Spectral density of a sample covariance matrix in a Gaussian Random Ensemble

Question

Let $N > 0$ and $T > N$ be integers and $C$ be a real, symmetric $N \times N$ matrix.The question is to compute the following integral: \begin{equation} U_{N,T}(t) := \frac{1}{N} \int\limits_{{\mathcal R}^{N T}}Tr\left[e^{\frac{\imath t}{T} {\bf X} \cdot {\bf X}^T}\right] e^{-\frac{1}{2} Tr\left[C^{-1} {\bf X} \cdot {\bf X}^T\right]} \frac{\prod\limits_{i=1}^N \prod\limits_{\xi=1}^T d X_{i,\xi}}{\sqrt{(2\pi)^{N T} \left(\det(C)\right)^T}} \end{equation} Here ${\bf X} := \left(X_{i\xi}\right)_{i=1,\xi=1}^{N,T}$. This integral is equal to the Fourier transform of the spectral density of the sample covariance matrix in a Gaussian random ensemble. I know that there exists a series expansion in terms of spectral moments of the matrix $C$ for this integral. Yet, I would like to know if there is some closed form expression for that integral.

Answer 1

I present a solution in case of $N=2$. Note, that the integrand is a function of the matrix elements of the sample covariance matrix only. Therefore we will reduce the $(N\cdot T)$-dimensional integral to a $N(N+1)/2$-dimensional integral over the elements of the sample covariance matrix. We insert the obvious identity: \begin{equation} 1 = \int\limits_{{\mathcal R} \times {\mathcal R}_+^2} \delta\left(c_{1,1} - \sum\limits_{j=1}^T X_{1,j}^2\right) \delta\left(c_{2,2} - \sum\limits_{j=1}^T X_{2,j}^2\right) \delta\left(c_{1,2} - \sum\limits_{j=1}^T X_{1,j} X_{2,j}\right) d c_{1,2} d c_{1,1} d c_{2,2} \end{equation} into the integrand, then we swap the order of integration, ie we integrate over the $X_{i,\xi}$ variables first and we get: \begin{eqnarray} &&U_{N,T}(t) = \\ &&\frac{1}{N} \int\limits_{{\mathcal R} \times {\mathcal R}_+^2} Tr\left[e^{\frac{\imath t}{T} \left(\begin{array}{cc}c_{1,1} & c_{1,2} \\ c_{1,2} & c_{2,2}\end{array}\right)}\right] \cdot {\mathcal A}_T \pi^{T-1} \left(\det(c)\right)^{\frac{T-3}{2}} \cdot e^{-\frac{1}{2} Tr\left[C^{-1} \left(\begin{array}{cc}c_{1,1} & c_{1,2} \\ c_{1,2} & c_{2,2}\end{array}\right) \right]} \frac{ d c_{1,2} d c_{1,1} d c_{2,2}}{\sqrt{(2 \pi)^{2 T} \left(\det(C)\right)^T}} \end{eqnarray} In here we have used the result given in Multidimensional integral involving delta functions . The constant ${\mathcal A}_T$ reads ${\mathcal A}_T := 2^{T-3}/(T-2)!$. Now, we perform an eigenvalue decomposition of the sample covariance matrix. \begin{equation} \left(\begin{array}{cc}c_{1,1} & c_{1,2} \\ c_{1,2} & c_{2,2}\end{array}\right) = O \cdot \left( \begin{array}{cc} s_1 & 0 \\ 0 & s_2 \end{array} \right)\cdot O^T \end{equation} where \begin{equation} O := \left( \begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \right) \end{equation} It is easy to show that the Jacobian of the appropriate transformation reads: \begin{equation} \frac{\partial\left(c_{1,1},c_{2,2},c_{1,2}\right)}{\partial\left(s_1,s_2,\theta\right)} = s_1-s_2 \end{equation} Therefore we get: \begin{eqnarray} &&U_{N,T}(t) = \\ && \frac{1}{2} \int\limits_{{\mathcal R}_+^2} \left(e^{\frac{\imath t s_1}{T}} + e^{\frac{\imath t s_2}{T}}\right) {\mathcal A}_T \pi^{T-1} \left(s_1 s_2\right)^{\frac{T-3}{2}} \left|s_1-s_2\right| \cdot \\ &&\int\limits_0^{2\pi} e^{-\frac{1}{2} Tr\left[C^{-1} O \left( \begin{array}{cc} s_1 & 0 \\ 0 & s_2 \end{array} \right) O^T \right]} d \theta \cdot \frac{d s_1 d s_2}{\sqrt{(2 \pi)^{2 T} \left(\det(C)\right)^T}} \end{eqnarray} The integral over the angle $\theta$ is the famous Itzykson-Zuber integral over the orthogonal group $O(2)$. Unfortunately there is no closed-form expression in terms of elementary functions, for that integral but the result can be expressed through a Bessel function. So the result is: \begin{eqnarray} &&U_{N,T}(t)= \frac{2^{T-2} \pi^{T}}{(T-2)!} \frac{1}{2} \int\limits_{{\mathcal R}_+^2} \left(e^{\frac{\imath t s_1}{T}} + e^{\frac{\imath t s_2}{T}}\right) \left(s_1 s_2\right)^{\frac{T-3}{2}} \left|s_1-s_2\right| \cdot \\ &&e^{-\frac{1}{4} \frac{C_{1,1}+C_{2,2}}{\det(C)}(s_1+s_2)} {\mathcal I}_0\left(\frac{1}{4} \frac{\sqrt{(C_{1,1}-C_{2,2})^2 + 4 C_{1,2}^2}}{\det(C)}(s_1-s_2)\right) \frac{d s_1 d s_2}{\sqrt{(2 \pi)^{2 T} \left(\det(C)\right)^T}} \end{eqnarray} I do not think that there exists any closed form solution for the integral above and therefore I leave it as it is. Yet, we will see that that result can be readily used to derive spectral moments of the eigenvalue density. \begin{equation} m_{p} := \frac{d^p U_{N,T}(t)}{d (\imath t)^p} = \int\limits_0^\infty \lambda^p \rho(\lambda) d\lambda \end{equation} for $p=0,1,2,3,\cdots$. We will derive those spectral moments in a future post.

