Past exam of the random matrix theory course of the University of Oxford / 2025 / 3

The characteristic polynomial of the rotation matrix isIts roots are $\cos U\pm i\sin U$, soFor example $(1,i{)}^T$ is an eigenvector for $e^{iU}$ and $(1,-i{)}^T$ for $e^{-iU}$. At $U=0$ or $\pi$ the two eigenvalues coincide; these events have probability zero under the uniform distribution.

The difference of the two eigenvalues is $2i\sin U$, hence $\Delta ={\sin }^{2}U$. Its expectation isFor $0\le x\le 1$, put $\alpha =\arcsin \sqrt{x}\in [0,\pi /2]$. The set of angles satisfying $|\sin u|\le \sqrt{x}$ consists of $[0,\alpha ]$, $[\pi -\alpha ,\pi +\alpha ]$ and $[2\pi -\alpha ,2\pi ]$, of total length $4\alpha$. Thus the full cumulative distribution function isDifferentiating on $(0,1)$ gives the probability density functionThis is the arcsine distribution, equivalently the beta distribution with both parameters $1/2$. It has no atoms at either endpoint.

Apply the same characteristic polynomial calculation with $U$ replaced by $B(t)$. Pathwise,The real-valued processes $\pm B(t)$ give continuous lifts of these angles. The eigenvalues coincide whenever $B(t)\in \pi \mathbb{Z}$; the exponential formulas remain valid at those times.

Here $\Delta (t)={\sin }^{2}B(t)$. For $f(b)={\sin }^{2}b$, $f^{\prime }(b)=\sin 2b$ and $f^{\prime \prime }(b)=2\cos 2b$. Since standard Brownian motion has quadratic variation $dt$, Itô's formula givesThereforeThe squared diffusion coefficient is $V(t{)}^2=4\Delta (t)(1-\Delta (t))$. Its sign depends on $B(t)$: one cannot replace $\sin (2B(t))$ by the nonnegative square root while retaining the same driving Brownian motion.

The characteristic function of $B(t)\sim N(0,t)$ gives $\mathbb{E}{e}^{2iB(t)}=e^{-2t}$. Taking real parts and using ${\sin }^{2}b=(1-\cos 2b)/2$,There is also a direct check from the stochastic differential equation. Its Itô integral has zero expectation because its integrand is bounded. Thus $m(t)=\mathbb{E}\Delta (t)$ satisfies $m^{\prime }(t)=1-2m(t)$ and $m(0)=0$, whose solution is the same expression.

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

Use Haar sampling by Gaussian QR decomposition:

The matrix $G$ is invertible almost surely. Indeed, successively sampled Gaussian columns have probability zero of lying in the proper linear subspace spanned by the previous columns. Thus the diagonal signs used above exist with probability one. Equivalently, apply the Gram-Schmidt process to the columns, always dividing each residual by its positive Euclidean norm; this produces $M$ directly.

To prove uniformity, fix $H\in O(n)$. Each column of $G$ has density proportional to $e^{-\Vert x{\Vert }^2/2}$, so orthogonal transformations preserve its probability distribution. Column independence therefore gives $HG\stackrel{\mathrm{d}}{=}G$.

The QR decomposition with positive diagonal is unique: if $G=Q_1{R}_1=Q_2{R}_2$, then $Q_2^T{Q}_1=R_2{R}_1^{-1}$ is both an orthogonal matrix and an upper triangular matrix, hence diagonal with entries $\pm 1$. Its diagonal entries are positive, so it is the identity. It follows that the orthogonal factor of $HG=HMR$ is $HM$. ThereforeBy uniqueness of normalized Haar measure on the compact group $O(n)$, the returned matrix has exactly the required probability distribution. Reflection invariance gives each determinant sign probability $1/2$, so the algorithm samples both components of the orthogonal group. The positive-diagonal correction is essential when a numerical QR decomposition chooses its own signs.