Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a probability distribution function $F(x)$, consider other probability distribution functions $F_1$ and $F_2$ such that $aF_1(x)+bF_2(x)=F(x)$ for some $a,b$ for all $x$. Under what conditions on $F_1$ and $F_2$ we have $F_1(x)(1-F_1(x))+F_2(x)(1-F_2(x)) \ge F(x)(1-F(x)) $?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

In order for $a F_1 + b F_2$ to be a probability distribution function, you need $a + b = 1$. I'll assume you're interested in the case $0 < a < 1$. If $F = a F_1 + (1 - a) F_2$, then $G = F_1 (1-F_1) + F_2 (1 - F_2) - F (1 - F) = (F_1 - F_2)^2 a^2 + (F_1 - F_2)(2 F_2 - 1) a + F_1 - F_1^2$. Now certainly the $a^2$ and constant terms are nonnegative. So one sufficient condition is that $(F_1 -F_2) (2 F_2 - 1) \ge 0$, i.e. either $F_1 \ge F_2 \ge 1/2$ or $F_1 \le F_2 \le 1/2$.
By symmetry, it also is true if $F_2 \ge F_1 \ge 1/2$ or $F_2 \le F_1 \le 1/2$. On the other hand, when $F_2 < 1/2 < F_1$ or $F_1 < 1/2 < F_2$, the minimum of $G$ is at $a = \frac{1/2 - F_2}{F_1 - F_2}$ where we get $G =F_1 - F_1^2 + F_2 - F_2^2 - 1/4$. Note that the curve $F_1 - F_1^2 + F_2 - F_2^2 = 1/4$ is a circle of radius $1/2$ centred at $(1/2,1/2)$. So the condition to have $G \ge 0$ for all $0 \le a \le 1$ is that $(F_1, F_2)$ avoids the two regions of the unit square outside that circle that are shown in red in this plot.

enter image description here

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.