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I have data from a random variable $X$:

$$\{x_{1},\dots,x_{N}\}$$

I am interested in the asymptotic distribution of the sample average $\bar{x}=\frac{\sum_{i=1}^N x_{i}}{N}$.

These draws are independent but not identically distributed. They have mean $\mu_{i}$ and variance $\sigma^2_{i}<\infty$. (sorry if notation is not strict enough)

I know the Lindeberg–Lévy Central Limit Theorem does not hold because moments are not identical. But assume the Lyapunov’s condition holds. Then,

$$ \frac{1}{s_{n}}\sum_{i=1}^N(x_{i}-\mu_{i}) \xrightarrow{d} N(0,1) $$

where $s_{n}^2 = \sum_{i=1}^N \sigma_{i}^2 $.

Is there anything I can be certain about the sample average? Some sufficient condition? Can I not infer some degree of asymptotic normality?

My data support normality of the average. I wonder if there is some theoretical basis behind that, or just mere coincidence.

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  • $\begingroup$ If you have Greene, theorem D.19 gives a variant of CLT that only requires independence. $\endgroup$ – VCG Sep 12 '16 at 20:25
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    $\begingroup$ Lindeberg CLT does not require the same first or second moments. It is a sufficient condition too. In essence, you only need a sequence of RVs whose don't get too big too fast (in the precise sense given in the theorem). $\endgroup$ – user75138 Sep 12 '16 at 21:11
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Note that if you have the sample average $\bar{x}=\frac{1}{n}\sum_{i=1}^nx_{i}$, then using the same "sample average" operator on the individual moments to define $\hat{\mu}=\bar{\mu}$ and $\hat{\sigma}^2=\overline{\sigma^2}$, your Lyapunov equation becomes $$\frac{\bar{x}-\hat{\mu}}{\hat{\sigma}/\sqrt{n}} \xrightarrow{d} N(0,1)$$ This is essentially the same as the i.i.d. result for the asymptotic sampling distribution of the sample mean.

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