gautam Shenoy Mar 5 at 9:01
  • $\begingroup$ @GautamShenoy no, I meant the central limit theorem. I think that the limit is a gaussian (at least in the unimodal case with tame $a_i$'s) but I cannot even begin proving this. $\endgroup$ – tst Mar 5 at 9:10
  • $\begingroup$ But if $a_i =1 $ you get $s_n = \frac{\sum_{i=1}^n X_i}{n}$. This converges to $E[X]$ the mean. I thought you may be missing a square root in denominator. $\endgroup$ – Gautam Shenoy Mar 5 at 9:24
  • $\begingroup$ @GautamShenoy ah, yes, you are right. I missed a square root. I corrected it, thank you. I also added the assumption that the mean is $0$, because I don't think this spoils anything anyway. $\endgroup$ – tst Mar 5 at 9:30

I think the convergence depends on how we choose $a_i$.

  • Say $a_i=A_i$ is a sequence of independent, identically distributed (i.i.d.) random variables which are also independent from $X_i$. Then, the random variables $Y_i=A_i X_i$ are a also an i.i.d. sequence (not independent of $A_i$ or $X_i$, but independent among themselves). Because of the independence between $A_i$ and $X_i$ we can factor out the moments: $$ E(Y^k)=E(A^k)E(X^k), $$ so we compute $$ E(Y)=E(A)E(X)=0, $$ as long as the mean of $A$ is finite, and $$ V(Y)=E(Y^2)-E(Y)^2=E(A^2)E(X^2) =(V(A)+E(A)^2)(V(X)), $$ which is finite if we assume $V(A)$ is finite as well. Now, CLT applies: $$ \sqrt{n}\,(S_Y) \xrightarrow[]{d} N(0,V(Y)), $$ since $E(Y)=0$, for $S_Y$ the sample mean of $Y$: $$ S_Y=\frac{1}{n}\sum_{i=1}^n Y_i =\frac{1}{n}\sum_{i=1}^n A_iX_i. $$

  • Say now $a_i$ are given instead by a measurable function of $X$, $a_i=a(X_i)$. Now we define $Z_i=a(X_i)X_i$ which are also i.i.d., and repeat the analysis. We cannot factor the moments anymore because $a(X_i)$ and $X_i$ are not independent, but if $E(Z)$ and $V(Z)$ are finite then we get the CLT limit again, this time $$ \sqrt{n}\,(S_Z-E(Z)) \xrightarrow[]{d} N(0,V(Z)). $$

  • What about the case of $a_i$ given by a known sequence? Then $W_i=a_iX_i$ are a sequence of independent random variables, but their distributions are no longer identical. There is a weaker version of the CLT which gives $$ \frac{1}{\gamma_n} \sum_{i=1}^n (W_i-E(W_i)) \xrightarrow[]{d} N(0,1), $$ where $$ \gamma_n^2=\sum_{i=1}^n V(W_i), $$ provided the quantity $$ \frac{1}{\gamma_n^{2+\delta}}\sum_{i=1}^n E[|X_i-E(X_i)|^{2+\delta}] $$ goes to zero as $n$ goes to infinity for some positive $\delta$.

I suspect the last case is the one you had in mind, I hope this helps.

  • $\begingroup$ Ah, this is certainly more information than what I hoped. Thank you. Is there any standard reference for these? $\endgroup$ – tst Mar 5 at 12:53
  • 1
    $\begingroup$ The first two are the classical CLT, and the last one is Lyapunov's CLT, I'm sure they're both on Wikipedia. $\endgroup$ – R_B Mar 5 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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