 Solution

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.