Now, we will compute the spectral moments. We rewrite the double integral over the eigenvalues by absorbing a constant into the $s$-values. We have: \begin{eqnarray} &&U_{N,T}(t) = \frac{2^{2 T-3}}{(T-2)!} \left(\frac{\sqrt{\det C}}{Tr C}\right)^T \\ &&\int\limits_{{\mathcal R}_+^2} \left( \sum\limits_{\xi=1}^2 e^{\imath \frac{t}{T} \frac{4 \det C}{Tr C} s_\xi} \right) \left(s_1 s_2\right)^{\frac{T-3}{2}}\left|s_1-s_2\right| {\mathcal I}_0\left(\frac{\sqrt{(C_{1,1}-C_{2,2})^2 + 4 C_{1,2}^2}}{Tr C}(s_1-s_2)\right) e^{-s_1-s_2} d s_1 d s_2 \end{eqnarray} Now we expand the Bessel function into a power series: \begin{equation} {\mathcal I}_0(x) = \sum\limits_{m=0}^\infty \frac{1}{(m!)^2} (\frac{x}{2})^{2 m} \end{equation} about the origin and we integrate term by term over the eigenvalues. In doing that we use the following identity: \begin{eqnarray} &&\int\limits_0^\infty \int\limits_0^\infty (s_1 s_2)^p \left(s_1^q + s_2^q\right) \left|s_1-s_2\right| (s_2-s_1)^{2 m} e^{-s_1-s_2} d s_1 d s_2 = \\ &&\frac{m!}{2^{2 p+q}} \cdot \frac{(p+q)!(2p+2 m+2+q)!}{(m+p+q+1)!} \cdot F_{3,2} \left[ \begin{array}{rrr} 1+m & \frac{1+q}{2} & \frac{q}{2} \\ \frac{1}{2} & 2+m+p+q \end{array}; 1 \right] \end{eqnarray} where $p \in {\mathbb R}_+$, $m\in {\mathbb N}$ and $q \ge -p$ and $q \in {\mathbb Z}$.

Now, having integrated over the eigenvalues we end up with an infinite series in $m$. We find a closed form for that series expansion using an identity derived in here Interesting combinatoral identity .This gives us a neat closed form expression for the $q$th spectral moment (where $q \ge 0$). For even values of $q$ we have: \begin{eqnarray} &&m_{2 q}= \frac{1}{T^{2q-1}} \sum\limits_{q_1=0}^{q} \left(\frac{2 q}{2 q- q_1}\right) \binom{2 q - q_1}{q_1} (-1)^{q_1} {\mathcal W}^{(2q-1)}\left(T\right) \cdot \left(Tr C\right)^{2q-2 q_1} \left(\det C\right)^{q_1} \end{eqnarray} and for odd values of $q$ we have: \begin{eqnarray} &&m_{2 q+1}= \frac{1}{T^{2 q}} \sum\limits_{q_1=0}^{q} \left(\frac{2q+1}{2 q+1-q_1}\right) \binom{2 q+1-q_1}{q_1} (-1)^{q_1} {\mathcal W}^{(2 q)}\left(T\right) \cdot \left(Tr C\right)^{2 q+1-2 q_1} \left(\det C\right)^{q_1} \end{eqnarray} In both the even and the odd case we have: \begin{equation} {\mathcal W}^{(q-1)}\left(T\right) := \sum\limits_{q_1=0}^{q} \prod\limits_{j=1}^{q-q_1-1} \left(T+ 2 j\right) \cdot \prod\limits_{j=q-q_1}^{q-1} \left(T+ 2 j-1\right) \end{equation} Using Mathematica we have finally checked that the following identities hold: \begin{eqnarray} \frac{1}{2} m_0 &=& 1\\ \frac{1}{2} m_1 &=& M_1 \\ \frac{1}{2} m_2 &=& M_2\left(1 + \frac{1}{T}\right) + r M_1^2\\ \frac{1}{2} m_3 &=& M_3\left(1 + \frac{3}{T} + \frac{4}{T^2}\right) + 3 r \left(1+\frac{1}{T}\right) M_2 M_1 + r^2 M_1^3\\ \frac{1}{2} m_4 &=& M_4\left(1+ \frac{6}{T} + \frac{21}{T^2} + \frac{20}{T^3}\right) + 4 r \left(1+\frac{3}{T}+\frac{4}{T^2}\right) M_3 M_1 + r \left(2+\frac{5}{T}+\frac{5}{T^2}\right) M_2^2 + 6 r^2 \left(1+\frac{1}{T}\right) M_2 M_1^2 + r^3 M_1^4 \end{eqnarray} where $M_p := Tr\left[ C^p \right]/N$ and $r := N/T$. The identities above match-- except for the multiplicative factor $1/2$ on the left hand side -- those known from literature and derived using a diagrammatical method based on Wick's theorem.

Update: Now we derive the negative spectral moments. Take $p\in {\mathbb N}$ and then from the very definition in the body of the question we have: \begin{equation} m_{-p} := \int\limits_0^\infty \frac{t^{p-1}}{(p-1)!} \cdot U_{N,T}(\imath t) dt \end{equation}