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?


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