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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.

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  • $\begingroup$ 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. $\endgroup$ – jbowman Mar 9 at 16:53
  • $\begingroup$ 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)? $\endgroup$ – user3285148 Mar 9 at 16:56

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