Some expectation values in the Gaussian orthogonal ensemble

I calculate some expectation values if the probability measure is given by \begin{equation*} P(H) dH = \mathcal{N} \prod_{ i \le j} dH_{ij} \ \exp \left( -\frac{1}{2}\ \mathrm{tr}(H^2)\right) \end{equation*} Hereby are \( H \) symmetric \( n \times n \) matrices and \(\mathcal{N}\) is the normalization factor. This is a special case of the Gaussian orthogonal ensemble (GOE).


If \( n =2 \) for example, the exponent is \begin{equation*} -\frac{1}{2} \sum_{i=1}^2 \sum_{j=1}^2 H_{ij} H_{ji} = -\frac{1}{2} \left( H_{11}^2 + 2 H_{12}^2 + H_{22}^2 \right) \end{equation*} Therefore, calculating the Gaussian integrals gives \begin{align*} \mathbb{E} [ H_{ij}] &= 0 \quad\text{for all } i,j\\ \mathbb{E} [ H_{11}^2] & = \mathbb{E} [ H_{22}^2] = 1\\ \mathbb{E} [ H_{12}^2] & = \mathbb{E} [ H_{21}^2] = \frac{1}{2} \end{align*} A similar calculation for \( n \times n \) matrices gives \begin{equation}\label{eq:20151029a} \mathbb{E} [ H_{ij}] = 0 \quad\text{for all } i,j \end{equation} \begin{align*} \mathbb{E} [ H_{ii}^2] & = 1\\ \mathbb{E} [ H_{ij}^2] & = \frac{1}{2}\quad\text{if } i\neq j \end{align*} I combine this to \begin{equation}\label{eq:20151029b} \mathbb{E} [ H_{ij} H_{kl} ] = \frac{1}{2} \left( \delta_{ik} \delta_{jl} + \delta_{il} \delta_{jk}\right) \end{equation} From \eqref{eq:20151029a} one can then for example calculate that \begin{equation*} \mathbb{E}[ \mathrm{tr} H ] =\mathbb{E}[ \sum_i H_{ii} ] = 0 \end{equation*} and \begin{equation*} \mathbb{E}[ \mathrm{tr} (H^2) ] = \mathbb{E}[ \sum_{ij} H_{ij}H_{ji} ] = \sum_i \mathbb{E} [ H_{ii}^2 ] + \sum_{i \neq j} \mathbb{E}[H_{ij}^2]= n + n (n-1) \frac{1}{2} \end{equation*} thus \begin{equation}\label{eq:20151029c} \mathbb{E}[ \mathrm{tr} (H^2) ] = \frac{1}{2} n (n+1) \end{equation} I can also calculate expectation values of matrix functions. For example \begin{equation*} \mathbb{E}[ (H^2)_{ij} ] = \mathbb{E}[ \sum_k H_{ik}H_{kj} ] = \sum_k \frac{1}{2} \left( \delta_{ik} \delta_{kj} + \delta_{ij} \delta_{kk}\right)= \frac{1}{2} \left( \delta_{ij} + n \delta_{ij} \right) = \frac{1}{2} (n+1 ) \delta_{ij} \end{equation*} thus \begin{equation}\label{eq:20151029d} \mathbb{E}[ H^2 ] = \frac{1}{2} (n+1 )1 \end{equation} where \( 1 \) is the \( n \times n \) unit matrix.

Remarks

• If I take the trace of equation \eqref{eq:20151029d} I get \begin{equation*} \mathbb{E}[ \mathrm{tr} (H^2) ] = \frac{n+1}{2} \mathrm{tr} (1) = \frac{n+1}{2} n \end{equation*} This agrees with \eqref{eq:20151029c}
• Because the GOE is invariant under orthogonal transformations, one has \begin{equation*} \mathbb{E}[ H^2 ] =\mathbb{E}[ ( R H R^T) ^2 ] = \mathbb{E}[R H R^T R H R^T ] = R \mathbb{E}[ H^2 ] R^T \end{equation*} this is indeed satisfied by expression \eqref{eq:20151029d}.

I can also calculate expectation values of scalar products. Suppose \( v \) is a vector, then \begin{equation*} \mathbb{E} [ v^T H v ] = v^T\mathbb{E} [H ] v = 0 \end{equation*} and similarly if \( m \) is an odd integer \begin{equation*} \mathbb{E} [ \left( v^T H v \right)^m ] = 0 \end{equation*} For even integers \(m \), I get for example \begin{equation*} \mathbb{E} [ \left( v^T H v \right)^2 ] = \mathbb{E} \left[ \left(\sum_{ij} v_i H_{ij} v_j \right)^2 \right] = \sum_{ijkl} v_i v_j v_k v_l \mathbb{E} [ H_{ij} H_{kl} ] = \frac{1}{2} \left( v^2 \cdot v^2 + v^2 \cdot v^2 \right) = || v || ^4 \end{equation*} With more work, using Wick's formulas for example, I find \begin{equation*} \mathbb{E} [ \left( v^T H v \right)^4 ] = 3 ||v ||^4 \end{equation*} I also want to calculate \begin{equation*} \mathbb{E} \left [ e^ {v^T H v} \right] \end{equation*} This is perhaps most easily done as follows. First observe that \(v^T H v \) is a Gaussian random variable. Then use some properties of Gaussian random variables to calculate \begin{equation*} \mathbb{E} \left [ e^ {v^T H v} \right] = \exp \left[ \mathbb{E} \left [ v^T H v \right] + \frac{1}{2} \mathrm{Var} \left [ v^T H v \right] \right] = \exp \left[\frac{1}{2} \mathbb{E} \left [ (v^T H v)^2 \right]\right] \end{equation*} thus \begin{equation}\label{eq:20151029e} \mathbb{E} \left [ e^ {v^T H v} \right] = e^{\frac{1}{2} ||v||^4} \end{equation}