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Suppose $X_1, \ldots, X_n$ are iid standard Cauchy random variables, does $\frac{1}{n}\sum_{i=1}^{n}X_i$ converge in probability or almost surely?

I know that by stable law, $\frac{1}{n}\sum_{i=1}^{n}X_i \sim Cauchy(0,1)$, but how can I find if it converges in probability or not by the definition of convergence in probability?

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2 Answers 2

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It's pretty clear that it can't converge (in either sense) to a number, since it always has the Cauchy(0,1) distribution. To see that it can't converge to anything, observe that $$ \frac{S_{2n}}{2n}-\frac{S_n}{n} = \frac{1}{2}\frac{S_{2n}-S_n}{n} -\frac{1}{2}\frac{S_n}{n}.$$ where $S_n = \sum_{k=1}^n X_k.$ $(S_{2n}-S_{n})/n$ and $S_n/n$ are independent Cauchys, so their difference is stable and the distance between $S_{2n}/2n$ and $S_n/n$ cannot go to zero as $n$ gets large.

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  • $\begingroup$ Thanks, I am curious about two things. 1) What is it about the Cauchy(0,1) distribution that it cant converge to a number? 2) What is it about a stable law such that it can't go to zero? Thanks!!!! $\endgroup$
    – user321627
    Jan 13, 2017 at 4:45
  • $\begingroup$ @user321627 First I should say I'm the upvote on the other answer which is essentially the same as mine but in better detail. By 'a number' I meant a fixed number (as opposed to a random variable). If its limiting distribution is Cauchy(0,1) then there's no fixed number that it could be going toward, since it has density everywhere. If it converged to a number it would converge in distribution to a number (i.e. a distribution where $P(X=\mu) = 1$ for some number $\mu.$) $\endgroup$ Jan 13, 2017 at 5:07
  • $\begingroup$ @user321627 By 'stable' I just meant the difference has the same distribution regardless of how large $n$ gets. So it can't obey the Cauchy sequence version of convergence in probability, which the other answerer shows means it can't converge in probability (same proof as convergence => cauchy sequence for non-random sequences) (Sorry there's two unrelated uses of 'cauchy' in this problem) $\endgroup$ Jan 13, 2017 at 5:12
  • $\begingroup$ @user321627 To the answer to the first question, I should add - in case it's still confusing- that there's nothing special about the cauchy distribution that means the sequence can't converge to a number. The same would be true if the limiting distribution were normal or any distribution that's not concentrated entirely on a single value. $\endgroup$ Jan 13, 2017 at 5:17
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Let $Z_n = \frac{1}{n} \sum_{i=1}^n X_i$.

Now $$Z_{2n} - Z_{n} = \underset{A_n}{\underbrace{-\frac{1}{2n} \sum_{i=1}^n X_i}} +\underset{B_n}{\underbrace{\frac{1}{2n} \sum_{i=n+1}^{2n} X_i}}.$$

Then $A_n$ and $B_n$ are independent identically distributed Cauchy $(0,\frac 12)$. Hence, $A_n + B_n $ is Cauchy $(0,1)$.

In particular,

$$P(|Z_n -Z_{2n}|>\epsilon )=c,$$

where $c$ is the probability that Cauchy $(0,1)$ has absolute value larger than $\epsilon$ (explicitly, $c= \frac{2}{\pi} \arctan\epsilon$). This is a constant independent of $n$.

From this we conclude that the sequence $(Z_n)$ does not converge in probability and therefore also not a.s. It is Cauchy distributed, but unfortunately, it is not a Cauchy sequence is probability...

Indeed, for any random variable $Z$ we have

$$|Z_n - Z_{2n}|= |Z_n -Z+ Z-Z_{2n}|\le |Z_n - Z| + |Z_{2n}-Z|.$$

Therefore, the event $\{|Z_n-Z_{2n}|>\epsilon\}$ is contained in the event $\{|Z_n -z|+|Z_{2n}-Z|>\epsilon\}$. But the latter event is contained in the event $\{|Z_n-Z|>\epsilon/2\}\cup \{|Z_{2n}-Z|>\epsilon/2\}$. Or:

$$c = P(|Z_n-Z_{2n}|>\epsilon) \le P(|Z_n -Z|>\epsilon/2)+P(|Z_{2n}-Z|>\epsilon/2).$$

Since the righthand side is bounded below by $c$, it follows that $Z$ is not the limit in probability of $(Z_n)$ (if it were, both summands on RHS would go to zero as $n\to\infty$).

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