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

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

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.

• I made the edit. Let me know if it is clear! Jun 7, 2020 at 9:04
• Can I ask what your definition of $X_n \to X$ in distribution is? Do you know that this means that $E(f(X_n)) \to E(f(X))$ for all continuous bounded functions? Jun 7, 2020 at 9:05
• This is the formula of the image measure. Suppose we have a measurable map between measurable spaces $T:(\Omega, \mathcal{F}) \to (\Omega', \mathcal{F}')$, and $\mu$ is a measure on $\mathcal{F}$ (here $\mu = \mathbb{P}, X= T, \Omega' = \mathbb{R}$), then we have an image measure $\mu_T$ on $\mathcal{F}'$ defined by $\mu_T(A) := \mu(T^{-1}(A))$ and the following holds: $\int_\Omega f \circ T d \mu = \int_{\Omega'} f d \mu_T$ for a measurable function $f: \Omega' \to \mathbb{R}$, in the sense that one side exists if and only the other one exists and then they are equal. Jun 7, 2020 at 9:16
• As always, glad to help! Jun 7, 2020 at 9:19
• If I may be honest, some of your questions seem to be more real analysis questions than measure theory questions (for example the $a_n^2 \to 0 \implies a_n \to 0$ one), so maybe start with that. If you feel confident with that, then you might have a look at: Billingsley "Probability and measure" (develops probability and measure simultaneously), Axler's "Measure theory" (freely available on Axler's homepage, book that looks like it will be a quick read) and more advanced and my personal favorite Cohn's "Measure theory". The last two books do not really have a probability orientation though. Jun 7, 2020 at 9:28