Prove or disprove statement about convergence of random variables 
Prove or disprove the following statement:
\begin{align*}
\text { If } X_{N} \stackrel{d}{\rightarrow} X \text { then } \frac{X_{N}}{\sqrt{N}} \stackrel{P}{\rightarrow} 0
\end{align*}
where $\stackrel{d}{\rightarrow}$ is convergence in distribution and $\stackrel{P}{\rightarrow}$ is convergence in probability.


Intuitively, if the sequence $X_1, \ldots, X_n$ is converging to a fixed distribution as $n \to \infty$, then dividing that random variable by larger and larger values of $\sqrt{n}$ should converge to zero. So I think that statement is true.
However, I feel like I don't have the mathematical language in probability theory to express this precisely.
Could someone help me with this problem?
 A: $\newcommand{\pr}{\mathbb P}$Let's formalize your intuition step by step. Consider this more an exercise in formalization of intuition, rather than the formation of the intuition itself (which seems to be sound). I'm going to first present a wrong way of doing things, and then the correct way of doing things, because I know students who, way too often, use the wrong way, get stuck and/or make wrong conclusions and don't get credit.

Your idea is the following (copied verbatim) :

Intuitively, if the sequence $X_1\ldots X_n$ is converging to a fixed distribution as $n\to\infty$, then dividing that random variable by larger and larger values of $n$ should converge to zero. So I think that statement is true.

Think about this in terms of resemblance. $\frac{X_n}{\sqrt n}$ looks like $\frac{X}{\sqrt n}$ for large $n$, which looks like $0$ for large $n$. So we expect $\frac{X_n}{\sqrt n}$ to look like $0$ for large $n$.
Now, the problem with this logic (and yes, there is!) is that one could be tempted to write down the following very natural looking equality (I'm mentioning this because I cannot tell you HOW many people have made this mistake from thinking merely intuitively, and I want to address the elephant in the room) :
$$
\frac{X_n}{\sqrt n} = \frac{X_n -X}{\sqrt n} + \frac{X}{\sqrt n}
$$
and then say something of the form : "both terms on RHS are usually small, so we can use the triangle inequality".
But there's trouble in this paradise : indeed, even if $X_n \to X$ in distribution, it is not true that $X_n -X \to 0$ in probability, or in fact even in distribution. That's basically because subtraction of potentially dependent random variables can have a devastating effect on the distribution functions.
In some sense, this explains the "intuition-mathematics" gap well, since the notion of what it means for two things to be close, actually fails in one particular kind of argument. Ideally speaking, in arguments regarding distribution it's extremely rare that you can add and subtract random variables from one another and have any structure remaining(unless random variables are degenerate , see Slutsky's theorem for example).
The basic reason is that when I look at, say $X_n-X$, I'm looking at it as the function $X_n(\omega) - X(\omega)$ for $\omega$ an element of the sample space. However, unlike convergence a.s. or convergence in probability, I don't actually get an almost sure $\omega$-by-$\omega$ statement anywhere in the definition of convergence in distribution. That's why I need to stop adding and subtracting random variables, and focus on mere probability comparison.
Example : Let $X_i$ be iid any standard random variable, like the Poisson r.v. or the uniform random variable (suggested by E-A below, and the idea of the example was from Bryden so I thank them as well) and $X$ be independent of the $X_i$ and have the same distribution. Now, $X_i$ converges to $X$ in distribution : this is obvious, because *the probabilities that $X_i,X$ lie in certain regions are close enough (I mean, they're literally the same) for all $i$, let alone just large enough $i$. However, because $X_i,X$ are independent, they are created in such a way that $X_n(\omega),X(\omega)$ will really look quite different. For example, if $X,Y$ are iid $N(0,1)$ then $X-Y$ is iid $N(0,2)$! The basic reason is that although we've linked $X_i$ and $X$ via a distribution, we haven't linked them in an $\omega$-to-$\omega$ way, or a way that allows one to depend upon the other w.r.t. the sample space elements themselves. So of course $X_i-X$ won't go in distribution to $0$, although $X_i$ has the same distribution as $X$.
Which means, that I'm going to have to say intuitively : "It's not $X_n$ and $X$ which are close, but rather the probabilities of $X_n$ and $X$ lying in certain regions, which are close". Note the subtlety and beauty of the statement, and how it applies to this particular situation.
So next time, when you say : "things are converging to a fixed distribution" and stuff like that, make sure you have additional intuition to make the argument the way it has to be made.

