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\begin{document}
    \section*{MTH6141 Random Processes, 2018, Exercise Sheet 8}
    Please drop your answers to \textbf{Questions 2 and 3} in the Random Processes coursework box by *12 noon on Wednesday 28th March* (note the earlier day and time to usual).\\
    You are strongly encouraged to attempt as many questions as possible. Please send comments and corrections to \href{d.ellis@qmul.ac.uk}{d.ellis@qmul.ac.uk}.
    %
    \begin{enumerate}
        \item 
        \begin{enumerate}
            \item I arrive at a bus stop at a random time. What is my expected waiting time under the following assumptions:
            \begin{enumerate}
                \item I am in Z\"{u}rich, and the buses are equally spaced in time, $10$ minutes apart?
                \item  I am in London, and the buses arrive according to a Poisson process with rate $6$ per hour?
            \end{enumerate}
            \item How would you explain the difference in your answers to the two parts to a non-mathematician? (Notice that the expected number of buses per hour is the same, namely $6$, in both cases!)
        \end{enumerate}
        \item  The teams Athletic and Borough play a football match of $90$ minutes. The match is divided into two ‘halves’ of $45$ minutes each. Athletic score goals according to a Poisson process of rate $1$ per hour. Borough score goals according to an independent Poisson process of rate $2$ per hour.
        \begin{enumerate}
            \item  What is the probability that the total number of goals scored during the match is $6$?
            \item What is the conditional probability that the final score is $3$ goals to Athletic and $3$ to Borough, given that the half-time score is $2$ goals to Athletic and $1$ goal to Borough?
            \item What is the conditional probability that the half-time score is $2$ goals to Athletic and $1$ goal to Borough, given that the final score is $3$ goals each?
            \item What is the conditional probability that Athletic is ahead of Borough at half time, given that the final score is $3$ goals each?
            \item Given that during the first $15$ minutes of the match, just one goal is scored, what is the probability that this goal was actually scored in the first two minutes of the match?
            \item Suppose that a total of three goals have been scored by half-time. What is the probability that all three of these goals were scored in the five minutes just before half-time?
        \end{enumerate}
        \item Particles arrive at a Geiger counter according to a Poisson process of rate $3$ per hour. Provided the Geiger counter is switched on, it detects every particle that arrives. Let $T_n$ denote the time (in hours) of the arrival of the $n$th particle, after the Geiger counter is switched on. Show directly (from the basic definition of the Poisson process) that the pdf of $T_1$ is given by
        $$
        f_{T_1}(t)
        =
        \begin{cases}
            3e^{-t} & \text{if}\ t \geq 0;\\
            0 & \text{if}\ t < 0,
        \end{cases}
        $$
        and that the pdf of $T_2$ is
        $$
        f_{T_2}(t)
        =
        \begin{cases}
            9te^{-3t} & \text{if}\ t \geq 0;\\
            0 & \text{if}\ t < 0.
        \end{cases}
        $$
        Hence calculate $\mathbb{E}[T_2]$.
        \item 
        \begin{enumerate}
            \item Show that for each $n \in \mathbb{N}$ and each $s > 0$, the probability density function of the first arrival time $T_1$, conditioned on $X(s) = n$, is
            $$
            f_{T_1|X(s)=n}(t) = \frac{n}{t}\left(1 - \frac{t}{s}\right)^{n-1}
            $$
            for $0 < t \leq s$.\\
            \textit{Hint}. first work out the cdf [conditioned on $X(t) = n$].
            \item Hence show that $\mathbb{E}[T_1\, | \, X(s)= n] = \frac{s}{n+1}$.
            \item  Evaluate $\mathbb{E}[T_1\, |\, X(s) \geq 1]$. (This part doesn’t really require any major new insights, but it is a little harder than usual and it does involve some
            calculation.)
        \end{enumerate}
        Note: pdf = probabiliity density function, cdf = cumulative distribution function.
        \item 
        \begin{enumerate}
            \item Let $(X(t) : t \geq 0)$ be a Poisson process of rate $\lambda$, and let $T_i$ denote the $i$th arrival time. Show that the probability density function of $T_2$ conditioned on $X(1) = 3$ is $f_{T_2\, |\, X(1)=3}(t) = 6t(1 − t)$.\\
            \textit{Hint}. First work out the cdf [conditioned on $X(1) = 3$].
            \item More generally, for each $n \in \mathbb{N}$, what is the pdf of $T_{n+1}$ conditioned on $X(1) = 2n + 1$?\\
            (Hint: Do something similar to what you did for part (a), but you need to persevere! The pdf has a simple expression.)
        \end{enumerate}
    \end{enumerate}
\end{document}