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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.

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  • $\begingroup$ What are $a_i$, and $p$? Where did this question arise? $\endgroup$
    – Carl
    Apr 18, 2020 at 17:50
  • $\begingroup$ $a_i, p$ are constants. $\endgroup$
    – John Smith
    Apr 18, 2020 at 18:02

2 Answers 2

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Necessary and sufficient conditions for a CLT of variables that are independent but not iid are known: the Lindeberg-Feller CLT.

The conditions for a CLT of an independent zero-mean, sequence $y_n$ (or triangular array $y_{in}$ to converge are

  • The 'Lindeberg' condition on the tails: with $s_n^2=\sum_{i=1}^n\mathrm{var}[Y_i]$, for any $\epsilon>0$, $$\lim_{n\to\infty}\frac{1}{s_n^2} \sum_{i=1}^n E[Y_i^2\mid |Y_i|>\epsilon s_n]=0$$
  • The 'uniform asymptotic negligibility' condition $$\max_{i\leq n} \mathrm{var}[Y_i]/s_n^2\to 0$$

The Lindeberg condition is implied by a bound on $E[|Y_i|^{2+\delta}]$ for any $\delta>0$ and that's often how it's proved. The Lindeberg condition is sufficient, adding the UAN condition tightens it to be necessary.

The basic idea behind the proof of the sufficiency half of theorem is to take a sequence of Normal random variables with the same means and variances as the $Y_i$. The sum of this is (trivially) Normal. Now replace them one at a time by the $Y_i$ and show that the expectation of a suitable set of functions of the partial sums doesn't change much, so that the limiting distribution is still Normal. The details are a bit annoying but widely available.

We will want to take $Y_i=a_iX_i$. Whether your conditions imply these conditions is not clear. For a start, I'm not sure whether you mean your variables $X_i$ to iid or just zero mean and constant variance. If they aren't iid, it's possible that the result fails even for $a_i\equiv 1$. For example, if the skewness of the $X_i$ increases fast enough with $i$ the skewness of the partial sums might not go to zero.

If your $X_i$ are iid, we'd only need to worry about the impact of the $a_i$. Your condition $$\frac{1}{n}\sum_{i\leq n} a_i^2\to p$$ is equivalent to $$\sum_{i\leq n} \frac{\mathrm{var}[Y_i]}{n\sigma^2}=\sum_{i\leq n} \frac{a_i^2\sigma^2}{n\sigma^2}\to p/\sigma^2$$ which implies $$\max_{i\leq n} \frac{\mathrm{var}[Y_i]}{n\sigma^2}\to 0$$ the UAN condition.

Again if your $X_i$ are iid $$E[Y_i^2\mid |Y_i|>\epsilon s_n]=a_i^2E[X_i^2\mid |X_i|>\epsilon \tau_n ]$$ where $\tau_n=s_n/a_n=\sigma\sqrt{\sum_{i=1}^na_i^2}$. Your condition on $a_i$ implies $\tau_n$ is close to $p$ (within an arbitrarily short interval except finitely often), so this implies Lindeberg's condition.

Note that convergence of $\frac{1}{n}\sum_i a_i^2$ was used; a $\limsup$ would not be enough.

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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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 at 21:24

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