up vote 6 down vote favorite
1

In Varadhan's lecture notes on Probability Theory (they are online here and on Amazon here) in Exercise 3.6 he writes:

The weak law may hold sometimes, even if the mean does not exist. If we dampen the tails of the Cauchy ever so slightly with a density $f(x) = \frac{c}{(1+x^2)\log(1+x^2)}$, show that the weak law of large numbers holds.

I get that the characteristic function here is differentiable at 0, hence we can get $\phi(\frac{t}{n})$ through the Taylor expansion. Since the characteristic function $\psi_n(t)$ of $\frac{S_n}{n}$ is given by $\phi_n(t) = [\psi(\frac{t}{n})]^n$, following the rest of the proof using characteristic functions in Varadhan's lecture notes shows us that $\frac{S_n}{n}$ converges in probability to $0$.

What I'm not sure about is how this translates to a truncation argument. I think my understanding of the various components of the truncation argument as they relate to properties of characteristic functions is very poor. Can someone elaborate as to the links between the two different proofs? What does it mean for a characteristic function to be differentiable at 0 in terms of what we can and cannot truncate?

As a followup, is it true then that if a characteristic function for the distribution of a random variable is differentiable at 0, then the weak law of large numbers holds? Or are there also other conditions that need to hold?

P.S. As a side question, is there a clean form (i.e. one without integrands) of the characteristic function for a r.v. with density $f(x) = \frac{c}{(1+x^2)\log(1+x^2)}$? I don't know any complex analysis, so the derivation of the characteristic function of the Cauchy distribution with density $\frac{1}{\pi(1+x^2)}$, $\phi(t) = e^{-|t|}$ flew over my head. I just want to know if the characteristic function for the density above could be derived through something like complex analysis as well, and if it would be in my interests to get a foundation in complex analysis for basic graduate probability theory.

share|cite|improve this question
  • I'm not sure if this is what you want, but it relates to truncation in the sense that you can get $\frac{S_n - nE(X_1 I(|X_1| \le n))}{n} \to 0$ in probability, provided that $nP(|X_1| \ge n) \to 0$ (which is just shy of being in $\mathscr L_1$); I think this is if-and-only-if but there might be another condition floating around. This is Feller's weak law. If $E(X_1 I(|X_1| \le n)) \to \mu$ then I'm guessing that should imply that $\varphi$ is differentiable at $0$. If only my notes were with me, I have a proof of Feller's weak law through the truncation approach. – guy Dec 29 '12 at 6:15
  • Let me see if I get this. So if $nP(|X_1| \geq n) \rightarrow 0$, then we can prove that $\frac{S_n - nE(X_1 I(|X_1| \leq n))}{n} \rightarrow 0$ because $P(X_i \neq X_i I(|X_i| \leq n)) \rightarrow 0$. Supposing $E(X_1 I(|X_1| \leq n)) \rightarrow \mu$ implies that $\int_{-n}^n xf(x) dx \rightarrow \mu$, which looks similar to $\phi'(t) = \int_{\mathbb{R}} ix e^{itx} f(x) dx$ but not exactly. What am I missing? – august Jan 3 '13 at 17:14
  • Maybe you could show that $\lim_{t \to 0} \int ixe^{itx} \ dF(x) = \lim_{n \to \infty} \int_{-n} ^ n x \ dF(x)$? I'm not sure if that's true, but it seems reasonable, and it seems like a nice explicit connection. The result I gave obviously implies that $\phi'(0) = \lim_{n \to \infty} \int_{-n} ^ n x \ dF(x)$ since Feller's result is if and only if. – guy Jan 3 '13 at 20:31

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.