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.

  • $\begingroup$ 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. $\endgroup$
    – guy
    Dec 29, 2012 at 6:15
  • $\begingroup$ 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? $\endgroup$
    – august
    Jan 3, 2013 at 17:14
  • $\begingroup$ 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. $\endgroup$
    – guy
    Jan 3, 2013 at 20:31

1 Answer 1


I'm having trouble with this exercise too. Someone on mse pointed me to the book of Feller where both the truncation and characteristic function proofs of the general weak law are given. Below is what I understand of the truncation argument for exercise $3.6$. I couldn't understand the characteristic function derivative argument. Can OP or someone else please write that down as well? Perhaps the connection between both methods will then become clear.

Let $\varepsilon>0$, and consider $$P\left\lbrace \omega : \left| \frac{S_n(\omega)}{n} \right| > \varepsilon\right\rbrace = P\left\lbrace \omega : \left| {S_n(\omega)} \right| > n\varepsilon\right\rbrace. $$ Let $X_i'$ be the truncation at $n$ and let $S_n'=\sum_1^n X_i'$. We can decompose the probability above into the two cases when $S_n=S_n'$ and otherwise. This latter case is a subset of the event when some $X_i \neq X_i'$. Using this and subadditivity, we estimate $$P\left\lbrace \omega : \left| {S_n(\omega)} \right| > n\varepsilon\right\rbrace \leq P\left\lbrace \omega : |S_n'(\omega)| > n\varepsilon\right\rbrace + nP\lbrace \omega : X_1(\omega) \neq X_1'(\omega)\rbrace. $$ Applying Chebyshev's inequality to the first term and rewriting the second as an integral, we see $$P\left\lbrace \omega : \left| {S_n(\omega)} \right| > n\varepsilon\right\rbrace \leq \frac{1}{n^2\varepsilon^2}\int \left( S_n'(\omega)\right)^2dP(\omega) + n \int \mathbb{1}_{[-n,n]^c}(X_1(\omega)) dP(\omega). $$ Using the independence of the $X_i$ (and hence the $X_i'$) and writing $\alpha$ to be the pushforward of the measure $P$, $$P\left\lbrace \omega : \left| {S_n(\omega)} \right| > n\varepsilon\right\rbrace \leq \frac{1}{n\varepsilon^2} \int (X_1'(\omega))^2dP(\omega) + n\int_{|x|>n} d\alpha(x) = \frac{1}{n\varepsilon^2}\int^{n}_{-n} x^2 d\alpha(x) + n\int_{|x|>n} d\alpha(x). $$ In the first inequality above we have used the fact that our distribution $d\alpha$ has density $\frac{c}{\left(1+x^2\right)\log\left(1+x^2\right)}$ (for $|x|>1$, say) so that the truncated distributions $X_i'$ have mean $0$. Now if we write $F(\cdot)$ for the cumulative density function of $d\alpha$, and write $\tau(n)$ for the tail estimate $n[1-F(n) + F(-n)] = n\int_{|x|>n}d\alpha$ then integration by parts shows that $$\frac{1}{n}\int_{-n}^n x^2d\alpha(x) = -\tau(n) + \frac{2}{n}\int_0^{n} \tau(x)dx $$ so that this term goes to $0$ as $n\to \infty$ if $\tau(t) \to 0$ as $t\to \infty$. Thus to show the weak law holds it remains to show that $\tau(t) \to 0$ for the specific distribution in exercise $3.6$. We compute by a change of variables $$t\int_{|x|>t} \frac{c}{(1+x^2)\log(1+x^2)}dx = 2t \int_{x>t} \frac{c}{(1+x^2)\log(1+x^2)}dx = 2 \int_{y>1}\frac{ct^2}{(1+t^2y^2)\log(1+t^2y^2)}dy $$ which goes to $0$ by the dominated convergence theorem.

Edit: Something common to both the differentiability of the characteristic function and to the estimates in the truncation argument is the quantitiy $$\tau(t) = t[1-F(t)+F(-t)]. $$ I found the paper (5 pages) On the derivatives of a characteristic function at the origin by E.J.G. Pitman very useful.


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