# Show that the intersection of two sets involving symmetric PMF is empty

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

• $$J<\infty$$

• $$\lambda\equiv (\lambda_1,...,\lambda_J)$$, $$\lambda_j\in [0,1]^J$$, and $$\sum_{j=1}^J \lambda_j=1$$.

• $$\Lambda\equiv \{\lambda\in \mathbb{R}^J: \text{\lambda_j\in [0,1]^J, and \sum_{j=1}^J \lambda_j=1}\}$$

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

• $$M\equiv \{\mu\in \mathbb{R}^d: \text{\mu_1<...<\mu_J}\}$$

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

Consider two independent random variables $$Y, Y'$$ respectively with CDF $$\Delta(\cdot; \lambda,\mu)$$ and $$\Delta(\cdot; \lambda',\mu')$$.

Let $$\Delta(\cdot; \lambda,\mu)\star\Delta^{-}(\cdot; \lambda',\mu')$$ denote the CDF of $$Y-Y'$$.

Let $$P(\cdot; \lambda,\mu)\star P^{-}(\cdot; \lambda',\mu')$$ denote the PMF of $$Y-Y'$$.

Consider the sets $$\mathcal{B}_J\equiv \{(\lambda, \mu)\in \Lambda\times M: P(\cdot; \lambda,\mu) \text{ is symmetric}\}$$ $$\Omega_J\equiv \Big\{(\lambda, \mu)\in \Lambda\times M: \\ \forall (\lambda', \mu')\in \Omega_J, P(\cdot; \lambda,\mu)\star P^{-}(\cdot; \lambda',\mu') \text{ symmetric around zero implies } \\(\lambda,\mu)=(\lambda', \mu')\Big\}$$

Question: I have to show that $$\mathcal{B}_j\cap \Omega_J=\emptyset$$.

However, I'm not actually sure that the claim is correct, given the "recursive" definition of $$\Omega_J$$. That is, $$\Omega_J$$ is obtained by selecting the elements of $$\Lambda\times M$$ such that, $$\forall (\lambda,\mu),(\lambda', \mu')$$ within such a selected set, $$P(\cdot; \lambda,\mu)\star P^{-}(\cdot; \lambda',\mu')$$ symmetric around zero implies $$(\lambda,\mu)=(\lambda', \mu')$$.

Take for example $$J=2$$.

Consider $$(\lambda_1,\lambda_2)=(\frac{1}{2},\frac{1}{2})$$ and $$(\mu_1,\mu_2)=(3,10)$$, so that $$(\lambda,\mu)\in \mathcal{B}_2$$.

Then, $$\Delta_2(\cdot; \lambda,\mu)\star \Delta^{-}_2(\cdot; \lambda',\mu')$$ can be made zero-symmetric by any $$(\lambda', \mu')\in \mathcal{B}_2$$ with $$\lambda_1'=\lambda_2'=\frac{1}{2}$$ and $$\mu_2'=13-\mu_1'$$. For example, one ca set $$(\lambda', \mu')\in \mathcal{B}^*_2$$ with $$\lambda_1'=\lambda_2'=\frac{1}{2}$$, $$\mu_2'=6$$, $$\mu_1'=7$$.

If we include into $$\Omega_2$$ $$(\lambda_1,\lambda_2)=(\frac{1}{2},\frac{1}{2})$$ and $$(\mu_1,\mu_2)=(3,10)$$, but we exclude every $$(\lambda', \mu')\in \{(\lambda,\mu)\in \Lambda\times M: \lambda_1'=\lambda_2'=\frac{1}{2}\text { and }\mu_2'=13-\mu_1' \text{ and } \mu_1'\neq 3\text{ and } \mu_2'\neq 10\}$$, then I think we are still consistent with the definition of $$\Omega_2$$ and we have that $$\Omega_2\cap \mathcal{B}_2 \neq \emptyset$$.

What is correct?