Now, $X_n \to X$ in distribution, means that for any $k$ (at which $F_X$ is continuous) we have $\pr(X_n \leq k) \to \pr(X \leq k)$. That is the kind of closeness we want to exploit.
Let's now look at the bad event, $\pr\left\{|X_n|>\epsilon\sqrt n\right\}$. Without having to look at $X_n -X$ , we can comfortably say this : This guy resembles $\pr\left\{|X| > \epsilon \sqrt n\right\}$ and that resembles $0$ for large enough $n$.
The point is, the resemblance is between the probabilities : thanks to convergence in distribution, we know this is the case, but because $\epsilon\sqrt n$ could be a point of discontinuity, we need to be slightly careful with our argument.
Well, we can prove the following statement fairly easily :

There exist sequences $x_i \to \infty$ and $y_i \to -\infty$ with $x_i ,y_i \in C(F_X)$ for all $i$.

What that means, is that I can go as far up and down the number line I want and still find continuity points of $F_X$. To prove this, if for the sake of contradiction there wasn't a continuity point beyond some $N$, then every point beyond $N$ is a point of discontinuity, which is a contradiction since that creates an uncountable set of discontinuities for a monotone function (which $F_X$ is), which can have only countably many discontinuities. (Do the same with $-N$ for the other side).
In particular, we know that $F_X(x_i) \to 1$ as $i \to \infty$ and $F_X(y_i) \to 0$ as $i \to \infty$. Furthermore, $F_{X_N}(x_i) \to F_{X}(x_i)$ for all $i$ (and similarly for $y_i$).
The advantage of this , is that even if I don't know about $F_X$ at a point, I can use monotonicity to ensure that it lies above/below some threshold by using these $x_i,y_i$ as reference points.
With that in mind, we are going to do the obvious , and write :
$$
\pr\{|X_n| > \epsilon \sqrt n\} = \pr\{X_n > \epsilon \sqrt n\} + \pr\{X_n < - \epsilon \sqrt n\} \leq 1 - F_{X_n}(\epsilon \sqrt n) + F_{X_n}(-\epsilon \sqrt n)
$$
(Note that $\pr\{X_n < - \epsilon \sqrt n\} \leq F_{X_n}(-\epsilon \sqrt n)$ but doesn't equal it necessarily) Let's mathematically argue our way out of this now. There's an order, a way of choosing things so that the mathematics is made easier. We know we have to bring in the following :

*

*$x_i,y_i$ to replace the $\epsilon\sqrt n$ terms somewhere.


