A set of equations for two (cumulative distribution) functions and their inverses

Question

I am trying to characterize a set of distributions that satisfy two conditions. It is easy to characterize distributions fitting each of those conditions separately, but I am unable to make progress on characterizing the intersection. Specifically:

Let's have $X_0,X_1$ independent random variables over the interval $[0, 1]$ and let $F_0, F_1 : [0,1] \to [0,1]$ be their cumulative distribution functions. We can assume that both $F_0$ and $F_1$ are invertible.

Condition 1 entails:

$$ \forall x \in [0,1]: F_1(x) = 2x - F_0(x) $$

This has implications for $F_0$ as not all valid $F_0$ will lead to valid $F_1$. Specifically $F_1$ is a valid invertible CDF as long as:

$$ \forall x \in [0,1]: 0 \leq 2x - F_0(x) \leq 1 \\ F_0'(x) < 2 $$

The problem is that Condition 2 works with the inverse CDFs:

$$ \forall x \in [0,1]: F^{-1}_1(x) = \sqrt{2x - 2 F^{-1}_0(x) + (F^{-1}_0(x))^2} $$

Similarly, this has additional implications - $F^{-1}_1$ will be a valid inverse CDF as long as $F^{-1}_0$ is valid and:

$$ \begin{align} \forall x, 0 \leq x \leq \frac{1}{2} : F^{-1}_0(x) &\leq 1 - \sqrt{1 - 2x} \\ \forall x, \frac{1}{2} \leq x \leq 1 : F^{-1}_0(x) &\geq 1 - \sqrt{2 - 2x} \\ (F^{-1}_0)^\prime(x) (F^{-1}_0(x) - 1) &\geq -1 \end{align} $$

I am unable to either re-express condition 1 in terms of the inverses - the closest I get is $x = F^{-1}_0(2x - F_1(x))$. I am similarly unable to transform condition 2 into a condition on the non-inverted CDFs.

This leaves me with equations that involve both a function and its inverse. Which I am unable to solve. I tried plugging that into Mathematica's RSolve but it couldn't solve it either.

Incorporating the implied conditions on $F_0$ into the solution is not necessary, they can stay separate, I am just sharing them in case they are helpful for making progress at combining condition 1 and condition 2.

From the way the criteria are constructed, I know for sure that there are at least two solutions:

$$ F_0(x) = F_1(x) = x $$

and

$$ F_0(x) = 2x - x^2, F^{-1}_0(x) = 1 - \sqrt{1-x} \\ F_1(x) = x^2, F^{-1}_1(x) = \sqrt{x} $$

It would surprise me if those were the only solutions.

Does this look hopeless or is there some strategy to attack this? Even finding additional solutions without claiming that all possibilities are exhausted would help.

Answer 1

So I figured it out.

first substituting $x = F_1(s)$ into condition 2:

$$ s = \sqrt{2 F_1(s) - 2 F_0^{-1}(F_1(s)) + (F_0^{-1}(F_1(s)))^2} \\ s^2 = 2 F_1(s) - 2 F_0^{-1}(F_1(s)) + (F_0^{-1}(F_1(s)))^2 $$

This is a quadratic equation for $F_0^{-1}(F_1(s))$ we solve it (only one branch of the solutions falls into $[0,1]$) and apply $F_0$ to both sides of the solution:

$$ F_0^{-1}(F_1(s)) = 1 - \sqrt{1 - 2 F_1(s) + s^2} \\ F_1(s) = F_0(1 - \sqrt{1 - 2 F_1(s) + s^2}| 0) $$

We now substitute condition 1 $F_1(s) = 2s - F_0(s)$ and thus obtain the equation:

$$ F_0(x) = 2x - F_0(1 - \sqrt{1 - 4x + 2 F_0(x) + x^2}) \tag{1} $$

Choosing $x_0$ and $v_0 = F_0(x_0)$, equation 1 implies a sequence of argument-value pairs that are fixed by that choice, i.e.:

$$ x_{i + 1} = 1 - \sqrt{1 - 4x_i + 2 v_i + x_i^2} \\ F_0(x_{i + 1}) = v_{i + 1} = 2x_i - v_i $$

We obtain that $x_i = x_{i+1} \iff x_i = v_i$, i.e. in this case the constraint on functional behaviour is local and does not induce any restrictions elsewhere. What about the other cases?

I plotted a bunch of those sequences and they looked quite quadratic. In fact it turns out that

$$ v_i = -x_i^2 + 2 x_i +c \implies v_{i + 1} = -x_{i + 1}^2 + 2 x_{i + 1} +c $$

So if the starting point lies on such a quad curve, the whole subsequence does. And any point on the plane lies on exactly one such quad curve.

Now any $F_0$ has to be a union of such sequences. If we assume that $F_0$ is continuous (which we do), it means that it can only "switch" between the $F_0(x) = x$ line and the quadratic curves at points where they intersect. Two quadratic curves differing only by constant never intersect each other, so we only need to care about intersections of the line and the quadratics which happens at:

$$ x = \frac{1}{2}\left( 1 \pm \sqrt{1 + 4c} \right) $$

We also need to have the intersection within the $[0,1]$ interval, so we get $-\frac{1}{4} \leq c \leq 0$. Finally, we need $F_0(0) = 0, F_0(1) = 1$. This gives us a family of solutions:

$$ F_0(s) = \begin{cases} -x^2 + 2x + c & \frac{1}{2}\left( 1 \pm \sqrt{1 - 4c} \right) < x < \frac{1}{2}\left( 1 + \sqrt{1 + 4c} \right) \\ x & \text{otherwise} \end{cases} $$

where at $c = -\frac{1}{4}$ we get one of the solutions I knew previously and at $c = 0$ we have the other solution. All the intermediate solutions do in fact satisfy all the constraints, so they are valid for the original question and (assuming continuity) those are the only solutions.