Showing that if the PMF of $W$ is symmetric around zero then some parameters entering it are equivalent

Summary: In what follows, I specify the probability mass function (PMF) of a random variable $$W$$, depending on some parameters $$(\lambda,\mu,\lambda',\mu')$$. I would like your help to show that $$\text{If the PMF of W is symmetric around zero then \mu=\mu' and \lambda=\lambda'}$$

More details on the structure of the problem:

Let $$P_Y:\mathbb{R}\rightarrow [0,1]$$ be the PMF of $$Y$$, prescribed by $$\begin{cases} P(\mu_1)=\lambda_1\\ P(\mu_2)=\lambda_2\\ P(\mu_3)=\lambda_3\\ P(\mu_4)=\lambda_4\\ P(d)=0 & \forall d\neq \mu_1,\mu_2,\mu_3,\mu_4 \end{cases}$$ where $$\mu\equiv (\mu_1,\mu_2,\mu_3,\mu_4)$$, $$\mu_1<\mu_2<\mu_3<\mu_4$$, $$\lambda\equiv (\lambda_1,\lambda_2,\lambda_3,\lambda_4)$$, $$\lambda_1\neq 0$$, $$\lambda_2\neq 0$$, $$\lambda_3\neq 0$$, $$\lambda_4\neq 0$$.

Let $$P_{Y'}:\mathbb{R}\rightarrow [0,1]$$ be the PMF of $$Y'$$ structured as above but using $$\mu'\equiv (\mu_1', \mu_2',\mu_3', \mu_4')$$ and $$\lambda'\equiv (\lambda_1', \lambda_2',\lambda_3', \lambda_4')$$, potentially different from $$\mu, \lambda$$. As above, $$\mu_1'<\mu_2'<\mu_3'<\mu_4'$$, $$\lambda_1'\neq 0$$, $$\lambda_2'\neq 0$$, $$\lambda_3'\neq 0$$, $$\lambda_4'\neq 0$$.

Let $$W\equiv Y-Y'$$.

Assume

1) $$\mu_2-\mu_1=\mu_2'-\mu_1'\equiv A$$

2) $$\mu_4-\mu_3\equiv c>A$$

3) $$\mu_4'-\mu_3'\equiv f>A$$

In what follows I will also use $$b\equiv \mu_3-\mu_2$$ and $$e\equiv \mu_3'-\mu_2'$$.

Claim to show: $$\text{If the PMF of W is symmetric around zero then \mu=\mu' and \lambda=\lambda'}$$

My attempts and thoughts: I have started to approach this problem by writing down in a matrix the potential support points of the PMF of $$W$$ together with some ordering considerations. $${\tiny \begin{pmatrix} \mu_1-\mu_1'& < & \mu_1-\mu_1'-A & < & \mu_1-\mu_1'-A -e & < & \mu_1-\mu_1'-A-e-f\\ \wedge & & \wedge & & \wedge & & \wedge\\ \mu_1+A-\mu_1'& < & \mu_1+A-\mu_1'-A & < & \mu_1+A-\mu_1'-A -e & < & \mu_1+A-\mu_1'-A-e-f\\ \wedge & & \wedge & & \wedge & & \wedge\\ \mu_1+A+b-\mu_1'& < & \mu_1+A+b-\mu_1'-A & < & \mu_1+A+b-\mu_1'-A -e & < & \mu_1+A+b-\mu_1'-A-e-f\\ \wedge & & \wedge & & \wedge & & \wedge\\ \mu_1+A+b+c-\mu_1'& < & \mu_1+A+b+c-\mu_1'-A & < & \mu_1+A+b+c-\mu_1'-A -e & < & \mu_1+A+b+c-\mu_1'-A-e-f\\ \end{pmatrix}}$$

Let $$\eta_1,...,\eta_m$$ be the support points of $$W$$ associated with non-zero probabilities $$p_1,...,p_m$$.

Under the assumptions above, we know that $$\begin{cases} \eta_1\equiv \mu_1-\mu_1'-A-e-f\\ p_1\equiv \lambda_1\lambda_4'\\ \eta_m\equiv \mu_1+A+b+c-\mu_1'\\ p_m\equiv \lambda_4\lambda_1'\\ \eta_2\equiv \mu_1+A-\mu_1'-A-e-f\\ p_2\equiv \lambda_2\lambda_4'\\ \eta_{m-1}\equiv \mu_1+A+b+c-\mu_1'-A \\ \lambda_{m-1}\equiv \lambda_4\lambda_2' \end{cases}$$ $$\overbrace{\Downarrow}^{\text{If PMF of W symmetric at 0}}$$ $$\begin{cases} \mu_1-\mu_1'-A-e-f=-(\mu_1+A+b+c-\mu_1')\\ \lambda_1\lambda_4'=\lambda_4\lambda_1'\\ \mu_1+A-\mu_1'-A-e-f=-(\mu_1+A+b+c-\mu_1'-A )\\ \lambda_2\lambda_4'=\lambda_4\lambda_2' \\ (\lambda_1+\lambda_2+\lambda_3+\lambda_4=1, \lambda'_1+\lambda'_2+\lambda'_3+\lambda'_4=1) \end{cases}$$

I tried to work on this system but I couldn't show that it implies $$\mu=\mu'$$ and $$\lambda=\lambda'$$ (that is, in other words, $$\lambda=\lambda'$$, $$\mu_1=\mu_1'$$, $$b=e$$, $$c=f$$).

I hence attempted to determine $$\eta_3, \eta_{m-2}$$ bu that requires to distinguish whether $$f>,=,< A+b$$ and $$c>,=,< A+e$$ and the resulting number of cases to analyse is huge.

I'm therefore thinking that the route I took is maybe not efficient and I was wondering whether you could see faster strategies.

• Given your specification of $\mu$, I don't see how it can be symmetric. Obviously for symmetry $\mu_1 < \mu_2 \leq 0 \leq \mu_3 < \mu_4$, and it must be that $\mu_4 = -\mu_1$ and similarly for $\mu_3$ and $\mu_2$. This implies that $\mu_4 - \mu_3 = \mu_2 - \mu_1$, which is contradicted by your assumptions. – jbowman Mar 9 at 16:53
• Your conditions ensure that $Y$ has a PMF symmetric around zero. Why is that necessary for $W\equiv Y-Y'$ to have a PMF symmetric around zero (which is instead what I'm focused on)? – user3285148 Mar 9 at 16:56