Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 / 2

Weak convergence of probability measures ${\mu }_n\to \mu$ on a metric space means thatfor every bounded continuous real function $f$. Weak convergence of random variables means weak convergence of their probability distributions; equivalently,for every bounded continuous $f$.

For a bounded measurable $f$, the dual estimate for total variation distance givesThe right-hand side tends to zero, so in particular the integrals converge for every bounded continuous $f$. Thus weak convergence of random variables follows.

The converse fails. On $\mathbb{R}$, let ${\mu }_n={\delta }_{1/n}$ and $\mu ={\delta }_0$. Continuity gives $f(1/n)\to f(0)$, so ${\mu }_n$ converges weakly to $\mu$. However, for $A=\{0\}$,for every $n$, so there is no convergence in total variation distance.

Assume first that $X_n\stackrel{d}{\to }X$. Because all variables take values in the compact interval $[-C,C]$, each monomial $x^k$ agrees there with a bounded continuous function on $\mathbb{R}$. The bounded moment criterion for weak convergence in this case begins withfor every natural number $k$.

Conversely, suppose all moments converge. Let $f$ be bounded and continuous. By the Weierstrass approximation theorem, for every $\epsilon >0$ there is a polynomial $P$ withMoment convergence implies $\mathbb{E}[P(X_n)]\to \mathbb{E}[P(X)]$, whileTaking the limit superior and then $\epsilon \downarrow 0$ proves $\mathbb{E}[f(X_n)]\to \mathbb{E}[f(X)]$, which is convergence in distribution.

Let $g:M^{\prime }\to \mathbb{R}$ be bounded and continuous. The composition $h=g\circ f$ is bounded and measurable, and every discontinuity point of $h$ is a discontinuity point of $f$. ThusThe Portmanteau theorem includes the null-discontinuity criterion: if $X_n\stackrel{d}{\to }X$ and a bounded measurable function is continuous at $X$ almost surely, then its expectations converge. HenceSince this holds for every bounded continuous $g$, it proves the continuous mapping theorem conclusion $f(X_n)\stackrel{d}{\to }f(X)$.