Solution or property to a functional equation

Question

I have a continuous and nondecreasing function $F:\mathbb{R}\to\mathbb{R}$ satisfying following conditions:

1. $F(x) = 0$ for any $x\in (-\infty, 1]$,
2. $\lim\limits_{x\to +\infty} F(x) = 1$,
3. for any $x\in \mathbb{R}$ we have $$F(x) \;=\; 0.8 F(2x - 1) + 0.2 F(x/2 - 1).$$

Is there any solution to this "functional equation" system? Is there any further property of $F$ like its lower bound or growing rate?

This problem actually comes from probability theory. Consider a sequence of i.i.d. Bernoulli variables $Y_1, Y_2, \dots$ following $$\mathbb{P}(Y_i = 1) = 1 - \mathbb{P}(Y_i = 0) = 0.8.$$ Then using $Y_1, Y_2, \dots$, we define a random series $S$ as follows $$S = \sum_{k=1}^\infty 2^{k - \sum_{i=1}^k Y_i} (1/2)^{\sum_{i=1}^k Y_i}.$$ I have known that $S$ converges almost surely. Now I want to find the cumulative distribution function (CDF) of $S$. It turns out that the CDF satisfies the properties I gave above.

Any comment helps.

I draw a picture for the empirical disrtibution as follows.

Answer 1

This answer is partial. Namely, we show that $F(7)\le \frac 1{1.12}=0.8728\dots$ and for each natural $n$ we have $$F(1+6\cdot 2^{-n})=0.8^nF(7)>0$$ and $$1-F(1+6\cdot 2^n)\le (1-F(7))4^{-n}<4^{-n}.$$

Substituting $x=\frac{y+1}2$ in Condition 3, we obtain the equality $$F(y)=1.25 F\left(\frac{y+1}2\right)-0.25F\left(\frac{y-3}4\right)$$ for each real $y$.

Put $\bar y=\sup\{y\in\mathbb R:F(y)=0\}$. Then $\bar y\ge 1$. Put $y_0=4\bar y+3$. Then $y_0\ge 7$. If $y\le y_0$ then $\frac{y-3}4\le \bar y$, so $F\left(\frac{y-3}4\right)=0$ and $F(y)=1.25F\left(\frac{y+1}2\right)$. For each natural $n$ put $y_n=\frac{y_{n-1}+1}2$. Then $(y_n)_{n\in\omega}$ is a decreasing sequence contained in the segment $[1,\bar y]$ and $F(y_n)=0.8F(y_{n-1})$ for each natural $n$. Let $y_\infty$ be the limit of the sequence. Then $y_\infty=\frac{y_\infty+1}2$, so $y_\infty=1$.

Suppose for a contradiction that $\bar y>1$. Then there exists natural $N$ such that $y_N<\bar y$. So $F(y_N)=0$. Since $F(y_n)=0.8F(y_{n-1})$ for each natural $n$, we have that $F(y_n)=0$ for each natural $n$. Then $F(y_0)=0$. But $y_0>\bar y$, a contradiction.

Thus $\bar y=1$ and $y_0=4\bar y+3=7$. It can be shown by induction on $n$ that $y_n=1+6\cdot 2^{-n}$ for each nonnegative integer $n$. So $$F(1+6\cdot 2^{-n})=0.8^nF(7)>0.$$

Moreover, we have $$F(13)=1.25F\left(\frac{13+1}2\right)-0.25F\left(\frac{13-3}4\right)=$$ $$1.25F(7)-0.25 F(2.5) =(1.25-0.25\cdot 0.8^2)F(7)=1.09F(7).$$

Since $F(13)\le 1$, $F(7)\le \frac 1{1.08}=0.9259\dots$. We can improve this bound a bit as follows.

