Cauchy random variables

Question

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?

Answer 1

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.

Answer 2

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.