Central limit theorem for weighted average

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.

• Did you mean to say law of large numbers instead of central limit theorem above? – Gautam Shenoy Mar 5 at 9:01
• @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. – tst Mar 5 at 9:10
• 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. – Gautam Shenoy Mar 5 at 9:24
• @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. – 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 $$\delta$$.