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.


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

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

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

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

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

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




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



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.

  • $\begingroup$ You are inserting values into equation that holds almost everywhere which has to be justified. $\endgroup$
    – ibnAbu
    May 15 '20 at 9:28
  • $\begingroup$ @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! $\endgroup$
    – boreca
    May 15 '20 at 9:37

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