solution to mixture CDF / inverse CDF of finite mixture

I am currently numerically solving the follwing equation, which is a convex combination/finite mixture of two marginal CDFs $F(x)$ and $G(x)$ (actually they are joint CDFs but all arguments but x are fixed) on the left hand side, for x: $$ωF(x)+(1−w)G(x)=a$$ where $a$ is fixed but I want to obtain solutions fo different $w∈[0,1]$. Furthermore are both $F(.)$ and $G(.)$ strictly increasing (because I am working with continous CDFs) and mapping from [0,1]. In my current example it seems that the solution $x0$ satisfies: $x_0=x_{0,ω=0}+ω(x_{0,ω=1}−x_{0,ω=0})$ where $x_{0,ω=0}$ is the solution for $ω=0$, that is the solution to $G(x)=a$, and $x_{0,ω=1}$ is defined accordingly. Both solutions exist. Is my observation true in general for the given assumptions? I have tried to show $ωF(x_0)+(1−w)G(x_0)=a$ but could not figure out how to. Alternatively, if I had an expressing for the inverse CDF of the mixture I could try to show that this equals $x_0$ for $a$.

Your relation $x_0=x_{0,ω=0}+ω(x_{0,ω=1}−x_{0,ω=0})$ will not hold in general. If $F(x)$ and $G(x)$ were linear over the same domain, but were different (which can not happen with a Uniform due to difference in domain if distributions aren't identical), then it would be true for all parameter values. The relationship might happen to work in particular instances of $F(x)$, $G(x)$, and $a$. What is your current example for which you believe the relation holds?

Here's a simple counterexample:

Let $F(x)$ be exponential with parameter value = 1, i.e., $F(x) = 1-exp(-x)$ for $x\ge 0$.

Let $G(x)$ be exponential with parameter value = 2, i.e., $G(x) = 1-exp(-2 x)$ for $x\ge 0$.

Let $a$ = 0.5

When $ω$ = 1, the solution is 0.6931.

When $ω$ = 0, the solution is half of that, namely 0.3466.

When $ω$ = 0.5, the solution is 0.4812, which is not halfway between the $ω$ = 0 and $ω$ = 1 solutions. This contradicts your assumed relation (as I am interpreting it).

• Thanks! Do you have an idea under which conditions it woould holds (at least approximately?) – InfiniteVariance Apr 18 '16 at 9:30
• @InfiniteVariance , the closer to linearity the CDFs, and therefore the inverse CDFs, are in the region of the solution, the better the approximation will hold. – Mark L. Stone Apr 18 '16 at 11:59