# Characterise the set of symmetric probability mass functions

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 $$\Delta(\cdot; \lambda,\mu)$$.

Consider the set $$\Omega_J\equiv \{(\lambda, \mu): P(\cdot; \lambda,\mu) \text{ is symmetric}\}$$

Question: is it possible to explicitly characterise the set $$\Omega_J$$ through conditions on $$(\lambda, \mu)$$ for any generic $$J$$?

For example, $$\Omega_2=\Big\{(\lambda,\mu): \lambda_1=1\text{ and }\lambda_2=0\text{, or }\lambda_1=0\text{ and }\lambda_2=1\text{, or } \lambda_1=1/2\text{ and }\lambda_2=1/2\Big\}$$

$$\Omega_3=\Big\{(\lambda,\mu): \lambda_1=\lambda_3 \text{ and } \mu_2-\mu_1=\mu_3-\mu_2\text{, or } \lambda_j=0 \text{ for some j\in \{1,2,3\}} \Big\}$$ I'm struggling to generalise these characterisations to any $$J$$. Any help?

Here are the required conditions: Symmetric mass functions (over a finite support) can be characterised by symmetry of support values around a mean/median and equality of probability values reflected around that mean/median. Thus, in order to have a symmetric mass function the vector $$\boldsymbol{\mu}$$ must satisfy the requirement:$$^\dagger$$

$$\mu_k - \mu_{k'} = \mu_{J-k'+1} - \mu_{J-k+1} \quad \quad \quad \text{for all } k,k' = 1,...,J,$$

and the vector $$\boldsymbol{\lambda}$$ must satisfy the requirement:

$$\lambda_k = \lambda_{J-k+1} \quad \quad \quad \text{for all } k = 1,...,J.$$

These two conditions exhaust all symmetric mass functions with a finite number of outcomes $$J$$. The mean/median of the resulting mass function is:

$$\mathbb{E}(X) = \text{Median}(X) = \frac{\mu_k + \mu_{J-k+1}}{2} \quad \quad \quad \text{for all } k = 1,...,J.$$

$$^\dagger$$ This first condition can be written alternatively as:

$$(\exists \mu_* \in \mathbb{R}) (\forall k = 1,...,J): \quad \mu_k + \mu_{J-k'+1} = 2 \mu_*.$$

In this alternative statement of the condition we use the value $$\mu_* = \mathbb{E}(X) = \text{Median}(X)$$, so the condition explicitly invokes an existence condition on the mean/median.

• Thanks. I think it is implicit in your statement but I want to be sure and double check: the conditions you provide are referred to the case in which all the $\lambda_j$ are different from zero; if one the $\lambda_j$ is equal to zero, then we "go back" to the conditions characterising $\Omega_{J-1}$.
– Star
Mar 4, 2019 at 11:40
• In other words, I think that \begin{aligned} \Omega^*_J=&\Big\{(\lambda,\mu): \lambda_j=0\text{ for some j\in \{1,...,J\} and (\lambda_{-j},\mu_{-j})\in \Omega^*_{J-1}} \Big\}\\ &\cup \\ &\Big\{ \{\lambda: \lambda_j=\lambda_{J-j+1} \text{ for j\in \{1,...,J\}}\}\times \{\mu: \mu_j-\mu_{j'}=\mu_{J-j'+1}-\mu_{J-j+1} \text{ for j,j'\in \{1,2,...,J\}}\} \Big\} \end{aligned} where $\lambda_{-j}$ is the vector $\lambda$ without the $j$th element and similar definition applies to $\mu_{-j}$.
– Star
Mar 4, 2019 at 11:57
• The conditions I have used are still okay if one or more of the $\lambda_i$ are zero. In this case the conditions would simply require that the other element reflected around the mean/median also has zero probability. So you needn't make any exception for that case.
– Ben
Mar 4, 2019 at 12:18
• Consider $J=3$. The set that you characterise is $A\equiv \{\lambda: \lambda_1=\lambda_3\}\times \{\mu: \mu_2-\mu_1=\mu_3-\mu_2\}$. Suppose that $\lambda_3=0$. Then $\lambda=(1/2,1/2,0)$ and $\mu=(1,100,1000)$ generates a symmetric PMF but it is not in the set $A$ that you have characterised. Where am I wrong?
– Star
Mar 4, 2019 at 12:35
• I think I have some basic confusion on the definition of symmetric PMF, which I tried to clarify here stats.stackexchange.com/questions/395753/…. Given the answer I received, I think it makes sense to take the union of the sets as I wrote. If instead you believe is redundant, could you clarify by considering the example $J=3$? Thanks.
– Star
Mar 5, 2019 at 12:53