0
$\begingroup$

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 $a$.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Thanks! Do you have an idea under which conditions it woould holds (at least approximately?) $\endgroup$ – InfiniteVariance Apr 18 '16 at 9:30
  • $\begingroup$ @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. $\endgroup$ – Mark L. Stone Apr 18 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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