###### Photo of Paul Lévy. Source: MacTutor and Ra-bird.

Using $latex {(ii)}&fg=000000$ of the Pormanteau lemma, it is possible to show convergence in distribution for a random vectors sequence via one “transformation”. The most important transform is the characteristic function

Definition  Let be $latex {X}&fg=000000$ one random vector of $latex {{\mathbb R}^k}&fg=000000$, $latex {t\in{\mathbb R}^k}&fg=000000$ and define the characteristic function of $latex {X}&fg=000000$ by

$latex \displaystyle \phi_X(t)=\mathbb E \left(\exp( \langle it,X \rangle)\right) . &fg=000000$

Theorem (Lévy’s continuity theorem) Let $latex {(X_n)_n}&fg=000000$ and $latex {X}&fg=000000$ random vectors of $latex {{\mathbb R}^k}&fg=000000$.

$latex {(i)}&fg=000000$ Then $latex {X_n \rightsquigarrow X}&fg=000000$ if and only if $latex {\phi_{X_n}(t)\rightarrow\phi_X(t),\ \forall t\in{\mathbb R}^k}&fg=000000$.

$latex {(ii)}&fg=000000$ If $latex {\phi_{X_n}(t)\rightarrow\phi(t)}&fg=000000$ , pointwise $latex {\forall t\in{\mathbb R}^k}&fg=000000$ and if $latex {\phi}&fg=000000$ is continue on 0, then $latex {\phi}&fg=000000$ is the characteristic function of $latex {X}&fg=000000$ and $latex {X_n\rightsquigarrow X}&fg=000000$.

Proof:

$latex {(i)}&fg=000000$ If $latex {X_n \rightsquigarrow X}&fg=000000$, by the Portmanteau’s lemma it enough to remark that for any $latex {t\in {\mathbb R}^k}&fg=000000$, $latex {x\mapsto \exp(i<t,x>)}&fg=000000$ is bounded and continue.
For the only if direction, it is enough to show $latex {(ii)}&fg=000000$ because the characteristic function is continue at 0.

$latex {(ii)}&fg=000000$ By now, suppose that $latex {X_n}&fg=000000$ is uniformly tight. Then by the Prohorov’s theorem, there exist a $latex {X_n}&fg=000000$ subset that converge weakly to a random variable $latex {Y}&fg=000000$. That means, $latex {\phi_{X_{n_k}}(t)\rightarrow\phi_Y(t),\ \forall t\in{\mathbb R}^k.}&fg=000000$ By the limit uniqueness, we deduce that $latex {\phi_Y=\phi}&fg=000000$. Furthermore, that implies that every $latex {X_n}&fg=000000$ subset converge to $latex {Y}&fg=000000$ in law. Hence, there exist one unique accumulation point in the convergence law sense. This implies that $latex {X_n}&fg=000000$ converge weakly to $latex {Y}&fg=000000$.

In fact, suppose that by contradiction that $latex {X_n}&fg=000000$ not converge in law to $latex {Y}&fg=000000$, there is one point $latex {x}&fg=000000$ of continuity of the $latex {Y}&fg=000000$’s distribution function, such that $latex {\mathbb P\left(X_n\leq x\right) \not\rightarrow\mathbb P \left(Y\leq x\right) }&fg=000000$.Then, there exist $latex {\epsilon>0}&fg=000000$ and one subset $latex {n_k}&fg=000000$ such that e $latex {\vert\mathbb P \left(X_{n_k}\leq x \right)-\mathbb P \left( Y\leq x\right)\vert\geq \epsilon}&fg=000000$. But as $latex {(X_n)}&fg=000000$ is uniformly tight, then $latex { X_{n_k}}&fg=000000$ it is tight also. We can use the Prohorov’s Theorem to extract one subset which converge in law to $latex {Y}&fg=000000$, what it is contradictory.

Let show that $latex {X_n}&fg=000000$ is uniformly tight. This will be a $latex {\phi}&fg=000000$’s continuity on 0 consequence. We can assume without loss of generality that $latex {X_n\in{\mathbb R}}&fg=000000$. Let be $latex {x}&fg=000000$ and $latex {\delta>0}&fg=000000$,

$latex \displaystyle 1_{|\delta x|>2}\leq 2 \left(1-\frac{\sin(\delta x)}{\delta x}\right) =\frac 1\delta\int_{-\delta}^\delta(1-\cos(tx))dt. &fg=000000$

We replace $latex {x}&fg=000000$ by $latex {X_n}&fg=000000$, take expectations and use the Fubini’s theorem to get

$latex \displaystyle \mathbb P\left(|\delta X_n|>2\right)\leq \frac1\delta\int_{-\delta}^\delta\mathbb E(1-\cos(tX_n)dt \leq \frac1\delta\int_{-\delta}^\delta \text{Re}(1-\mathbb E\left(\exp(itX_n)\right) dt. &fg=000000$

By hypothesis, the second integrate converge pointwise to $latex {\text{Re}(1-\phi(t))}&fg=000000$. By the dominated convergence the whole expression goes to

$latex \displaystyle \frac 1\delta\int_{-\delta}^\delta \text{Re}(1-\phi(t))dt.&fg=000000$

Let be $latex {\epsilon>0}&fg=000000$, by continuity of $latex {\phi}&fg=000000$ in 0, there exist $latex {\delta>0}&fg=000000$, such that $latex {|t|\leq\delta}&fg=000000$ implies $latex {|1-\phi(t)|\leq\epsilon}&fg=000000$. For this $latex {\delta}&fg=000000$ the limit integral is smaller than $latex {2\epsilon}&fg=000000$. There exist $latex {N}&fg=000000$ such that for every $latex {n\geq N,}&fg=000000$ $latex {\mathbb P\left(|\delta X_n|>2\right) \leq 2\epsilon}&fg=000000$, whence the sequence $latex {X_n}&fg=000000$ is uniform tight. $latex \Box&fg=000000$