$X_n\overset{p}{\rightarrow}0\Leftrightarrow E(g(X_n))\rightarrow0$ for $g(x)=\frac{x^2}{1+x^2}$

Question

If $g(x)=\frac{x^2}{1+x^2}$ then, show that $X_n\overset{p}{\rightarrow}0\Leftrightarrow E(g(X_n))\rightarrow0$.

I haven't been able to go too far with the proof. What I know is: $g(0)=0, |g(x)|\le\frac{1}{2}$ and $g(x)$ is continuous. We have a result that states for continuous $g$, if $X_n\overset{p}{\rightarrow}X$ then $g(X_n)\overset{p}{\rightarrow}g(X)$ and it follows for the "only-if" part that $g(X_n)\overset{p}{\rightarrow}0\Leftrightarrow P(|g(X_n)|>\epsilon)\rightarrow0$. How do I proceed from this? I'm pretty lost with the "if" part too.

I don't see much use for the fact that $g(x)$ is bounded as $E(g(X_n))\le\frac{1}{2}$ doesn't get me anywhere. To show that $E(g(X_n))\le\epsilon$ for large enough $n$ for any $\epsilon$, what can I possibly do?

Help would be appreciated!

Answer 1

A hint for the 'only-if' part: you can write for every $\epsilon$ $$E[g(X_n)] = E[g(X_n)I(g(X_n)>\epsilon)] + E[g(X_n)I(g(X_n)<\epsilon)] \le P(g(X_n)>\epsilon) + \epsilon$$ since $0\le g(x)\le 1$ for every $x$.

A hint for the 'if' part: Since $g$ is increasing, and $g(|x|)=g(x)$, we have for every $\epsilon$ $$P(|X_n|>\epsilon) \le P(g(|X_n|)>g(\epsilon))=P(g(X_n)>g(\epsilon)).$$

Answer 2

Direct part

First proof: Convergence in probability implies convergence in distribution and $g$ is a bounded continuous function. Hence, $Eg(X_n) \to Eg(0)=0$.

Second proof. DCT holds with convergence in probability in place of a.e. convergence. Since g(X_n) is dominated by $1$ the conclusion follows.

For the converse part use the fact that $g(X_n) \to 0$ in probability and $|X_n|>\epsilon$ implies $g(X_n) >\frac {\epsilon^{2}}{1+\epsilon^{2}}$.