Past exam of the random matrix theory course of the University of Oxford / 2025 / 2

Let $P$ be the $3$-by-$3$ evaluation matrix $P_{i,\ell +1}=p_{\ell }({\lambda }_i)$. The kernel evaluation matrix is the Gram matrix $P{P}^T$, so multiplicativity of the determinant givesBecause the polynomials are monic of consecutive degrees, their coefficient matrix relative to $1,x,x^2$ is a triangular matrix with unit diagonal. It has determinant one, so the Vandermonde determinant invariance under monic basis change givesSquaring removes the ordering sign. Thus the Gaussian unitary ensemble joint probability density function has precisely the kernel formNo orthogonality of the polynomials is needed for this step.

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are

Gaudin's lemma is the projection kernel determinant integration identity. Suppose an integral kernel $K$ satisfieswith the relevant integrals absolutely convergent. Define $D_k(x_1,\dots ,x_k)=\det [K(x_i,x_j){]}_{i,j=1}^k$ and $D_0=1$. ThenIn particular, a finite-rank projection kernel of rank $N$ has $\int D_{N}d{\mu }^N=N!$.

One can see the factor directly from the permutation expansion of the determinant. Terms whose permutation fixes $k$ integrate to $N{D}_{k-1}$. Every permutation on the other $k-1$ indices admits $k-1$ insertions of $k$ into a cycle. Each insertion reverses its sign, and the reproducing identity contracts its two adjacent kernel factors into one. These terms integrate to $-(k-1)D_{k-1}$, proving the formula.

The squared norm of $p_{\ell }(x)e^{-x^2/4}$ in $L^2(\mathbb{R},dx)$ is $\sqrt{2\pi }{h}_{\ell }$, where $(h_0,h_1,h_2)=(1,1,2)$. Choose positive normalizing constantsThen ${\varphi }_0,{\varphi }_1,{\varphi }_2$ are an orthonormal set for Lebesgue measure. Expanding the finite sums givesThus this is an orthogonal polynomial projection kernel, explicitlyThe positive constants above are forced by the reproducing identity: before normalization, its convolution multiplies each independent term ${\varphi }_{\ell }(x){\varphi }_{\ell }(z)$ by $\sqrt{2\pi }{h}_{\ell }/{\kappa }_{\ell }^2$. Each multiplier must equal one. Changing the sign of a real ${\kappa }_{\ell }$ leaves the kernel unchanged.

Use the symmetric joint probability density function of exchangeably labelled eigenvalues, normalized over all of ${\mathbb{R}}^3$. The finite-rank projection kernel hasSuccessive applications of Gaudin's lemma therefore giveIt follows thatAs a normalization check, the weighted Vandermonde determinant yieldsIf instead one uses ordered eigenvalues on ${\lambda }_1<{\lambda }_2<{\lambda }_3$, the normalized density on that chamber is six times the labelled density. In that convention the determinant prefactor on the chamber is one.

The correlation functions of a point process use factorial normalization:Start with ${\rho }_3=D_3/6$. By Gaudin's lemma, integrating $D_3$ once gives $D_2$, and integrating $D_2$ once gives $2D_1$. The factorial factors cancel, so the eigenvalues form a determinantal point process withandThe last equality uses the real symmetry of this kernel. In particular, $\int R_3^{(1)}=3$ and $\iint R_3^{(2)}=6$, as required for three points and six ordered pairs of distinct points.

Evaluate the orthogonal polynomial projection kernel on its diagonal:The ${\lambda }^2$ terms cancel. Integrating against Lebesgue measure gives $(\mathbb{E}{Z}^4+3)/2=3$ for a standard normal random variable $Z$, checking the correlation function of a point process normalization.

The correlation function of a point process is even and tends to zero as $|\lambda |\to \infty$. Its derivative factors asConsequently its local maxima occur at $0,\pm \sqrt{3}$, and its local minima at $\pm 1$. The heights are

For an interval $I$, ${\int }_I{R}_3^{(1)}(\lambda )d\lambda$ is the expected value of the number of eigenvalues in $I$. Thus equally short intervals have the highest expected counts nearwith the central peak highest. Each exchangeably labelled eigenvalue has the same marginal probability density function $R_3^{(1)}/3$. The three peaks should not be interpreted as three separate marginal densities for ordered eigenvalues.

In this case the most likely joint configuration also lies at these three positions. On the ordered chamber, the logarithm of the joint probability density function, up to a constant, isIts stationary equations are ${\lambda }_i=2{\sum }_{j\ne i}({\lambda }_i-{\lambda }_j{)}^{-1}$, satisfied by $(-\sqrt{3},0,\sqrt{3})$. For every nonzero real vector $v$, the second directional derivative isThus $L$ is strictly concave on this convex chamber, proving that the stationary configuration is its unique global maximum. Permuting the labels gives the other maxima on ${\mathbb{R}}^3$.