4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\endgroup$
0
2
+100
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

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.