Limit of $E\left(\frac{X_1+\ldots+X_n}{\vert X_1\vert + \ldots + \vert X_n\vert}\right)$ for $(X_k)$ i.i.d. with $P(X_1=0)=0$

Question

Let $X$ be a real-valued random variable, and let $X_1, X_2, X_3,\dots$ be an infinite sequence of independent and identically distributed copies of $X$. Suppose that the first moment ${\Bbb E}|X|$ of $X$ is finite and $P(X_1=0)=0.$

Consider $S_n := \frac{1}{n}(X_1 + \ldots + X_n)$ the empirical averages and $T_n:=\frac{1}{n}(\vert X_1\vert + \ldots + \vert X_n\vert).$

I prove that the probability of $T_n$ being strictly positive is $1.$

We define $R_n(w):=\frac{S_n(w)}{T_n(w)}$ if $T_n(w)>0$ and $0$ if $T_n(w)\le 0$.

By the strong law of numbers we have $R_n\to \frac{{\Bbb E}X_1}{{\Bbb E}|X_1|}$ almost surely.

The question is to find the limit of ${\Bbb E}R_n$; I have no clue for this one.

Answer 1

Note that

$$|R_n| = \frac{|X_1+\ldots+X_n|}{|X_1|+\ldots+|X_n|} \leq \frac{|X_1|+\ldots+|X_n|}{|X_1| + \ldots + |X_n|}=1.$$

Since you already know that $$R_n \to \frac{\mathbb{E}(X_1)}{\mathbb{E}(|X_1|)} \qquad \text{almost surely},$$ the dominated convergence theorem yields $$\mathbb{E}(R_n) \to \frac{\mathbb{E}(X_1)}{\mathbb{E}(|X_1|)}.$$