In this paper, in proposition 2, the covariance matrix of the periodogram estimator is computed for the Gaussian process. However, at the end of proof says "upon substituting (52) and (54) into (51), the results follows". I do not know how they obtain that result Eq. (15).
May someone elaborate on this derivation?
Splitting the summation in $(51)$ into the terms where $r\neq r'$ and those where $r=r'$ gives that \begin{align*} \Sigma_{pg}&=\mathbb E[\hat{p}_{pg}\hat{p}_{pg}^H]-pp^H\\&=\frac1{R^2}\sum_{r=1,r'=1}^R\mathbb E[\text{diag}(V^Hx_rx_r^HV)\text{diag}(V^Hx_rx_r^HV)^H]-pp^H\\&=\frac{1}{R^2}\left(\sum_{r\neq r'}\mathbb E[\text{diag}(V^Hx_rx_r^HV)\text{diag}(V^Hx_rx_r^HV)^H]+\sum_{r=r'}\mathbb E[\text{diag}(V^Hx_rx_r^HV)\text{diag}(V^Hx_rx_r^HV)^H]\right)-pp^H\\&=\frac1{R^2}\left(\sum_{r\neq r'}pp^H+\sum_{r=r'}2(\text{diag}^2(p)+pp^H)\right)-pp^H\\&=\frac1{R^2}\left((R^2-R)pp^H+R(2\text{diag}^2(p)+pp^H)\right)-pp^H\\&=\frac2R\text{diag}^2(p), \end{align*} which is exactly $(15)$.