Expected Values probalem

Question

The problem is:

In a world without gravity, a very small gun shooting point-like balls is located at the lower left end $(0, 0)$ of a $2D$ corridor. The corridor has length $L = 100\thinspace m$ and height $h = 5\thinspace m$. The gun shoots with initial velocity $v = 1\thinspace m/s$ at a random angle $\alpha$, distributed uniformly between 0 and $\pi/4$. Each time the ball bounces off the wall, its velocity component perpendicular to the wall is reduced by a fraction $\eta/(1 + \eta)$, \begin{equation} \upsilon_{vertical}\rightarrow -\left(1-\frac{\eta}{1+\eta}\right)\upsilon_{vertical}, \end{equation} where $\eta$ has an exponential distribution with mean $E(\eta) = 0.15$ and is drawn independently for every bounce

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What is the expected flight time (and number of bounces). What approximation can you use to get 95% confidence interval?

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I calculated it to be 250.83599 (8.32808). Using CLT, interval would be E(X)+-1.96*stddev/n where n is the number of simulations. Are those corrects? Are my derivations correct?

The paper is available at: http://www.filedropper.com/quanttestv2

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Below is an extract for $E[T]$.

We quickly find that: \begin{equation} T=\frac{5}{sin(\alpha)}\sum_{k=0}^{N-1}\prod_{i=0}^{k}\left( 1+\eta_i \right)+T_N \end{equation} with \begin{equation} T_N=(\frac{100}{cos(\alpha)}-(\lfloor 20tan(\alpha) \rfloor)\frac{5}{sin(\alpha)})\prod_{i=0}^{N}\left( 1+\eta_i \right) \end{equation}

\begin{align} E[T]&=E[\frac{100}{cos(\alpha)}|\alpha<=\arctan(\frac{1}{20})]+E[\frac{5}{sin(\alpha)}\sum_{k=0}^{N-1}\prod_{i=0}^{k}\left( 1+\eta_i \right)+T_N|\alpha>arctan(\frac{1}{20})] \end{align} \begin{align} E[\frac{100}{cos(\alpha)}|\alpha<=\arctan(\frac{1}{20})]&=\int_{\arctan(0)}^{\arctan(\frac{1}{20})}\frac{4}{\pi} \frac{100}{cos(x)} \thinspace dx\\ &=\frac{400}{\pi}\left[ ln|tan(\frac{x}{2}+\frac{\pi}{4})|\right]_{\arctan(0)}^{\arctan(1/20)} \end{align}

\begin{align} E[T|\alpha,\alpha>arctan(\frac{1}{20})]&=\frac{5}{sin(\alpha)}\sum_{k=0}^{N-1}E[\prod_{i=0}^{k}\left( 1+\eta_i \right)]+(\frac{100}{cos(\alpha)}-N \frac{5}{sin(\alpha)})E[\prod_{i=0}^{N}\left( 1+\eta_i \right)] \end{align}

The $\eta_i$ are independent from each other, and so are $(1+\eta_i)$.

\begin{align} E[T|\alpha,\alpha>arctan(\frac{1}{20})]&=\frac{5}{sin(\alpha)}\sum_{k=0}^{N-1}1.15^k+(\frac{100}{cos(\alpha)}-N \frac{5}{sin(\alpha)})1.15^N \end{align}

\begin{align} E[T|\alpha>arctan(\frac{1}{20})]&=\int_{arctan(\frac{1}{20})}^{\pi/4} f_\alpha(x) \left( \frac{5}{sin(x)}\sum_{k=0}^{N-1}1.15^k +(\frac{100}{cos(x)}-N \frac{5}{sin(x)})1.15^N \right) \thinspace dx \\ &=\frac{20}{\pi} \sum_{j=1}^{19} \left( \sum_{k=0}^{j-1}1.15^k \right) \left[ ln|tan(\frac{x}{2})| \right]_{\arctan(\frac{j}{20})}^{arctan(\frac{j+1}{20})}\\ &+\frac{4}{\pi}\sum_{j=1}^{19} 1.15^j \left( 100\left[ ln|tan(\frac{x}{2}+\frac{\pi}{4})| \right]_{\arctan(\frac{j}{20})}^{\arctan(\frac{j+1}{20})} - 5j\left[ ln|tan(\frac{x}{2})| \right]_{\arctan(\frac{j}{20})}^{\arctan(\frac{j+1}{20})} \right) \end{align}

\begin{align} E[T]&=6.36355+219.90387+24.56857\\ &=250.83599 \end{align}

Answer 1

Assuming that $L=Nh$ for some integer $N\geqslant1$, the number $B$ of bounces before exiting the corridor is such that, for every $k$, $$[B\geqslant k]=[\tan\alpha\geqslant k/N],$$ hence $$E(N)=\sum_{k\geqslant1}P(B\geqslant k)=\sum_{k\geqslant1}P\left(\tan\alpha\geqslant \frac{k}N\right)=\sum_{k=1}^N\left(1-\frac4\pi\arctan\left(\frac{k}N\right)\right).$$ For $N=20$, one finds $$E(B)\approx8.328.$$ For every $N$, the sum is $N$ times a Riemann sum hence, when $N\to\infty$, $$E(B)\sim N\int_0^1\left(1-\frac4\pi\arctan x\right)\mathrm dx=\frac{\log4}\pi N.$$ For $N=20$, this approximation reads $$E(B)\approx8.825.$$ A similar approach can be applied to the time $T$ to exit the corridor since $$N\tan\alpha=B+C,$$ with $C$ in $[0,1)$ and the $k$th part of the flight, between abscissae $(k-1)h/\tan\alpha$ and $kh/\tan\alpha$ has length $h/\sin\alpha$ and is done at a speed $v_k$ such that $E(1/v_k)=(E(1+\eta))^{k-1}=\delta^{k-1}$ with $\delta=1.15$, hence $$E(T)=E\left(\sum_{k=1}^B\frac{h}{\sin\alpha}\delta^{k-1}+\frac{hC}{\sin\alpha}\delta^B\right).$$ This is equivalent to $$E(T)=L\,\tau_0+h\,\tau_{\delta,N},$$ where $$\tau_0=E\left(\frac1{\cos\alpha}\right),\qquad\tau_{\delta,N}=(\delta-1)\,\sum_{k=1}^{N}E\left(\frac{k\delta^{k-1}}{\sin\alpha}\mathbf 1_{N\tan\alpha\geqslant k}\right).$$ The term $L\tau_0$ corresponds to purely elastic bounces, that is, to $\delta=1$, described by the factor $$\tau_0=\frac4\pi \log(1+\sqrt2)\approx1.122.$$ The term $h\tau_{\delta,N}$ is the correction due to the inelasticity of the bounces, described by the factor $$\tau_{\delta,N}=\frac4\pi (\delta-1)\sum_{k=1}^Nk\delta^{k-1}\,\left(\log(1+\sqrt{1+(k/N)^2})-\log(k/N)-\log(\sqrt2+1)\right).$$ For $N=20$ and $\delta=1.15$, $\tau_{\delta,N}=55.879$ hence, for $(L,h)=(100,5)$, $$E(T)\approx L\,1.122+h\,55.879\approx391.6.$$