Suppose $x_i, i=1, ..., n$ are independently distributed with mean $0$ and variance $\sigma^2$. Then $\frac{1}{\sqrt n}\sum_{i=1}^n a_ix_i$ converges to a normal distribution with mean 0 and variance $\sigma^2 p$ if $\frac{1}{n}\sum_{i=1}^n a_i^2 \to p > 0$.

How to prove? What I learned was that central limit theorem was for independently identically distributed random variables. But this is not the case here.

  • $\begingroup$ What are $a_i$, and $p$? Where did this question arise? $\endgroup$ – Carl Apr 18 at 17:50
  • $\begingroup$ $a_i, p$ are constants. $\endgroup$ – John Smith Apr 18 at 18:02

What you have is what is called the weighted sum CLT. As far as I know, the exact conditions for it to hold is that $\sup_{1\leq i \leq n} \frac{a_i^2}{\sum_{j=1}^n a_j^2} \xrightarrow{n\to \infty} 0$.

This theorem states that for $\{W_n\}$ an i.i.d sequence of real random variables wih common mean 0 and variance 1,

$$ \frac{1}{\sqrt{\sum_{j=1}^n a_j^2}} \sum_{i=1}^n a_i W_i \xrightarrow{d} \mathcal{N}(0,1) \tag{1}$$

You can prove it using Lyapunov CLT which is for independent but not identically distributed variables. You should check the answer to this question in this same forum).

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  • $\begingroup$ So, if $\frac{1}{n}\sum_{i=1}^n a_i^2 \to p > 0$ implies $\frac{\max a_i^2}{\sum_{i=1}^n a_i^2} \to 0$, then a CLT can be applied to prove the original assertion. But I am stuck with this now. $\endgroup$ – John Smith Apr 18 at 19:57
  • $\begingroup$ I was just stating the exact minimum conditions. In your case this of course holds. Note that any $a_i$ must be finite. Say $\sup_i a_i=c<\infty$. Then of course $\frac{c}{np}\to 0$ as $n\to \infty$. $\endgroup$ – Oriol B Apr 18 at 20:27
  • $\begingroup$ Could you please elaborate more on the derivations? I don't see why $\sup_i a_i=c$ is a constant, how do you get $\frac{c}{np}$. $\endgroup$ – John Smith Apr 18 at 20:35
  • $\begingroup$ What I am saying is that the series must be convergent and hence each $a_i$ must be finite. Since you are dividing by something that goes to infinity (since you have a $n$ term) this limit goes to 0. I only defined the supremum as $c$ to further clarify (you can obviate this). Also if your sequence $\{a_i\}$ is bounded (probably the case) $\sup \equiv \max$ (I just used supremum to be more precise). Was the confusion about this? $\endgroup$ – Oriol B Apr 18 at 21:07
  • $\begingroup$ Thanks although I wish I could understand your arguments better. "the series" means what here? what limit is "this limit"? Where did you get the result that the sequence $\{a_i\}$ is bounded? $\endgroup$ – John Smith Apr 18 at 21:24

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