Let $X$ be a non-negative random variable with cdf $F$ and define
$$G(s) = E[\max(0,u(X)-sX)],$$ where $u$ is some real function.
Let $s\_0$ be the unique fixed point of $G$.


Now let $X\_1,\dots,X\_t$ be a sequence of samples drawn independantly from $F$.


Let
$$G\_t(s) = \frac{1}{t}\sum\_{i=1}^t \max(0,u(X\_i)-sX\_i)$$
and let $s\_t$ be the unique fixed point of $G\_t$.


Is it true that $s\_t$ converges almost surely to $s\_0$, and is it possible to say something about the speed of convergence? If not, how could I construct a converging estimator of $s\_0$?



Convergence of estimator given by a fixed point
$\newcommand\N{\{1,2,\dots\}}
\newcommand\NN{\{1,2,\dots,\infty\}}
\newcommand\si{\sigma}
\newcommand{\ep}{\varepsilon}$
Suppose that $E u(X)\_+<\infty$, where $x\_+:=\max(0,x)$. Then the convergence takes place, with the rate $O(1/\sqrt t)$ (as $t\to\infty$) if we also assume that $EX<\infty$, $E u(X)\_+^a<\infty$ for some $a>2$, and  

$$\si(s):=\sqrt{Var\,(u(X)-s X)\_+}\ne0$$ 
for any real $s$. 


Indeed, let 
$$G\_\infty(s):=G(s)=E(u(X)-sX)\_+,$$
and then let 
$$H\_t(s):=G\_t(s)-s$$
for all $t\in\NN$, so that $EH\_t(s)=H\_\infty(s)$. For each $t$, the function $G\_t$ is nonincreasing and hence $H\_t$ is strictly decreasing. Moreover, by dominated convergence, $G\_\infty$ is continuous and hence $H\_\infty$ is continuous. Also, $H\_t$ is obviously continuous for $t\in\N$. Further, $H\_t(0)\ge0>-\infty=H\_t(\infty-)$. So, for each $t\in\NN$, the function $H\_t$ has a unique zero in $[0,\infty)$, that is, the function $G\_t$ has a unique fixed point, say $s\_t$, in $[0,\infty)$.


Also, using the condition $E u(X)\_+^a<\infty$ with $a>2$ and, again, the dominated convergence theorem, we see that, for each $b\in[0,a]$, $E(u(X)-sX)\_+^b$ is continuous in $s$ and hence bounded in $s$ in any compact interval. 
So, by the Lyapunov--Bikelis inequality (see e.g. [inequality (1)](https://arxiv.org/pdf/2001.01123.pdf)), 
\begin{equation}
 P(Z\_t(s)>z)-P(Z>z)\to0\tag{1}
\end{equation}
uniformly over all real $z$ and all $s$ in any compact interval, where $Z\sim N(0,1)$ and 
$$Z\_t(s):=\frac{H\_t(s)-H\_\infty(s)}{\si(s)/\sqrt t}.$$
So, for any real $z$ and any $t\in\N$
\begin{align\*}
P(s\_t>s\_\infty+z/\sqrt t)&=P(H\_t(s\_t)<H\_t(s\_\infty+z/\sqrt t)) \\
&=P(H\_t(s\_\infty+z/\sqrt t)>0) \\
& =P\Big(Z\_t(s\_\infty+z/\sqrt t)>\frac{-H\_\infty(s\_\infty+z/\sqrt t)}{\si(s\_\infty+z/\sqrt t)/\sqrt t}\Big) \\ 
& =P\Big(Z>\frac{-H\_\infty(s\_\infty+z/\sqrt t)}{\si(s\_\infty+z/\sqrt t)/\sqrt t}\Big)+o(1), 
\end{align\*}
by (1). 
Next, again by dominated convergence, the right derivative of $H\_\infty$ at any real $s$ is $-\mu^+(s)-1$, where 
$$\mu^+(s):=EX1\_{u(X)-sX>0}.$$  

So, for any real $z>0$, $H\_\infty(s\_\infty+z/\sqrt t)=H\_\infty(s\_\infty+z/\sqrt t)-H\_\infty(s\_\infty)\sim-(1+\mu^+(s\_\infty))z/\sqrt t$. 
Also, using again the mentioned continuity of $E(u(X)-sX)\_+^b$ in $s$ for each $b\in[0,a]$, we see that $\si(s\_\infty+z/\sqrt t)\to\si(s\_\infty)$. 


We conclude that for any real $z>0$
\begin{align\*}
P(s\_t-s\_\infty>z/\sqrt t)&\to P\Big(Z>\frac{(1+\mu^+(s\_\infty))z}{\si(s\_\infty)}\Big). 
\end{align\*}
Similarly, for any real $z>0$
\begin{align\*}
P(s\_t-s\_\infty<-z/\sqrt t)&\to P\Big(Z<-\frac{(1+\mu^-(s\_\infty))z}{\si(s\_\infty)}\Big), 
\end{align\*}
where $\mu^-(s):=EX1\_{u(X)-sX\ge0}$. 


Thus indeed, $s\_t$ converges to $s\_\infty$ at the rate $O(1/\sqrt t)$. 




---


Let us also show that $s\_t\to s\_\infty$ almost surely (a.s). Take any real $\ep>0$. Since the function $H\_\infty$ is strictly decreasing, we have 
\begin{equation\*}
 H\_\infty(s\_\infty-\ep)>H\_\infty(s\_\infty)>H\_\infty(s\_\infty+\ep). 
\end{equation\*}
By the strong law of large numbers, $H\_t(s\_\infty\pm\ep)\to H\_\infty(s\_\infty\pm\ep)$ a.s. So, for some random variable $T\_\ep$ such that $P(T\_\ep\in\N)=1$ and all natural $t$, on the event $\{t\ge T\_\ep\}$ we have 
\begin{equation\*}
 H\_t(s\_\infty-\ep)>H\_\infty(s\_\infty)=0=H\_t(s\_t)=H\_\infty(s\_\infty)>H\_t(s\_\infty+\ep),
\end{equation\*}
which yields
\begin{equation\*}
 s\_\infty-\ep<s\_t<s\_\infty+\ep,
\end{equation\*}
because the function $H\_t$ is strictly decreasing. Thus indeed, $s\_t\to s\_\infty$ a.s.


