I want to solve the following exercise but i am unsure if my ideas are correct or not.

Let $(X_n)_{n\in\mathbb{N}}$ be i.i.d. random variables with probability density $$ f_a(x)=\frac{1}{a\pi}\frac{1}{1+(x/a)^2}, $$ i.e. each $X_n$ is Cauchy-distributed.

a) For which $\gamma$ does $n^{-\gamma}(X_1 + \cdots + X_n)$ converge in distribution?

b) Does $( n^{-1}(X_1 + \dots + X_n) )_{n\in\mathbb{N}}$ converge almost surely? If not, why not?

My ideas so far:

a) Define $S_n:=X_1 + \dots + X_n$. Because of the $X_i$'s independency $S_n$ is Cauchy-distributed with parameter $na$. I tried to use Lévy's continuity theorem and characteristic functions. The characteristic function of $S_n$ is $\varphi_{S_n}(t) = e^{-na|t|}$ and the characteristic function of $n^{-\gamma}S_n$ is $\varphi_n(t):=\varphi_{n^{-\gamma}S_n}(t) = e^{-n^{1-\gamma} |t|}$.

Now i look at several cases:

Let $\gamma = 1$. Then $\varphi_n(t) = e^{-|t|}$ and thus $n^{-1}S_n$ is Cauchy-distributed with parameter $1$ (independent of $n$) which implies that the sequence converges in distribution.

Now let $\gamma > 1$. Then $$ \lim_{n\rightarrow \infty} \varphi_n (t) = \lim_{n\rightarrow \infty} e^{-n^{1-\gamma} |t|} \underset{1-\gamma < 0}{=} 1 $$ for all $t\in\mathbb{R}$. I'm not sure about this though. What kind of distribution has a constant characteristic function? Based on my knowledge about Fourier transform i guess that this has something to do with the Dirac-Delta.

Finally let $\gamma < 1$. Then $$ \lim_{n\rightarrow \infty} \varphi_n(0) = \lim_{n\rightarrow \infty} e^0 = 1 $$ and for $t\neq 0$ $$ \lim_{n\rightarrow \infty} \varphi_n(t) = \lim_{n\rightarrow \infty} e^{-n^{1-\gamma} |t|} \underset{1-\gamma > 0}{=} 0. $$

The characteristic function converges point-wise but the limit function isn't continuous at $0$ and by Lévy's continuity theorem we conclude that the sequence doesn't converge in distribution.

Now part b): My intuiion says that the sequence ($n^{-1} S_n$ is the arithmetic mean) does not converge and that it has something to do with the fact that a Cauchy-distributed random variable does not have an expected value.

Thanks in advance for any tips or help in general.

  • $\begingroup$ Your "Dirac delta" observation precisely means that if $E[e^{itX}]=1$ then $X=0$ a.s. What you are seeing is that $\sum_{n=1}^N X_n$ diverges faster than $N$ but slower than $N^c$ for any $c>1$. By comparison, if the $X_n$ had finite, nonzero expectation, then $\sum_{n=1}^N X_n$ would diverge at exactly the speed of $N$ (by the law of large numbers). $\endgroup$ – Ian Jul 11 '15 at 23:56

Once you've shown that $n^{-1} S_n$ has the same distribution as $X_1$, then it's pretty much immediate that $n^{-\gamma} S_n$ does not convergence in distribution for $\gamma < 1$. If $\gamma > 1$ the characteristic function argument shows that $n^{- \gamma} S_n \stackrel{\text{d}}{\to} 0$. So $n^{-\gamma} S_n$ converges in distribution for $\gamma \geq 1$.

For the second part, since $\text{E}(|X_1| / k) = \infty$ for every $k \in \mathbb{N}$,

\begin{align} \sum_{n=1}^{\infty} P(|X_n| / k > n) &= \sum_{n=1}^{\infty} P(|X_n| / n > k) \\ &= \infty , \end{align}

which by Borel-Cantelli and independence implies $P(|X_n| / n > k \,\, \text{i.o.}) = 1$ for every $k$. Therefore $\limsup_{n \to \infty} |X_n| / n \stackrel{\text{a.s.}}{=} \infty$, which in turn implies $\limsup_{n \to \infty} |S_n| / n \stackrel{\text{a.s.}}{=} \infty$, so $n^{-1} S_n$ does not converge almost surely.

  • $\begingroup$ You mean $\gamma<1$. For $\gamma>1$ you have the distribution of $n^{\epsilon} X_1$ for $\epsilon<0$. $\endgroup$ – Ian Jul 12 '15 at 1:12
  • $\begingroup$ I meant $\gamma \neq 1$. For $\gamma > 1$ what I said is still true. $\endgroup$ – dsaxton Jul 12 '15 at 1:20
  • $\begingroup$ Can you explain why? That makes very little sense to me, since the absolute value of the Cauchy distribution "just barely" has infinite expectation ($\int_{-M}^M \frac{|x|}{1+x^2} dx = \ln(1+M^2)$.) $\endgroup$ – Ian Jul 12 '15 at 1:23
  • $\begingroup$ So to me that means that $\sum_{n=1}^N |X_n|$ should be on the order of $N \log(N)$ for large $N$. This is consistent with some quick simulations I just did. If that's correct, then by the triangle inequality $n^{-\gamma} S_n$ is at most on the order of $n^{1-\gamma} \log(n)$, which goes to zero for $\gamma>1$. $\endgroup$ – Ian Jul 12 '15 at 1:37
  • $\begingroup$ Ok, I see what you mean, let me fix my answer. $\endgroup$ – dsaxton Jul 12 '15 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.