0

X has a discrete distribution with support $x1, x2, ...$ in $ {]}0,1{[}$. You have the right to change only one of the $xi$ to lead to the highest increase in variance (or, at least, a systematic increase in variance). How do you describe this change?

Background of the question: I was trying to describe variance-increasing transformations to prove Popoviciu's inequality: X has maximal variance when it takes only two values, 0 and 1, with probability $\frac{1}{2}$. I hoped I could find a proof by defining a suite of transformations leading to such a distribution and increasing variance every time. With this in mind, I initially thought of this function, changing $xp$ only, as described:

\begin{equation} f(xi)= \begin{cases} x & \text{if }i \neq p \\ 1 & \text{if }E(X) < \frac{1}{2} \text{ & } i = p \\ 0 & \text{if }E(X) > \frac{1}{2} \text{ & } i = p\\ \end{cases} \] \end{equation}

However, this does not increase the variance every time if xp is chosen randomly...

GabCaz
  • 1
  • 2
  • 1
    I would guess taking the value closest to the mean and moving it as far away from the mean as possible would be a good starting point. If the probabilities aren't equal, you'd want to involve them in the choice of which point to move as well. – jbowman Jan 15 '20 at 16:34

1 Answers1

0

Thank you for your advice! Denoting the support of random variable X with {x1, x2, ...}, the function I found is defined with, for each p: S(xi) = xi if i is not p (this means we only change one element indexed p at a time). S(xp) = 1 if E(X) < x(p) * [1 + f(xp)] / 2 ; S(xp) = 0 otherwise

The way I found this function is by writing Var(S(X)) - Var(X) as a function of S(xp), E(X) and xp, and then by defining S for Var(S(X)) to be always greater than Var(X).

GabCaz
  • 1
  • 2