Consider the following stochastic dynamical system. 


Fix $a > 0$, $b > 0$, and $v > 0$, and let $\mathbf{r}(t)=(x(t),y(t))$ be the position at time $t$ of a point which moves in the rectangle $R=\{ (x,y) \in \mathbb{R}^2: 0 \leq x \leq a, 0 \leq y \leq b \}$ with velocity of constant magnitude $v$ a according to the following rules:


(i) in the interior of $R$ the point is subject to no force, so that is moves with constant velocity $(v\_x(t), v\_y (t)) = \mathbf{v}(t)=\frac{d\mathbf{r}}{dt}(t)$;


(ii) when the point reaches the vertical side of $R$ corresponding to $x=0$ it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[-\pi/2,\pi/2]$;


(iii) if the point reaches any other side of $R$, then it is reflected elastically, that is in the case of the vertical side corresponding to $x=a$ the $y$-component of $\mathbf{v}$ is preserved, while the $x$-component of $\mathbf{v}$ changes sign, while in the case of either of the two horizontal sides the $x$-component of $\mathbf{v}$ is preserved, while the $y$-component of $\mathbf{v}$ changes sign;


(iv) finally, if the point reaches one of the vertices of $R$, then the reflection must be thought as the combination of two reflections by the the two edges involved; so e.g. if the point hits the vertex $(0,0)$ then it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[0,\pi/2]$; if instead the point hits e.g. the vertex $(a,0)$ then its velocity $\mathbf{v}$ before the collision is changed in $-\mathbf{v}$.


Consider a time $T> 0$, and let $N(T)$ be the number of time the point touches the vertical side of $R$ corresponding to $x=a$ (you can compute or not in $N(T)$ the times the point touches one of the vertices of this side: it should make no essential difference for what we want to prove, I think). Let $\theta\_j$ be the angle that the velocity $\mathbf{v}$ just before the collision makes with $\mathbf{i}$ the $j$-th time the point touches the vertical sides of $R$ corresponding to $x=a$ (or one of its vertices, if you have considered also them in the computation of $N(t)$), and form the random sum
\begin{equation}
\sum\_{j=1}^{N(T)} cos \theta\_j.
\end{equation}
I would like to prove that for any initial conditions $(\mathbf{r}(0), \mathbf{v}(0))$, with $v\_{x}(0) \neq 0$, the following equation holds
\begin{equation}
\lim\_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum\_{j=1}^{N(T)} cos \theta\_j \right] = \frac{v}{4 a}.
\end{equation}


Any help is welcome. For now, I have no idea about a possible proof.


NOTE. This problem has been suggested to me by an interesting proof of a physical law called Wien's displacement law given by Richtmyer, Kennard and Cooper in their book "Introduction to Modern Physics", Sixth Edition, Appendix of Chapter 5 (see in particular p. 145). I formulated a first model of dynamic stochastic billiard in my post [Stochastic Dynamic Billiard](https://mathoverflow.net/questions/282145/a-stochastic-dynamical-billiard), but it was considerably more complicated. Then I realized that the simplest model above could be a good formulation of the physical system as well. In the obvious three-dimensional version of this model, in which the point moves inside a parallelepiped, whose face corresponding to $x=0$ reflect diffusely (that is $v$ remains unchanged and $\mathbf{v}/v$ is uniformly distributed over a hemisphere) while all other faces (and all the edges and vertices) reflect elastically, we should get


\begin{equation}
\lim\_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum\_{j=1}^{N(T)} cos \theta\_j \right] = \frac{v}{6 a},
\end{equation}


which has a deep thermodynamical meaning (see the reference above and also ter Haar and Wergeland, Elements of Thermodynamics, $\S 5.3.3$).



A Simple Stochastic Dynamic Billiard
To avoid trivialities, assume that the initial position of the particle is in the interior of the rectangle. 
Then the limit is actually $0$. This is trivial in the case when the $x$-coordinate of the initial velocity is $0$. 


Otherwise, in a nonrandom finite time, depending only on the initial position and velocity, the particle will hit the side $x=0$. So, without loss of generality let us assume the movement begins right after that hitting time, with the initial velocity of magnitude $v$ at a random angle $\theta\_1\in[-\pi/2,\pi/2]$ with the $x$-axis. Note that the horizontal coordinate of the velocity may change in magnitude only right after hitting the side $x=0$; except for this, it may only change in sign right after hitting the side $x=a$. 
Also, the cosine of the angle between the velocity and the $x$-axis does not vary while the particle is traveling from the side $x=0$ toward the side $x=a$. 
Therefore, introducing $Y\_j:=\cos\theta\_j$, $X\_i:=\frac{a}{vY\_i}$ -- the $i$th traveling time from the side $x=0$ to the side $x=a$, $T\_j:=\sum\_1^j X\_i$, $N\_t:=\sum\_{j=1}^\infty I\{T\_j\le t\}$ (with $I\{\cdot\}$ denoting the indicator function), we see that $N(t)\le N\_t$ for real $t>0$; here, we understand $N(t)$ as the number of times the particle hits the vertical side $x=a$ before or at time $t$. 
So,  

\begin{equation}
 \sum\_{j=1}^{N(t)}\cos\theta\_j\le R\_t:=\sum\_{j=1}^{N\_t}Y\_j. 
\end{equation}
By the [Renewal Reward Theorem](http://www.randomservices.org/random/renewal/Reward.html), we have 
\begin{equation}
 \frac1t\,ER\_t\underset{t\to\infty}\longrightarrow\frac{EY\_1}{EX\_1}=\frac{2/\pi}\infty=0. 
\end{equation}
Thus, 
\begin{equation}
 0\le\frac1t\,E\sum\_{j=1}^{N(t)}\cos\theta\_j\le
 \frac1t\,ER\_t\underset{t\to\infty}\longrightarrow0. 
\end{equation}


