How to calculate math expectation of recurrent function? [closed]

Question

I try to figure martingale strategy mathematically. Let us say we start with bet $x$ and designate obtained value as $f(x)$

I come up with the following equation $$ \mathbb{E}f(x) = \frac{18}{37} x + \frac{19}{37}(\mathbb{E}f(2x) - x) = \frac{19}{37}\mathbb{E}f(2x) - \frac{1}{37}x $$ I want to compute $\mathbb{E}f(x)$ in two scenarios:

1. In case $x$ can grow infinitely. Since limit of probability to win is 1 with number of games tending to infinity I expect that in that case $\mathbb{E} f(x)$ should be $x$. Well, function $\mathbb{E}f(x) \equiv x$ satisfies equation above for all $x$. Is it correct to conclude that $\mathbb{E}f(x) = x$ and there are no other solutions?
2. How to calculate $\mathbb{E} f(x)$ in case that $x$ is limited? I can think of something like defining $f(2^Mx):= 0$ and then backwards expanding all $f(2^kx$) but it seems ugly. Can it be done differently?

And is recurrent equation like this is good in general for such problems?

Answer 1

Let $g(x):=Ef(x)$. For real $x\ne0$, let $$u(x):=\frac{g(x)-x}{x^p},$$ where $p:=\log_2\frac{37}{19}$, so that $$Ef(x)=g(x)=x+x^p u(x). \tag{0}\label{0}$$ Then your equation, $$Ef(x)=\frac{19}{37}\,Ef(2x)-\frac1{37}\,x, \tag{1}\label{1}$$ can be rewritten for real $x\ne0$ simply as $$u(2x)=u(x). \tag{2}\label{2}$$ Assuming that there is a finite limit $$c:=\lim_{x\to0}u(x),\tag{3}\label{3}$$ we conclude that $u(x)=c$ for all real $x\ne0$ and hence, by \eqref{0}, $$Ef(x)=x+cx^p \tag{4}\label{4}$$ for real $x\ne0$. Formula \eqref{4} holds also for $x=0$, according to your equation \eqref{1}.

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If the condition of the existence of a finite limit $c$ in \eqref{3} is dropped, then equation \eqref{2} and thus equation \eqref{1} will have infinitely many irregular solutions, of them infinitely many measurable ones. Indeed, all the solutions of \eqref{3} (and thereby all the solutions of \eqref{1}) can be described as follows:

For distinct nonzero real $x$ and $y$, let us write $x\sim y$ if $x=2^k y$ for some integer $k$. Then the relation $\sim$ is an equivalence on the set $\mathbb R\setminus\{0\}$. For each equivalence class $\bar x$, pick any real $r(\bar x)$ and let the function $u$ be equal $r(\bar x)$ on $\bar x$. Then $u$ is a solution of equation \eqref{2}.

Moreover, any solution of equation \eqref{2} can be obtained this highlighted way.