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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 \begin{equation} \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}} \end{equation} 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 \begin{equation} \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}} \end{equation} 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?

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  • $\begingroup$ Are your two quotations from the same book? $\endgroup$ May 28, 2020 at 22:55
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    $\begingroup$ 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 $Z$. $\endgroup$
    – Michh
    May 28, 2020 at 22:58
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    $\begingroup$ 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? $\endgroup$
    – Michh
    May 28, 2020 at 23:08
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    $\begingroup$ 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. $\endgroup$
    – Michh
    May 28, 2020 at 23:15
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    $\begingroup$ 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. $\endgroup$
    – Michh
    May 28, 2020 at 23:26

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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.

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  • $\begingroup$ Thank you especially for the last remark, which was the part I was missing before :) $\endgroup$ Jun 1, 2020 at 13:50
  • $\begingroup$ I figured by reading the comments above. $\endgroup$ Jun 1, 2020 at 13:50

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