Considering the random variables $X_1,\ldots,X_n$, i.i.d, with Cauchy distribution, and the random variable $Y_n=\frac{X_1+\cdots+X_n}{n}$

Determinate the characteristic function of $Y_n$ and show that the Central Limit Theorem does not check for i.i.d Cauchy random variable.

If $X_1, \ldots, X_n$ are i.i.d. Cauchy$(0, 1)$ then we can show that $\bar{X}$ is also Cauchy$(0, 1)$ using a characteristic function argument:

\begin{align} \varphi_{\bar{X}}(t) &= \operatorname{E} \left (e^{it \bar{X}} \right ) \\ &= \text{E} \left ( \prod_{j=1}^{n} e^{it X_j / n} \right ) \\ &= \prod_{j=1}^n \operatorname{E} \left ( e^{it X_j / n} \right ) \\ &= \operatorname{E} \left (e^{it X_1 / n} \right )^n \\ &= e^{- |t|} \end{align}

How can I show that the Central Limit Theorem fails?

  • 1
    $\begingroup$ Hint: To show the central limit theorem does not hold you need to show that the characteristic function does not to converge to the one for the standard normal. $\endgroup$ – Gregory Jun 22 '18 at 14:21
  • $\begingroup$ You have in fact shown that the average of $n$ i.i.d. standard Cauchy random variables has the distribution of a standard Cauchy random variable $\endgroup$ – Henry Jun 22 '18 at 14:25

You have largely shown it. The conclusion of CLT, which states $$\frac{\sqrt{n}(\bar X_n-\mu)}{\sigma}\to_D N(0,1)$$ implies the weak law of large numbers $\bar X_n\to_P \mu.$ But by showing the sample mean converges in distribution to a Cauchy distribution, not a constant, you have shown that the weak law of large numbers does not hold.

Also, we typically have as part of the conclusion of CLT that $\mu$ is the mean of the parent distribution, so really in this case the conclusion of the CLT does not even make sense here since the mean of the Cauchy distribution is undefined. But inasmuch as we can make sense of it, the above argument shows it does not hold.

  • $\begingroup$ "The CLT implies a law of large numbers" In which sense? It is easy to find some sequence $(V_n)$ such that $V_n/\sqrt n$ converges in distribution to the standard normal distribution while $V_n/n$ diverges almost surely. $\endgroup$ – Did Jun 22 '18 at 15:04
  • $\begingroup$ @Did I don't think that's the point. It is always true that $\sqrt{n}(\bar{X}_n - \mu) \leadsto N(0, \sigma_2)$ implies $\bar{X}_n \to \mu$ in probability. $\endgroup$ – Galton Jun 22 '18 at 15:15
  • $\begingroup$ @Galton This is exactly the point, which is why I am asking in which sense the OP is declaring that CLT $\implies$ LGN. $\endgroup$ – Did Jun 22 '18 at 15:41
  • $\begingroup$ @Did Good point, I should have specified weak. $\endgroup$ – spaceisdarkgreen Jun 22 '18 at 17:47

You should notice that the charateristic function $t\mapsto e^{-|t|}$ that you found is the same as the characteristic function of the distribution of $X_1.$ Thus any measure of dispersion that you use for the distribution of the sample mean (for example, the interquartile range) does not get smaller as $n$ grows.

For the central limit theorem to hold, you'd need to have the characteristic function approaching that of the normal distribution.


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