Past exam of the random matrix theory course of the University of Oxford / 2025 / 1

Taking the transpose gives $M^T=(X^T+X)/\sqrt{2}=M$, so $M$ is a symmetric matrix. For $i<j$, $M_{ij}=(X_{ij}+X_{ji})/\sqrt{2}$ has the standard normal distribution, whereas $M_{ii}=\sqrt{2}{X}_{ii}$ has the normal distribution $N(0,2)$. Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.

Consequently the joint probability density function, relative to ${\prod }_{i\le j}d{m}_{ij}$, isHere $\mathrm{tr}(m^2)={\sum }_i{m}_{ii}^2+2{\sum }_{i<j}{m}_{ij}^2$ because $m$ is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix $M/\sqrt{n}$ has the corresponding $1/n$ off-diagonal variance.

We will also use its entry covariance. Expanding in the independent entries of $X$, for which $\mathbb{E}[X_{ij}{X}_{kl}]={\delta }_{ik}{\delta }_{jl}$, givesThe symbols ${\delta }_{ab}$ are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.

The joint probability distribution of all entries of $X$ is unchanged by $X\mapsto -X$, so $M$ and $-M$ have the same probability distribution. For the specified cyclic product $P=M_{i_1{i}_2}\cdots M_{i_k{i}_1}$, this transformation multiplies $P$ by $(-1{)}^k$. Every such product is an integrable random variable, since normal random variables have finite moments of all orders. Thus, for odd $k$,Repeated indices cause no difficulty: all factors change sign simultaneously. It would be incorrect to assume that these factors are independent.

The probabilistic form of Wick's theorem is Isserlis's theorem. If $(Y_1,\dots ,Y_k)$ is a centered random vector with a multivariate Gaussian distribution and covariance matrix $C$, its odd product moments vanish, and, for even $k$,Here ${\mathcal{P}}_2(k)$ is the set of perfect matchings of the $k$ labelled positions. The formula allows repeated coordinates and singular covariance matrices. To see the pairing rule, differentiate the moment-generating function $\exp (t^{T}Ct/2)$ once in each coordinate and set $t=0$: only terms pairing all differentiated coordinates survive, each with the displayed product of covariances.

Take $Y_a=M_{i_a{i}_{a+1}}$, with $i_{k+1}=i_1$. These variables are linear combinations of the original independent normal random variables, hence are jointly Gaussian. They have zero means, soSubstituting the entry covariance from part (a)(i) identifies each parenthesis with $\mathbb{E}[M_{i_a{i}_{a+1}}{M}_{i_b{i}_{b+1}}]$, giving the requested identity.

The spectral theorem for real symmetric matrices identifies the sum of the $k$th powers of the eigenvalues with the trace of the $k$th matrix power. Therefore the moment of the empirical spectral measure isFor odd $k$, every summand has zero expectation by part (a)(ii), so $\boxed{\mathbb{E}[m_k(n)]=0}$ for every $n$.

For the second moment, use $M_{ji}=M_{ij}$. There are $n$ diagonal entries with variance two and $n(n-1)$ ordered off-diagonal entries with variance one. Their means vanish, givingHence $\boxed{\underset{n\to \infty }{\lim }\mathbb{E}[m_2(n)]=1}$.

Write the four cyclic factors as $A=M_{ij}$, $B=M_{jk}$, $C=M_{kl}$ and $D=M_{li}$. The Gaussian moment pairing theorem gives three terms, corresponding to $(AB)(CD)$, $(AC)(BD)$ and $(AD)(BC)$. We sum each term over $i,j,k,l$.

For $(AB)(CD)$, the entry covariance givesIndeed the four products leave respectively three, two, two and one free indices. A cyclic relabelling gives exactly the same sum for $(AD)(BC)$.

For the crossing pairing $(AC)(BD)$,The first product allows $i=k$ and $j=l$ freely, whereas each other product forces all four indices to coincide. Combining the three pairings and the trace normalization,

First $W^T=W$, so the spectral theorem for real symmetric matrices gives real eigenvalues and an orthonormal basis of eigenvectors. Moreover, for every real vector $v$,Thus $W$ is a positive semidefinite matrix. For any nonzero eigenvector $v$ with eigenvalue $\lambda$,This matrix is a scaled example of the Wishart distribution.

Let $g_i=(X_{i1},\dots ,X_{in})$ be row $i$ of $X$, viewed as a real vector. These are independent standard Gaussian random vectors, and $(X{X}^T{)}_{ij}=g_i\cdot g_j$. The trace expression for the empirical spectral measure givesWe need the norm momentsFor completeness, $S=\Vert g_i{\Vert }^2$ has the chi-squared distribution with $n$ degrees of freedom. By independence and the Gaussian integral, its moment-generating function is $\mathbb{E}{e}^{sS}=(1-2s{)}^{-n/2}$ for $s<1/2$. Differentiating at zero once, twice and three times proves these three identities.

For the first moment,For the second moment, if $i\ne j$, conditional expectation given $g_i$ gives $\mathbb{E}[(g_i\cdot g_j{)}^2\mid g_i]=\Vert g_i{\Vert }^2$, hence an unconditional value $n$. Separating equal and unequal row indices yields

For the third moment, the summand is $(g_i\cdot g_j)(g_j\cdot g_k)(g_k\cdot g_i)$. There are three disjoint cases.

ConsequentlyThe three limiting expected values are therefore