Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 27J (Probability and Measure)

For every $a\in \mathbb{R}$,After squaring and taking expectations, the cross term vanishes because $\mathbb{E}[X-\mathbb{E}X]=0$. ThereforeThe second term is nonnegative and vanishes exactly at $a=\mathbb{E}X$, so the variance as the minimum mean squared error of a constant identity is

Using the conventionthe Fourier transform of an interval indicator isThe removable value at the origin is obtained by taking the limit. Thus

For $f=1_{[-1,1]}$, part (b) gives $\hat{f}(u)=2\sin u/u$. The Plancherel theorem in the stated Fourier convention givesSince the left side is two,The integrand is even, so the sinc-squared integral is

Set $Z=e^{iuX}$. Then $|Z|=1$ almost surely andThus $Z=\mathbb{E}Z$ almost surely. Its constant value has modulus one, so it equals $e^{i\theta }$ for some $\theta \in \mathbb{R}$. Consequently the unit modulus of a characteristic function criterion gives

Let $X^{\prime }$ be an independent copy of $X$. Part (i), applied simultaneously to $X$ and $X^{\prime }$, givesalmost surely. Since $u/v$ is irrational, the lattices $(2\pi /u)\mathbb{Z}$ and $(2\pi /v)\mathbb{Z}$ intersect only at zero. Hence $X=X^{\prime }$ almost surely.

If two independent variables with the same law agree almost surely, that law is a point mass: for every Borel set $A$, independence givesso every such probability is zero or one. Applying this to half-lines identifies a constant $c$ with $X=c$ almost surely. Therefore the irrational unit-modulus frequencies force a point mass result is