I came across a question regarding the WLLN. Suppose for $X \geq 0$ , $\mathbb{E}[X] = \infty $ , $S_n = \sum_{i \leq n} X_i$, $X_i$ are iid copies of $X$ , and

$\frac{\mathbb{E}[X \mathbf{1} _{X \leq s} ] }{s(1 - F_X(s))} $ $\rightarrow$ $\infty$ .

Let $\mu(s) = \mathbb{E}[X \mathbf{1} _{X \leq s} ] $ , then for $n$ large enough we can choose $b_n$ such that $n\mu(b_n) = b_n$ .

I am stuck at the existence of $b_n$. Basically, the problem goes onto to ask us to prove that $\frac{S_n}{b_n} \rightarrow 1$ , which I am able to prove if I assume the existence of $b_n$.


1 Answer 1


For every $s\gt0$, let $\nu(s)=\mu(s)/s+P(X\gt s)$, then $\nu(s)=E(u_s(X))$ where, for every $x\geqslant0$, $u_s(x)=\min\{1,x/s\}$. Assume that $P(X=0)\ne1$. Then, for every $x\geqslant0$, $u_s(x)\leqslant1$, $u_s(x)\leqslant u_t(x)$ for every $s\geqslant t$, $u_s(x)\to0$ when $s\to+\infty$, and, for every $x\gt0$, $u_s(x)\to P(X\gt0)$ when $s\to0$.

Thus, the function $\nu$ is nonincreasing and, by dominated convergence, $\nu(s)\to P(X\gt0)$ when $s\to0$ and $\nu(s)\to0$ when $s\to+\infty$.

Furthermore, for every $s\leqslant t$, $\|u_s-u_t\|_\infty=u_s(s)-u_t(s)=(t-s)/t$ hence the function $\nu$ is continuous.

If the distribution of $X$ is continuous, the function $s\mapsto\mu(s)/s=\nu(s)-P(X\gt s)$ is continuous, with limits $0$ when $s\to0$ and when $s\to+\infty$. Furthermore, there exists some $s^*\gt0$ such that $P(0\lt X\lt s^*)\gt0$, hence $\mu(s^*)\gt0$. Let $M=\mu(s^*)/s^*$, then, for every $x$ in $(0,M)$ there exists some $\beta(x)$ in $(s^*,+\infty)$ such that $\mu(\beta(x))/\beta(x)=x$.

For every $n$ large enough, $1/n\lt M$ hence $b_n=\beta(1/n)$ solves the question in this case.

If the distribution of $X$ has some atoms, the continuity argument used above fails (the function $\nu$ is still continuous but not the function $\mu$) and actually, one doubts that the result holds in full generality.

  • $\begingroup$ Hi, Thanks for the answer. I just need one clarification, isn't $E(u_s(X)) = \frac{\mu(s)}{s} + \mathbb{P}(X > s) $ ? I think the proof method still works though, that is $\frac{\mu(s)}{s} \rightarrow E(u_s(X)) $ as $s \rightarrow \infty$ and it tends to $\mathbb{P}(X > 0)$ as s goes to 0. $\endgroup$
    – rajatsen91
    Oct 15, 2014 at 16:19
  • $\begingroup$ Excellent remark, the previous version was faulty, sorry about that. See revised version, and the caveat at the end about the result itself. $\endgroup$
    – Did
    Oct 15, 2014 at 16:48
  • $\begingroup$ Yes, the continuity assumption is needed otherwise the exact equality will not hold I presume. However, we can take $b_n = inf \lbrace x : n.\mu(x) < x \rbrace $, and the argument would still work order-wise I guess. However, I am not sure about this. Thanks, for the detailed and nice answer. $\endgroup$
    – rajatsen91
    Oct 15, 2014 at 20:35

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.