# Application of Levy's Continuity Theorem. How is it applied herebelow?

THEOREM (Levy's Continuity Theorem)

Let $$(\mu_n)_{n\geq1}$$ be a sequence of probability measures on $$\mathbb{R}^d$$, and let $$(\hat{\mu}_n)_{n\geq1}$$ denote their characteristic functions (or Fourier transforms).

i) If $$\mu_n$$ converges weakly to a probability measure $$\mu$$, then $$\hat{\mu}_n(u)\rightarrow\hat{\mu}(u)$$ for all $$u\in\mathbb{R}^d$$;
ii) If $$\hat{\mu}_n(u)$$ converges to a function $$f(u)$$ for all $$u\in\mathbb{R}^d$$, and if in addition $$f$$ is continuous at $$0$$, then there exists a probability $$\mu$$ on $$\mathbb{R}^d$$ such that $$f(u)=\hat{\mu}(u)$$, and $$\mu_n$$ converges weakly to $$\mu$$.

Let $$(X_n)_{n\geq1}$$ be a sequence of random variables, $$i$$ the imaginary unit, $$S_n=\sum\limits_{i=1}^nX_i$$ and $$u\in\mathbb{R}$$. For a certain constant $$L$$, it holds that $$$$\Bigg|\mathbb{E}\Big(e^{iu\frac{1}{\sqrt{n}}S_n}\Big)-\Big(1-\frac{u^2}{2n}\Big)^{n}\Bigg|\leq L\frac{|u|^3}{6\sqrt{n}}$$$$ At this point, since the r.h.s of the above inequality tends to $$0$$ as $$n\rightarrow\infty$$ and since $$\begin{equation*} \begin{split} \lim\limits_{n\to\infty}\Big(1-\frac{u^2}{2n}\Big)^{n}=e^{-\frac{u^2}{2}} \end{split} \end{equation*}$$ recalling that $$\lim\limits_{x \to \infty} |f(x) + g(x)| = \lim\limits_{x \to \infty} f(x) + \lim\limits_{x \to \infty} g(x)$$, I have that $$$$\lim\limits_{n\to\infty}\mathbb{E}\Big(e^{iu\frac{S_n}{\sqrt{n}}}\Big)=\lim\limits_{n\to\infty}\Big(1-\frac{u^2}{2n}\Big)^{n}=e^{-\frac{u^2}{2}}$$$$ Now, I read the following statement:

By Levy's Continuity Theorem, we have that $$\frac{S_n}{\sqrt{n}}$$ converges in law to $$Z$$, where the characteristic function of $$Z$$ is $$e^{-\frac{u^2}{2}}$$

My question is: HOW EXACTLY is the above-quoted Levy's Continuity Theorem applied to get the above conclusion?

• Are your two quotations from the same book? May 28 '20 at 22:55
• A direct consequence of the Lévy continuity theorem as you stated it is that: if $\mu_n$ and $\mu$ are probability measures such that $\hat{\mu}_n(u)$ converges to $\hat{\mu}(u)$, then $\mu_n$ converges weakly to $\mu$. In your setting, this is what you apply since you know that $e^{-u^2/2}$ is the characteristic function of $Z$. May 28 '20 at 22:58
• The probability measures are implicit here: these are simply the laws of the random variables involved. For example, $e^{-u^2/2}$ is the ch.f. of $Z$ but this is equivalent to saying that it is the ch.f. of the probability measure $P_Z$ (which is the law of $Z$). Does this make any sense? May 28 '20 at 23:08
• Perhaps a simpler way of viewing how the Lévy continuity theorem is applied in this context would be to translate the statement in terms of random variables: if the ch.f. of $X_n$ converges to the ch.f. of $X$, then $X_n$ converges to $X$ in distribution. This is just another way to formulate the same result. May 28 '20 at 23:15
• Yes, I am. This is just a question of juggling between the definitions. Again, the ch.f. of a random variable $X$ is nothing but the ch.f. of its law $\mathbb{P}_X$ (recall that $\mathbb{P}_X$ is the pushforward measure of $\mathbb{P}$ by $X$). Similarly, for random variables, convergence in distribution is defined to be weak convergence of their laws. May 28 '20 at 23:26

You have $$\lim_n \mathbb{E}(\exp(iu S_n/\sqrt{n})) = \exp(-u^2/2) = \mathbb{E}(\exp(iuZ))$$for all $$u$$ and thus the characteristic function of $$S_n/ \sqrt{n}$$ converges to the characteristic function of $$Z$$. The Lévy continuity theorem says that this is equivalent with $$S_n/ \sqrt{n} \stackrel{d}\to Z$$.
Recall, that by definition for a sequence of random variables on a $$p$$-space $$(\Omega, \mathcal{F}, \mathbb{P})$$ we have
$$X_n \stackrel{d}\to X \iff \mathbb{P}_{X_n} \stackrel{w}\to \mathbb{P}_X$$
where the $$d$$ is convergence in distribution and the $$w$$ is weak convergence.