Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 201 / 6

The Lévy continuity theorem states that if characteristic functions ${\phi }_n$ of probability distributions ${\mu }_n$ converge pointwise on $\mathbb{R}$ to $\phi$, and $\phi$ is continuous at zero, then $\phi$ is the characteristic function of a unique probability distribution $\mu$ andHere the arrow denotes weak convergence of probability measures, equivalently convergence in distribution of associated random variables. Conversely, weak convergence of probability measures implies pointwise convergence of their characteristic functions to the characteristic function of the limit. The continuity-at-zero condition is essential to ensure that the pointwise limit is a characteristic function.

Write $A_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and $\phi (t)=\mathbb{E}{e}^{it{X}_1}$. The given two-sided limit $|f(u)|/u\to 0$ implies $|f(u)|/|u|\to 0$, so $f(u)=o(|u|)$. For each fixed real $t$, independence and the characteristic function of a sum of independent variables giveSet $z_n=iat/n+f(t/n)$. For $t\ne 0$, $n{z}_n\to iat$ and $z_n=O(n^{-1})$. The local Taylor expansion of the complex logarithm at $1$ givesThus $\mathbb{E}{e}^{it{A}_n}\to e^{iat}$; at $t=0$ the identity is immediate. The limit is continuous at zero and is the characteristic function of the constant random variable $a$. The Lévy continuity theorem proves the weak law from a characteristic-function expansion:No integrability hypothesis has been used. The conclusion is also convergence in probability by (d)'s convergence in distribution to a constant implies convergence in probability. If the word “constant” were to allow complex $a$, the characteristic function symmetry $\phi (-t)=\overline{\phi (t)}$ forces $a=\overline{a}$, so it is necessarily real.

Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

Let $F_n$ be the distribution function of $X_n$. The limiting constant zero has distribution function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$. For every $\epsilon >0$, both $-\epsilon$ and $\epsilon$ are continuity points of $F$. Thus convergence in distribution implies $F_n(-\epsilon )\to 0$ and $F_n(\epsilon )\to 1$. ThereforeThis provesThe same proof after subtracting a fixed real $c$ proves convergence in distribution to a constant implies convergence in probability. A deterministic limit fixes the coupling automatically; the counterexample in (c) exploits a nonconstant limit whose probability distribution alone does not fix that coupling.