For each real $x$ we have

$$F(2x-1)-F(x)\ge 0.25(F(x)-F(x/2-1))\ge 0.25(F(x)-F(x/2+1)).$$

Therefore if we put $x_0=x$ and $x_n=2x_{n-1}-1$ for any natural $n$ (that is $x_n=1+2^n(x_0-1)$) we obtain that $$F(x_{n+1})-F(x_n)\ge 0.25(F(x_n)-F(x_{n-1})).$$ It follows $$F(x_n)\ge F(x_0)+(F(x_1)-F(x_0))\sum_{i=0}^n 0.25^i=$$ $$F(x_0)+(F(x_1)-F(x_0))\frac 43(1-0.25^{n+1}).$$ Therefore $$F(x_0)+(F(x_1)-F(x_0))\frac 43\le 1.$$

That is $$1-F(2x-1)=1-F(x_1)\ge \frac {1-F(x_0)}4=\frac {1-F(x)}4.$$

Iterating, we obtain for each natural $n$

$$1-F(1+2^n(x_0-1))=1-F(x_n)\le (1-F(x_0))4^{-n}.$$

Now we can bound the growth rate of the function $F$ as follows.

Putting $x_0=7$ we obtain $F(7)+0.09F(7)\frac 43\le 1,$ so $F(7)\le \frac 1{1.12}=0.8728\dots$ and for each natural $n$ $$1-F(1+6\cdot 2^n)\le (1-F(7))4^{-n}< 4^{-n}.$$

Answer 2

Your functional equation comes from the self consistent equation $$ S=2^{1-2Y}(1+S') $$ with $Y,S'$ independent, $Y$ distributed as your Bernoulli distribution and $S,S'$ identically distributed via the unknown series' distribution.

You can use this to prove the existence and uniqueness of $F$ as a bounded continuous function. The idea is to view the equation as a fixed point equation for the affine functional (in your case $p=.8$): $$ \Phi: F\to \big(x\to pF(2x-1)+(1-p)F(x/2-1)\big) $$ You might think that it is even linear, but you are interested in its action on the affine space of bounded, non-decreasing, continuous functions whose limit is $1$ at $+\infty$. It is complete, and $\Phi$ is contracting (of ratio $\max(p,(1-p))$) you can apply the Banach fixed point theorem. In fact, by iterating $\Phi$, you get better and better approximations of $F$ and gives you a numerical method to compute it.

In general you cannot say much about $F$, but, you can view it as the distribution of an IFS for the two mappings of $[1,+\infty)$: $$ \phi_1(x) = \frac{1+x}{2} \\ \phi_2(x) = 2(1+x) $$ weighted respectively by $p,1-p$. The issue is that there is some significant overlap in the region $[4,+\infty)$, so the usual methods do not directly apply. You can however expect $F$ to not be absolutely continuous (no pdf) and rather to be multifractal.

You can estimate the asymptotic behaviour for $x\to1^+$, the attracting fixed point of $\phi_1$. Indeed, you are only influenced by $\phi_1$, so you recover self similarity. For $x\to1^+$, let $x=1+h$, $F(x)=f(h)$. In this limit, you rigorously have $x/2-1<1$, so you only have the first term: $$ f(h) = pf(h/2) $$ By iteration, you obtain the local Hölder exponent: $$ f(h) = O(h^{-\ln p/\ln2}) $$ i.e.: $$ F(x) = O((x-1)^{-\ln p/\ln2}) $$

For $x\to+\infty$, the attraction fixed point of $\phi_2$, things are a bit harder. You always will have the influence of $\phi_1$. by considering $F_>(x) = 1-F(x)$, assuming that the second term will dominate, you get approximately: $$ F_>(h) = (1-p)F_>(x/2-1) $$ By the same argument, you expect the scaling: $$ F_>(x) = O(x^{\ln (1-p)/\ln2}) $$ i.e.: $$ F(x) = 1-O(x^{\ln (1-p)/\ln2}) $$ You could try to compute the multifractal spectrum, since the maps are affine, it should be doable by hand.

Hope this helps.