*$X_n$ to be replaced by $X$ somewhere.
The point is , which to do first? Let's think about it this way : we can't replace $X_n$ by $X$ unless we know the points at which the replacements are to be made! So the order is clear : choose the points first, then do the replacement.
Now let's choose the $x_i,y_i$. Fix a $\delta > 0$ which we will use for deciding the $x_i,y_i$ carefully.
Find $N$ such that $1-F_X(x_N) > \delta , F_X(y_N)<\delta$. This can be done using the limit conditions and the property of distribution functions at $\pm \infty$.
With that, where were we? Oh yes, we were here :
$$
\pr\{|X_n| > \epsilon \sqrt n\} = \pr\{X_n > \epsilon \sqrt n\} + \pr\{X_n < - \epsilon \sqrt n\} \leq 1 - F_{X_n}(\epsilon \sqrt n) + F_{X_n}(-\epsilon \sqrt n)
$$
Since $\epsilon \sqrt n\to \infty$ as $n \to \infty$, let's consider $n>N_0$ large enough , such that $\epsilon \sqrt n  > x_N$ and $-\epsilon \sqrt n < y_N$ for $n>N_0$. We chose this to ensure that:
$$
1 - F_{X_n}(\epsilon \sqrt n) + F_{X_n}(-\epsilon \sqrt n) \leq 1 - F_{X_n}(x_N) + F_{X_n}(y_N)
$$
Now, because $N$ doesn't depend upon $n$ (whereas the earlier term inside the brackets very much did), we are in pole position to drag $n$ to infinity, in the inequality:
$$
0 \leq \pr\{|X_n| > \epsilon \sqrt n\} \leq 1 - F_{X_n}(x_N) + F_{X_n}(y_N)
$$
except there's a small problem : the RHS has a limit, but the LHS does not! So one cannot use monotonicity of limits directly. Instead, one uses the monotonicity of the limit superior and limit inferior, which are stronger notions than that of a limit which coincides with it for convergent sequences. Using the first notion , which is also monotonic, gives :
$$
0 \leq \limsup_n \pr\{|X_n| > \epsilon \sqrt n\} \leq \limsup_n 1 - F_{X_n}(x_N) + F_{X_n}(y_N) = 1 - F_X(x_n) + F_{X}(y_n) < 2\delta 
$$
for all $\delta>0$. In particular, we must have $\limsup_n \pr\{|X_n| > \epsilon \sqrt n\}  = 0$. It is easy to see from monotonicity for the $\liminf$ that $\liminf 0 \leq \liminf \pr\{|X_n| > \epsilon \sqrt n\}$. In particular, the $\limsup$ and $\liminf$ for the given sequence coincide and equal $0$ : it's easy to prove that the limit is , in fact, zero, from this.
This, of course, concludes $\frac{X_n}{\sqrt n} \to 0$ in probability.

So the key idea while formalizing intuition into argument is to go further into your investigation and see what key object needs approximations, what subapproximations have to be made, and what order they are to be made in. Thinking linearly and in an orderly fashion can be greatly beneficial while tackling simple problems.
A: $\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}$Consider a fixed $ε_0 > 0$. For any $ε > 0$, since $X_n \stackrel{\mathrm{d}}{→} X$, there exist $A_1$ and $A_2$ such that$$
P(X_n \leqslant A_k) → P(X \leqslant A_k) \quad (n → ∞,\ k = 1, 2)
$$
and $P(X \leqslant A_1) < ε$, $P(X \leqslant A_2) > 1 - ε$. Note that for $n \geqslant \dfrac{1}{ε_0^2} \max(A_1^2, A_2^2)$,\begin{align*}
P\paren{ \abs{ \frac{X_n}{\sqrt{n}} } > ε_0 } &\leqslant P(X_n > ε_0\sqrt{n}) + P(X_n < -ε_0\sqrt{n})\\
&\leqslant P(X_n > A_2) + P(X_n < A_1) < 2ε.
\end{align*}
Thus $\varlimsup\limits_{n → ∞} P\paren{ \abs{ \dfrac{X_n}{\sqrt{n}} } > ε_0 } \leqslant 2ε$, and making $ε → 0+$ yields $\lim\limits_{n → ∞} P\paren{ \abs{ \dfrac{X_n}{\sqrt{n}} } > ε_0 } = 0$. Therefore, $\dfrac{X_n}{\sqrt{n}} \stackrel{P}{→} 0$.
A: I would like to repeat two standard lemmas which are apparently learnt by heart by most of probabilists. These two will help you to translate better your intuition into probabilistic language.
Lemma 1 If $X_n \xrightarrow[]{(d)}X$ and $ Y_n  \xrightarrow[]{(d)}c $ for some constant $c$ then $(X_n,Y_n) \xrightarrow[]{(d)} (X,c)$.
Lemma 2 If $X_n \xrightarrow[]{(d)} c $  for some constant $c$ then $ X_n  \xrightarrow[]{P}c $
Back to your question, applying lemma 1 above, you can see that:
$$g(X_N,\frac{1}{\sqrt{N}}) \xrightarrow[]{(d)} g(X,0)$$
For any continuous function $g \in \mathcal{C}( \mathbb{R}^2,\mathbb{R})$.
Choosing $g(x,y)= \frac{x}{\max(y,1)}$, we obtain
$$\frac{X_N}{\sqrt{N}} \xrightarrow[]{(d)} 0$$
Using lemma 2, we imply the desired result.
