Why $\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$?

Question

Consider the following theorem regarding convergence in distribution

THEOREM 1: Let $(X_n)_{n\geq1}$, $X$ be $\mathbb{R}^d$-valued random variables. Then $X_n$ converges to $X$ in distribution if and only if $\lim\limits_{n\to\infty}E\{f(X_n)\}=E\{f(X)\}$ for all continuous, bounded functions $f$ on $\mathbb{R}^d$.

Then consider the following theorem and its proof

THEOREM 2 Let $(X_n)_{n\geq1}$, $X$ be defined on a given fixed probability space $(\Omega\text{, }\mathcal{A}\text{, }\mathbb{P})$. If $X_n$ converges to $X$ in distribution, and if $X$ is a r.v. a.s. equal to a constant, then $X_n$ converges to $X$ in probability as well.

PROOF: Suppose that $X$ is a.s. equal to a constant $a$ (that is $\mathbb{P}(X=a)=1)$. The function $f(x)=\frac{|x-a|}{1+|x-a|}$ is bounded and continous. Therefore, $\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$, hence $X_n$ converges to $a$ in probability by a result assuring that.

I have a doubt about the proof.
In particular, since $X_n$ converges to $X$ in distribution, given THEOREM 2, $\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=E\Big\{\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}\Big\}$. At this point, since in the proof it is directly stated that "$\lim\limits_{n\to\infty}E\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$", I guess it holds that $\lim\limits_{n\to\infty}\Big\{\frac{|X_n-a|}{1+|X_n-a|}\Big\}=0$, but I cannot figure out why this is true.


So, the question is: how can I show that $$\lim\limits_{n\to\infty}\Bigg\{\frac{|X_n-a|}{1+|X_n-a|}\Bigg\}=0$$?

Answer 1

We have $X_n \stackrel{d}\to a$ (convergence in distribution). Thus $$\lim_n\mathbb{E}(f(X_n)) = \mathbb{E}(f(a))= f(a)$$ for every bounded continuous function $f$.

With your choice of $f$, the result immediately follows.

The following result/definition was used:

$$X_n \stackrel{d}\to X$$ $$\iff$$ $$\mathbb{P}_{X_n} \stackrel{w}\to \mathbb{P}_X$$ $$\iff$$ $$\int fd\mathbb{P}_{X_n }= \int f(X_n) d \mathbb{P}=\mathbb{E}(f(X_n))\to \mathbb{E}(f(X)) = \int f(X) d \mathbb{P} = \int f d\mathbb{P}_X$$ for all $f \in C_b$.

Your claim that $$\lim_n \frac{|X_n-a|}{1+|X_n-a|}=0$$ need not be true.