2

Take two distributions $F_B(x)$, $F_A(x)$ with the same support. Assume that B is a mean-preserving spread of A.

What I want to understand is whether $E_{A}[x | x \leq t] \geq E_{B}[x | x \leq t]$, but I'm struggling. From the definition of mean-preserving spread:

$\int_{-\infty}^{t} F_B(x) dx \geq \int_{-\infty}^{t} F_A(x) dx \quad \leftrightarrow \quad \int_{-\infty}^{t} 1 - F_B(x) dx \leq \int_{-\infty}^{t}1 - F_A(x) dx$

$ \quad \qquad \qquad \qquad \qquad \qquad \qquad \quad \leftrightarrow \quad E_B [x | x\leq t]F_B(t) \leq E_A [x | x\leq t]F_A(t)$.

Unfortunately $F_B(t)$, $F_A(t)$ cannot generally be ordered. Do I need some more stringent conditions (eg: $t< E[x]$)?

user313975
  • 21
  • 1
  • Update: Since B is a mean-preserving spread of A, $F_B$ and $F_A$ must cross at least once. Take the smallest crossing point and call it $t_1$: then $\forall t \leq t_1$ we'll have $F_A(t) \leq F_B(t)$ so at least for $t \leq t_1$ the statement must be true. – user313975 Mar 12 '21 at 02:10

0 Answers0