# When is the pmf of the difference of two independent random variables symmetric in zero?

Consider the stepwise cumulative distribution function $$\Delta(x; \lambda, \mu)=\sum_{j=1}^J \lambda_j 1\{x\geq \mu_j\} \hspace{1cm} \forall x \in \mathbb{R}$$ where

• $$J<\infty$$

• $$\lambda\equiv (\lambda_1,...,\lambda_J)$$

• $$\mu\equiv (\mu_1,...,\mu_J)$$

• $$\mu_1<...<\mu_J$$, $$\mu_j\in \mathbb{R}$$ $$\forall j$$

• $$\lambda_j\in [0,1]^J$$ and $$\sum_{j=1}^J \lambda_j=1$$

Let $$P(\cdot; \lambda,\mu)$$ denote the probability mass function associated with the CDF $$\Delta(\cdot; \lambda,\mu)$$.

Take a random variable $$Y$$ with cdf $$\Delta(\cdot ; \lambda, \mu)$$ and the variable $$Y'$$ with cdf $$\Delta(\cdot; \tilde{\lambda}, \tilde{\mu})$$ with $$Y$$ stochastically independent of $$Y'$$.

Consider the random variable $$W\equiv Y-Y'$$.

Let $$P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$$ be the probability mass function of $$W$$.

We know that if $$\lambda=\tilde{\lambda}$$ and $$\mu=\tilde{\mu}$$, then $$P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$$ is symmetric around zero.

Question: Suppose that $$P(\cdot; \lambda, \mu)$$ is symmetric (not necessarily around zero). Show that $$P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$$ can be symmetric around zero for some $$\tilde{\lambda}\neq \lambda$$, $$\tilde{\mu}\neq \mu$$.

My attempt: Naively, I think that we can find some step CDF with $$\tilde{\lambda}\neq \lambda$$, $$\tilde{\mu}\neq \mu$$ that is "shifting" $$P(\cdot; \lambda, \delta)$$ towards zero but I would like your help to formalise this.

• Hint: Consider the case when both $\Delta(\cdot; \lambda,\mu)$ and $\Delta(\cdot; \tilde\lambda,\tilde\mu)$ are symmetric around the same point $\xi$. – Xi'an Feb 22 at 20:41