Show that if $a_n(X_n-X) \overset{\mathcal{D}}\to Z$ then $X_n \overset{P}\to X$.

Question

Let $(a_n)\subseteq \Bbb{R}$ be a sequence such that $a_n \to \infty$. Let $(X_n)$ be a sequence of random variables such that $a_n(X_n-X) \overset{\mathcal{D}}\to Z$ fore some random variables $X$ and $Z$.

Show that $X_n \overset{P}\to X$.

So I have some sort of Idea why should this happen. Since $a_n \to \infty$ for $a_n(X_n - X)$ to have a "finite" limit, say $Z$, then $X_n - X \to 0$ somehow. (The excercise says that the convergence must be in probability.) But Im having a lot of trouble in making this ideas more precise. I've tried with characteristic functions for the convergence in distribution. Got to nothing. Any hint?

Answer 1

Let $Z_n=a_n(X_n-X)$. We are told that $\frac{1}{a_n}\to 0$ so, in particular, $\frac{1}{a_n}\overset{P}{\to}0$. So by Slutsky's Theorem, $X_n-X=\frac{1}{a_n}\times Z_n\overset{d}{\to}0\times Z=0$. But convergence in distribution to constant is equivalent to convergence in probability to that same constant, so we conclude that $X_n-X\overset{P}{\to}0$.

Answer 2

$a_n(X_n-X)\stackrel{\mathcal{D}}{\to}Z$ implies that $\mathbb{P}\left(|Z|<\infty\right)=1$ - $Z$ should be a proper random variable.

If the claim would not hold for some $\varepsilon$, then: $$ \exists\varepsilon>0: \lim_{n\to\infty}\mathbb{P}\left(|X_n-X|>\varepsilon\right)\neq 0\Rightarrow \limsup_{n}\mathbb{P}\left(|X_n-X|>\varepsilon\right)=\delta(\varepsilon)>0 $$ Thus, $\exists\sigma:\mathbb{N}\to\mathbb{N}: \sigma(i)<\sigma(i+1)$ and $$ \mathbb{P}\left(|X_{\sigma(i)}-X|>\varepsilon\right)=\delta(\varepsilon)\left(1-\frac{1}{i}\right) $$ Since $a_n\to\infty$, thus $a_{\sigma(n)}\to\infty$ as well. Choose $\theta:\mathbb{N}\to\mathbb{N}: a_{\theta(\sigma(n))}<a_{\theta(\sigma(n+1))}$ (an inscreasing subsequence of $a_{\sigma(n)}$ - this must exist, as $a_n\to\infty$. $$ \mathbb{P}\left(|a_{\theta(\sigma(n))}||X_{\theta(\sigma(n))}-X|>|a_{\theta(\sigma(n))}|\varepsilon\right)>\delta(\varepsilon)\left(1-\frac{1}{\theta(n)}\right)\geq \delta(\varepsilon)\left(1-\frac{1}{n}\right) $$ Since $a_n(X_n-X)$ converges to $Z$ in distribution, so does any subsequence of it. Let $Y_n=a_{\theta(\sigma(n))}(X_{\theta(\sigma(n))}-X)$. Then by the previous claim, $$ F_{Y_n}(-|a_{\theta(\sigma(n))}|\varepsilon)+1-F_{Y_n}(|a_{\theta(\sigma(n))}|\varepsilon)>\delta(\varepsilon)\left(1-\frac{1}{n}\right)\geq\frac{\delta(\varepsilon)}{2} $$ Since $a_{n}\to \infty$ and $a_{\theta(\sigma(n))}$ is an increasing sequence, $\forall x\in\mathbb{R}\exists N_x\in\mathbb{N}:\forall n>N_x$ $$ |x|<\varepsilon|a_{\theta(\sigma(n))}|\Rightarrow $$ $$ \forall n>N_0: F_{Y_n}(-|x|)+1-F_{Y_n}(|x|)\geq \frac{\delta(\varepsilon)}{2} $$ But as $Z$ is a proper random variable, $\lim_{x\to-\infty}F_Z(x)=0$ and $\lim_{x\to\infty}F_Z(x)=1$, and $F$ is a non-decreasing function, thus: $$\forall \varepsilon'>0:\exists x(\varepsilon')\in\mathbb{R}: \forall x>|x(\varepsilon')|: F(-x)+1-F(x)<\varepsilon'$$ Now choose $\varepsilon'=\frac{\delta(\varepsilon)}{3}$, thus by the convergence in distribution: $$ \frac{\delta(\varepsilon)}{2}\leq\lim_{n\to\infty}F_{Y_n}(-|x(\delta(\varepsilon)/3)|)+1-F_{Y_n}(|x(\delta(\varepsilon)/3)|)\leq\frac{\delta(\varepsilon)}{3} $$ Which is a contradiction.