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I am using Terry Tao's book on Random Matrix Theory. I came across Definition 1.1.28 on page 37 of the book. There, Tao defines various modes of convergence in decreasing order of strength. I am having a hard time understanding items (ii) and (iii) in this definition which talk about convergence in probability and convergence in distribution respectively. Briefly speaking I want to ask

Why does convergence in probability imply convergence in distribution? In particular, I am not sure if I really need the functions $F$ to be both bounded and continuous as the definition of convergence in distribution insists (and so far it seems to me that $F$ being bounded suffices)

I am certain that I am missing something crucial. To see this better, I will first state the definitions as given in Tao's book. And then, I will show how I use these definitions (possibly, I might have misunderstood them). I will make a simplification and assume that all random variables are real valued.

Def: [Convergence in probability] A sequence of Random Variables $X_n$ converges to a random variable $X$ in probability if for every $\varepsilon > 0$, one has $$ \lim_{n \rightarrow \infty } \inf \Pr(|X_n - X| \leq \varepsilon) = 1$$

The way I parse this statement is the following

For every $\delta > 0$, for every $\varepsilon > 0$, there exists $n_0 = n_0(\varepsilon, \delta) \in \mathbb{N}$ such that for all $n > n_0$ we have

$$\Pr(|X_n - X| \leq \varepsilon) \geq 1 - \delta.$$

Now, here is the definition of convergence in distribution (I think I parse this one just fine)

Def:[Convergence in distribution] $X_n$ converges in distribution to $X$ if for every bounded and continuous function $F : \mathbb{R} \to \mathbb{R}$ one has $$\lim_{n \to \infty} \mathbb{E}(F(X_n)) = \mathbb{E}(F(X)).$$

Here is my attempt at proving (ii) implies (iii). I want to see why my "proof" does not seem to require that $F$'s have to be continuous.

Suppose $-M \leq F \leq M$ everywhere. Also, suppose we denote the underlying space common to all random variables by $(\Omega, \mathcal{F}, \mu)$

I want to show that if $X_n$ converges to $X$ in probability, then for any $\beta > 0$, there exists $n_{\beta} \in \mathbb{N}$ such that for every $n > n_{\beta}$ we have $| \mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) | \leq \beta$.

To this end, I write $$\mathbb{E}(F(X_n)) = \displaystyle \int_{\Omega} F(X_n(\omega))\ d\mu(\omega)$$

And similarly, let us write $$ \mathbb{E}(F(X)) = \displaystyle \int_{\Omega} F(X(\omega))\ d\mu(\omega)$$

Let us now go back to convergence in probability which we are told holds. Let us denote by $G \subseteq \Omega$ the 'good' set of all those outcomes for which $|X_n - X| \leq \epsilon$ holds (where $n > n_0({\varepsilon, \delta})$). We know that $\mu(G) \geq 1 - \delta$. What is left over is the set $B = \Omega \setminus G$ of bad outcomes with measure at most $\delta$. Now, we upper bound the difference $| \mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) |$ by considering the absolute differences on the sets $G$ and $B$. That is,

$\mathbb{E}(F(X_n)) - \mathbb{E}(F(X)) = \left| \displaystyle \int_{\Omega} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{\Omega} F(\color{red}{X}(\omega))\ d\mu(\omega) \right|$

$$\leq \left| \displaystyle \int_{G} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{G} F(\color{red}{X}(\omega))\ d\mu(\omega) \right| + \left|\displaystyle \int_{B} F(\color{red}{X_n}(\omega))\ d\mu(\omega) - \displaystyle \int_{B} F(\color{red}{X}(\omega))\ d\mu(\omega) \right|$$

This is seen to be $$\leq \varepsilon \cdot \mu(G) + \sup_{\omega \in B}|F(\color{red}{X_n}(\omega)) - F(\color{red}{X}(\omega))| \cdot \mu(B) \leq \varepsilon + 2M \cdot \delta$$

Now i choose $\varepsilon = \frac{\beta}{2}$ and $\delta = \frac{\beta}{2M}$ and use $n_0(\varepsilon, \delta)$ (in my unpacking of convergence in probability) as the desired $n_{\beta}$.

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I am aware that I am making a mistake, please help me narrow that down.

Thanks

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You need continuity in order to ensure that $F (X_n)-F (X) $ is small when $X_n-X $ is small. In other words you need to invoke continuity to handle the integral over $G $.

Also, your $n_\beta$ will depend on $F $ as a result of this step.

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  • $\begingroup$ Thanks Ian. Embarassed to see what i missed :-( $\endgroup$ – Akash Kumar Sep 22 '16 at 19:13

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