# Solving a functional equation in CDF's of probability distributions

Reading a paper I've come across the following functional equation for unknown CDF's $$F_1, F_2$$ of centered probability distributions $$\mu_1, \mu_2$$ with variance $$1$$: $$F^{-1}(G_2(x+y)) = F_1^{-1}(G_1(x))+ F_2^{-1}(G_1(y))\qquad \text{for all} \ (x,y) \in \mathbb{R}^2$$ where $$G_i$$ is the CDF of a centered gaussian with Variance $$i$$ and $$F$$ is the CDF of the convolution $$\mu_1 \ast \mu_2$$. The unique solution is actually $$F_1 = F_2 = G_1$$ but I've not been able to proof that. I (think I) can show that $$F_1 = F_2$$: $$F_1^{-1}(G_1(x))+ F_2^{-1}(G_1(y)) = F^{-1}(G_2(x+y)) = F^{-1}(G_2(y+x)) = F_1^{-1}(G_1(y))+ F_2^{-1}(G_1(x))$$ so $$F_1^{-1}(G_1(x)) - F_2^{-1}(G_1(x)) = F_1^{-1}(G_1(y)) - F_2^{-1}(G_1(y))$$ which means, that $$F_1^{-1}(G_1(x)) - F_2^{-1}(G_1(x))$$ is constant, since the right-hand side does not depend on $$x$$. If the difference was not $$0$$ then either $$\mu_1$$ or $$\mu_2$$ is not centered since, $$\mathbb{E}[\mu_i] = \int_0^1 F_i^{-1}(y)dy$$, so $$F_1 = F_2$$.

Is this argument correct? How can I proceed to show uniqueness of solution?

You can find the paper here - the functional equation is part of the proof of Theorem 2 on page 49.

$$F^{-1}(G_2(x+y)) = F_1^{-1}(G_1(x))+ F_2^{-1}(G_1(y))$$

$$\text{let } h_1(x)=\int_0^tF_1^{-1}(G_1(xt))dt$$

$$\text{let } h_2(y)=\int_0^tF_2^{-1}(G_1(yt))dt$$

$$\text{let } h(x+y)=\int_0^tF^{-1}(G_1(xt+yt))dt$$

$$\text{it is easily seen that h_i(x)=\frac{\int_0^xF_i^{-1}(G_1(u))du}{x}} \text{ ,it is continuous}$$

$$h_1(x)+h_2(y)=h(x+y) \text{ holds everywhere}$$

$$h_1(x)+h_2(0)=h(x)$$

$$h_1(0)+h_2(x)=h(x)$$

$$h_1(x)+h_2(0)=h_1(0)+h_2(x)$$

$$\text{The derivative: } h_1'(x)=h_2'(x)$$

$$xh_1'(x)+h_1(x)=F_1^{-1}(G_1(x))$$

$$xh_2'(x)+h_2(x)=F_2^{-1}(G_1(x))$$

The equations imply $$h_1(x)-h_2(x) \text{ is constant}$$

$$F_1^{-1}(G_1(x))-F_2^{-1}(G_1(x)) \text{ is constant}$$

$$h(x)-h_1(x) \text{ is constant}$$

$$h(x)-h_2(x) \text{ is constant}$$

Getting a (big) hint from a professor I was able to solve the equation:

In the following we'll write $$h:=F^{-1} \circ G_2$$, $$f:=F_1^{-1} \circ G_1$$, $$g:=F_2^{-1} \circ G_2$$. Inserting $$(x,0)$$ and $$(0,x)$$ in the equation yields $$f(x) + g(0) = h(x+0)=h(0+x) = f(0) + g(x)$$ therefore $$f(x) = h(x) - g(0)\quad \text{and} \quad g(x) = h(x) - f(0)$$ Reinserting this in the function equation, we find $$h(x+y) = h(x) - g(0) + h(y) - f(0) = h(x) + h(y) - h(0)$$ Hence $$\phi(x):=h(x)-h(0)$$ satisfies Cauchy's functional equation $$\phi(x+y)=h(x+y)-h(0) = h(x) + h(y) - h(0) - h(0) = \phi(x) + \phi(y)$$ which only admits one monotone solution, $$\phi(x) = ax$$. From here, using that $$\mu$$ and $$\nu$$ are centered with variance on we can show that $$a=1$$, $$h(0)=0$$ and finally $$h=f=g=id_\mathbb{R}$$, arriving at the desired conclusion.

• You are inserting values into equation that holds almost everywhere which has to be justified. May 15 '20 at 9:28
• @ibnAbu I checked, and the paper actually requires the equation to be satisfied for all $x,y \in \mathbb{R}$. I'll edit this in my question! May 15 '20 at 9:37