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

Take a random variable $Y$ with cdf $\Delta(\cdot ; \lambda, \mu)$ and the variable $Y'$ with cdf $\Delta(\cdot; \tilde{\lambda}, \tilde{\mu})$ with $Y$ stochastically independent of $Y'$.

Consider the random variable $W\equiv Y-Y'$.

Let $P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$ be the probability mass function of $W$.

We know that if $\lambda=\tilde{\lambda}$ and $\mu=\tilde{\mu}$, then $P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$ is symmetric around zero.

Question: Suppose that $P(\cdot; \lambda, \mu)$ is symmetric (not necessarily around zero). Show that $P(\cdot; \lambda,\tilde{\lambda},\mu,\tilde{\mu})$ can be symmetric around zero for some $\tilde{\lambda}\neq \lambda$, $\tilde{\mu}\neq \mu$.

My attempt: Naively, I think that we can find some step CDF with $\tilde{\lambda}\neq \lambda$, $\tilde{\mu}\neq \mu$ that is "shifting" $P(\cdot; \lambda, \delta)$ towards zero but I would like your help to formalise this.

  • 1
    $\begingroup$ Hint: Consider the case when both $\Delta(\cdot; \lambda,\mu)$ and $\Delta(\cdot; \tilde\lambda,\tilde\mu)$ are symmetric around the same point $\xi$. $\endgroup$ – Xi'an Feb 22 at 20:41

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