"Deconvoluting" the sum of independent random variables

Question

I encountered the following question in my research. See here for some background, but this post should be self-contained. Suppose

• $X$ follows a discrete uniform distribution on $m$ points $\{X_1, X_2, \cdots, X_m\}$;
• $Z$ is a random variable independent of $X$;
• $Y = X + Z$ has a standard Gaussian distribution.

What's the density function of $Z$?

Attempt:

It is easy to write $p_Y(\cdot)$ as a function of $p_Z(\cdot)$, but what's next?

\begin{equation} \begin{aligned} p_Y(y) =& \frac{1}{m}\sum_{i=1}^{m} p_{Y|X}(y|x_i) \\ =& \frac{1}{m}\sum_{i=1}^{m} p_{X+Z|X}(y|x_i) \\ =& \frac{1}{m}\sum_{i=1}^{m} p_{Z}(y-x_i) \\ \end{aligned} \end{equation}

Answer 1

Lévy–Cramér Theorem tells that if $Y$ is a gaussian random variable and $Y = X + Z$ for some independent random variables $X$ and $Z$, then each of $X$ and $Z$ is also gaussian. In other words, a gaussian distribution cannot be decomposed into the convolution of non-gaussian distributions.

In summary, there is no such $Z$ in your case.