Question:
Let $(X_n)$ be a sequence of i.i.d. Cauchy random variables with density $\frac{1}{ π(1+x^2)}$. Use the characteristic function $φ(t) = e^{−|t|}$ of the Cauchy distribution to find the distribution of the average $$\frac{1}{n}\sum\limits_{i=1}^nX_i$$
How do the averages behave as n → ∞? Why does not this violate the Strong Law of Large Numbers?
Answer
The characteristic function of the average is $φ(t/n)^n = e^{−|t|}$, that is, the same as for the Cauchy distribution. Hence each average is a Cauchy random variable. In particular, this implies that they converge weakly to a Cauchy random variable. This does not violate the Strong Law of Large Numbers since $E|X_n| = ∞$
I only understand the last part why it doesn't violate the strong law of large numbers but could any kind soul actually explain the answer to me? I dont even understand the characteristic function's average is $φ(t/n)^n = e^{−|t|}$ part. A detailed explanation would be sooo helpful as my exam is tomorrow. Thank you so much !
The general property of characteristic functions that you need follows from properties of exponents and independence:
$$E[e^{it \frac{\sum_{i=1}^n X_i}{n}}]=E[e^{it/n X_1} e^{it/n X_2} \cdots e^{it/n X_n}] = E[e^{it/n X_1}] \cdots E[e^{it/n X_n}] = \phi(t/n)^n$$
where $\phi$ is the characteristic function of each of the $X_i$. Now you just need to check that $\phi(t/n)^n = \phi(t)$ if $\phi$ is the characteristic function of a Cauchy variable.