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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 $b_n$.

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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.

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  • $\begingroup$ 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

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