$E(h(X_n)) \to E(h(X)).$

Question

Let $X,X_n(n=1,2,\ldots)$ random variable such that $X_n \xrightarrow{a.s.} X$, $E(|X_n|) \leq K < \infty$ $(n=1,2,\ldots)$ and let $h : \mathbb{R} \to \mathbb{R}$ continuos function such that $$\left|\frac{h(x)}{x}\right| \to 0 \hspace{1cm} \text{when $|x| \to \infty$}.$$ Prove that
$$E(h(X_n)) \to E(h(X)).$$

I have some problems with this statement. First of all I have to prove that for $n$ sufficiently large for all $\epsilon >0$

$$|E(h(X_n)) - E(h(X))| \leq \epsilon.$$

Then

\begin{align*} |E(h(X_n)) - E(h(X))| &= |E(h(X_n)) - E(h(X))1_{|X_n-X|\leq \delta}|+ |E(h(X_n)) - E(h(X))1_{|X_n-X|>\delta}|\\ \end{align*}

The first part I think I could bound it since $X_n \to X$ almost sure. But I could not advance from this nor apply the other hypothesis.

Any hint? Thanks in advance.

Answer 1

(Completed solution)

Split differently. Observe that you have also $E[|X|] \le K$ by Fatou's lemma.

Fix $\epsilon>0$.

Last assumption on $h$ yields a real number $R \ge 1$ such that $|h(x)| \le \epsilon|x|/K$ whenever $|x| \ge R$.

Then, uniform continuity of the restriction of $h$ on $[-R+1,R+1]$ yields a positive real number $\alpha \le 1$ such that $|h(x)-h(y)| \le \epsilon$ whenever $|x|,|y| \le R$ and $|x-y| \le \alpha$. And it yields also a bound $M = \sup\{|h(x)| : x \in [-R+1,R+1]\}$.

The event $[|X_n-x| > \alpha]$ has small probability. On this event, you use $$|h(X_n)-h(X)| \le |h(X_n)|+|h(X)|$$ and the upper bounds on $|h|$ (one is valid on $[-R-1,R+1]$, the other on is valid out of $(-R,R)$.

On the event $[|X_n-X| \le \alpha]$, since $\alpha<1$, you know that both $X_n$ and $X$ are in $[-R-1,R+1]$ (here you use uniform continuity) or both of them are out of $(-R,R)$ (here, you use the upper bound on $h$).

Let us put the pieces together \begin{eqnarray*} |h(X_n)-h(X)| &\le& \epsilon \mathbf{1}_{[|X_n-X| \le \alpha~;~|X_n| \le R+1~;~|X| \le R+1]} \\ & & \quad + \big(|h(X_n)|+|h(X)|\big)~\mathbf{1}_{[|X_n-X| \le \alpha~;~|X_n| \ge R~;~|X| \ge R]} \\ & & \quad + \big(|h(X_n)|+|h(X)|\big)~\mathbf{1}_{[|X_n-X|>\alpha]} \\ &\le& \epsilon + \frac{\epsilon}{K}\big(|X_n|+|X|\big)\mathbf{1}_{[|X_n-X| \le \alpha} \\ & & \quad + \Big(2M+ \frac{\epsilon}{K}(|X_n|+|X|)\Big)~\mathbf{1}_{[|X_n-X| > \alpha]} \\ &\le& \epsilon + \frac{\epsilon}{K}\big(|X_n|+|X|\big) + 2M~\mathbf{1}_{[|X_n-X| > \alpha]} \end{eqnarray*} Taking expectations yields $$E[|h(X_n)-h(X)|] \le 3\epsilon + 2MP[|X_n-X| > \alpha].$$ Hence, since $X_n \to X$ in probability, $$\limsup_{n \to +\infty} E[|h(X_n)-h(X)|] \le 3\epsilon.$$ Since $\epsilon>0$ is arbitrary, the limsup above is $0$, hence $h(X_n) \to h(X)$ in $L^1(P)$.

Answer 2

Let $Y_n=h\left(X_n\right)$. Continuity of $h$ combined with $X_n\to X$ a.s. shows that $Y_n\to h(X)=:Y$ almost surely. In order to get the convergence of the expectations, it suffices to show that $\left(Y_n\right)$ is uniformly integrable. To do so, define for $x>0$ the quantity $\tau(x):=\sup_{t:\lvert t\rvert\geqslant x}\lvert h(t)/t\rvert$. By assumption, $\tau(x)\to 0$ as $x$ goes to infinity. We have $$ \mathbb E\left[\lvert Y_n\rvert \mathbf{1}_{\{\lvert Y_n\rvert>R\}}\right] =\mathbb E\left[\lvert h(X_n)\rvert \mathbf{1}_{\{\lvert h(X_n)\rvert>R\}}\right] =\mathbb E\left[\lvert h(X_n)\rvert \mathbf{1}_{\{\lvert h(X_n)\rvert>R\}} \mathbf{1}_{\{\lvert X_n\rvert\leqslant x\}}\right]+\mathbb E\left[\lvert h(X_n)\rvert \mathbf{1}_{\{\lvert h(X_n)\rvert>R\}} \mathbf{1}_{\{\lvert X_n\rvert\gt x\}}\right]. $$ Since $$ \lvert h(X_n)\rvert \mathbf{1}_{\{\lvert h(X_n)\rvert>R\}} \mathbf{1}_{\{\lvert X_n\rvert\gt x\}}\leqslant \lvert X_n\rvert \tau(x) $$ and letting $M(x):=\sup_{s\in[-x,x] }\lvert h(s)\rvert$ $$ \lvert h(X_n)\rvert \mathbf{1}_{\{\lvert h(X_n)\rvert>R\}} \mathbf{1}_{\{\lvert X_n\rvert\leqslant x\}} \leqslant M(x)\mathbf{1}_{R<\lvert h(X_n)\rvert\leqslant M(x)} \leqslant M(x)\mathbf{1}_{R\lt M(x)}, $$ we get that for each $R,x>0$ and $n\geqslant 1$, $$ \mathbb E\left[\lvert Y_n\rvert \mathbf{1}_{\{\lvert Y_n\rvert>R\}}\right] \leqslant M(x)\mathbf{1}_{R\lt M(x)}+\sup_{k\geqslant 1}\mathbb E\left[\lvert X_k\rvert\right]\tau(x). $$ Therefore, $$ \limsup_{R\to\infty}\sup_{n\geqslant 1}\mathbb E\left[\lvert Y_n\rvert \mathbf{1}_{\{\lvert Y_n\rvert>R\}}\right] \leqslant \sup_{k\geqslant 1}\mathbb E\left[\lvert X_k\rvert\right]\tau(x), $$ proving uniform integrability.