While discussing the central limit theorem, my text makes the following remark:

The independence of the random variables $X_1, X_2, \ldots$ is essential. To see this, take $X_1 = X_2 = \ldots = X$ where $X$ is not normally distributed. Then $$\frac{X_1+\ldots+X_n-nE(X)}{\sigma\sqrt{n}}\xrightarrow{\text{in dist.}}X$$

I don't see why would the LHS converge to $X$. Does the quoted passage make sense?

What if we show the significance of independence in the following way: we take $X_1 = X_2 = \ldots = X\sim N(0,1)$ (contrary to the quoted above). Then $$\frac{X_1+\ldots+X_n-nE(X)}{\sigma\sqrt{n}}=\sqrt{n}X$$ and $\sqrt{n}X$ doesn't converge to $X$ in distribution, since $$F_{\sqrt{n}X}(1)=F_{X}(\frac{1}{\sqrt{n}})\xrightarrow{n\to\infty}F_X(0)\neq F_X(1)$$

Does my proof make sense?

What is the best way to demonstrate the importance of independence in the central limit theorem?

  • $\begingroup$ @Did Thank you, sorry, I have no idea what I was thinking. $\endgroup$ – snar Mar 16 '14 at 20:44

I. The quote doesn't make sense to me. First, it should be $X_1 + \dots X_n - n E[X]$, not $X_1 + \dots X_n + n E[X]$. Then $$\frac{X_1 + \dots X_n - n E[X]}{\sqrt{n}}$$ diverges when $X_1= \dots = X_n = X$. This example shows that if $X_i = X$ then the Central Limit Theorem doesn't hold for $X_i$.

We can also let $X\sim {\cal N}(0,1)$ and let $X_k = (-1)^k X$. Then each $X_k$ is distributed as ${\cal N}(0,1)$. But $$\frac{X_1 + \dots X_n}{\sqrt{n}} \to 0$$ in distribution.

II. Your proof is correct.


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