Variance of the logarithm of the mixed Rademacher and complex Gaussian distribution

Question

Consider the scenario where $X$ is a Rademacher random variable taking values $\{−1,+1\}$ with equal probability, and $Z$ is a complex Gaussian random variable with a mean of $0$ and a variance of $\sigma^2$. Let's assume that $Y=X+Z$. Is there any closed-form or approximate relationship available for the variance $\mathrm{Var}_Y\left[\log(p(Y))\right]$, where \begin{align} p(y)=\frac{1}{2\pi\sigma^2}\left(\exp\left(-\frac{(y-1)^2}{\sigma^2}\right)+\exp\left(-\frac{(y+1)^2}{\sigma^2}\right)\right). \end{align}

Answer 1

I mean the procedure would be as follows. Although I am eager to explore other potential approaches.

First, I provide an upper bound for the moment generating function (MGF) $M_{\log p(Y)}(t)$. It is well-known that $\mathbb{E}[X]=M'(t)$ and $\mathbb{E}[X^2]=M''(t)$. \begin{align} M_{\log p(Y)}(t)&=\mathbb{E}_Y\left[\exp\left(t\log\left(\frac{1}{2\pi\sigma^2}\left(\exp\left(-\frac{(Y-1)^2}{\sigma^2}\right)+\exp\left(-\frac{(Y+1)^2}{\sigma^2}\right)\right)\right)\right)\right]\\ &=\mathbb{E}\left[\left(\frac{1}{2\pi\sigma^2}\right)^t\left(\exp\left(-\frac{(y-1)^2}{\sigma^2}\right)+\exp\left(-\frac{(y+1)^2}{\sigma^2}\right)\right)^t\right] \end{align} Since we are interested in the limit behavior of the MGF around $0$, we can consider $0\leq t\leq 1$. Inequality $(x+y)^t\leq x^t+y^t$ yields the following upper bound: \begin{align} &=\left(\frac{1}{2\pi\sigma^2}\right)^t\left(\mathbb{E}\left[\exp(-\frac{t(Y-1)^2}{\sigma^2})\right]+\mathbb{E}\left[\exp(-\frac{t(Y+1)^2}{\sigma^2})\right]\right), \end{align} where each integral is computable using the distribution $p(y)$. This computation can be easily extended to the case where $X$ is randomly chosen from any finite set.