Functional equation for distribution function I have next functional equation for some distribution:
$$\overline F_\xi^2(T) = \overline F_\xi(2T) \forall T&gt;0$$
If suggest, that it differentiable, we can do something like this:
$$2\overline F_\xi(T) \cdot \overline F'_\xi(T) = 2\overline F'_\xi(2T)$$
$$\overline F_\xi(T) \cdot \overline F'_\xi(T) = \overline F'_\xi(2T)$$
But how to solve this differential functional equation?
where $$\overline F_\xi = 1 - F_\xi = P(\xi &gt; T)$$
Also, I have $\mathbb E\xi=1.$
For brevity, put $f=\bar F_{\xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $f\le 0$ everywhere, which contradicts $\Bbb E[\xi]=1$. Hence, $f&gt;0$ and we can define $g(t)=\ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $\Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)\Rightarrow g^{(n)}(0)=0$ for $n&gt;1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $\xi\sim \text{Exp}(-a)$, but $\Bbb E[\xi]=-\frac 1a=1$, so $a=-1$.
Hint: divide both sides by F(2T)