 Solution

For an integrable discrete random variable, its expectation and variance are $\mu ={\sum }_{n\in \mathbb{Z}}n{p}_n$ and ${\sum }_{n\in \mathbb{Z}}(n-\mu {)}^2{p}_n$, respectively. The first formula requires $\sum |n|p_n<\infty$ for a finite ordinary mean; the variance may be infinite. A probability density function gives $P(\alpha <Y<\beta )={\int }_{\alpha }^{\beta }f(y)dy$.

Conditioning on $X$ and using independence givesThe interchange is justified by Tonelli's theorem, since all integrands are nonnegative. Consequently $Z$ has an absolutely continuous distribution, withIts integral is ${\sum }_n{p}_n=1$. This is the convolution of independent random variables, specifically a Gaussian smoothing of an integer-valued distribution for the given normal density. The specified weights are normalized because $2(3/{\pi }^2){\sum }_{n\ge 1}{n}^{-2}=1$.