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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?

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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.

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  • $\begingroup$ 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 $\Omega_{J-1}$. $\endgroup$ – user3285148 Mar 4 at 11:40
  • $\begingroup$ 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 $\mu_{-j}$. $\endgroup$ – user3285148 Mar 4 at 11:57
  • $\begingroup$ 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. $\endgroup$ – Ben Mar 4 at 12:18
  • $\begingroup$ 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? $\endgroup$ – user3285148 Mar 4 at 12:35
  • $\begingroup$ 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. $\endgroup$ – user3285148 Mar 5 at 12:53

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