Spring 2018, Math 171
Week 10

Poisson Process Conditioning

Let \(N(t)\) be a Poisson process with rate \(\lambda\). Fix \(0 \le n\), \(s \le t\).

Compute \(\mathbb{P}(N(s)=m \mid N(t) = n)\) for \(m \le n\). Give the name and parameters of the distribution.
 (Answer) Binomial(\(n,\frac{s}{t}\))

Compute \(\mathbb{E}[N(s) \mid N(t) = n]\)
 (Answer) \(n\frac{s}{t}\)


Let \(N(t)\) be a Poisson process with rate \(\lambda\). Fix \(0 \le m \le n\), \(r \le s \le t\).

Compute \(\mathbb{P}(N(s)N(r)=k \mid N(t) = n, N(r)=m)\). Give the name and parameters of the distribution.
 (Answer) Binomial(\(nm,\frac{sr}{tr}\))

Compute \(\mathbb{E}[N(s) \mid N(t) = n, N(r)=m]\)
 (Answer) \((nm)\frac{sr}{tr}\)

Compute the expected amount if time between the first and last arrivals in \([r, t]\) given that \(N(t) = n\) and \(N(r)=m\).
 (Solution) For convenience of writing, I will omit the condition on \(N(t) = n\) and \(N(r)=m\) in the expectations.
 Let \(L\) be the time of the last arrival in the interval, and \(F\) be the time of the first arrival in the interval. Then we seek \[\begin{aligned}\mathbb{E}[LF] &= \mathbb{E}[L]  \mathbb{E}[F]\\&=t  \mathbb{E}[tL]  r  \mathbb{E}[Fr] \\&= (tr)  \mathbb{E}[tL]  \mathbb{E}[Fr].\end{aligned}\] By symmetry, we can see that \(tL\) (the time between the last arrival and the end of the interval) and \(Fr\) (the time between the start of the interval and the first arrival) have the same distribution, so \[\mathbb{E}[LF] = (tr)  2\mathbb{E}[Fr].\]
 Since we have conditioned on \(N(t) = n\) and \(N(r)=m\), we know that there are \(nm\) arrivals uniformly distributed in the interval by the conditioning property of the poisson process. Therefore, letting \(U_1, \dots U_{nm}\) be i.i.d uniform\([0, tr]\) random variables, we have \[\begin{aligned}\mathbb{E}[Fr] &= \mathbb{E}[\min(U_1, \dots U_{nm})]\\&= \int_0^{tr}P(\min(U_1, \dots U_{nm}) > s) ds\\&=\int_0^{tr}P(U_1>s, \dots U_{nm}>s) ds\\&=\int_0^{tr}P(U_1>s)^{nm} ds\\&=\int_0^{tr}\left(\frac{trs}{tr}\right)^{nm} ds\\&=(tr)\frac{\left(\frac{trs}{tr}\right)^{nm+1}}{nm+1} \Big\vert_0^{tr}\\&=\frac{tr}{nm+1}\end{aligned}\]
 So finally, we have \[\begin{aligned}\mathbb{E}[LF] &= (tr)  2\frac{tr}{nm+1}\\&=(tr)\left(1\frac{2}{nm+1}\right)\end{aligned}\]
 As a sanity check, we check that if \(nm=1\) we have \(\mathbb{E}[LF]=0\), and if \(nm=\infty\) we have \(\mathbb{E}[LF]=(tr)\)



Renewal Process

Cars queue at a gate. The lengths of the cars are i.i.d with distribution \(F_L\) and mean \(\mu\). Let \(L \sim F_L\). Each successive car stops leaving a gap, distributed according to a uniform distribution on \((0, 1)\), to the car in front (or to the gate in the case of the car at the head of the queue). Consider the number of cars \(N(t)\) lined up within distance t of the gate. Determine \(\lim_{t \to \infty} \frac{\mathbb{E}[N(t)]}{t}\) if

\(L = c\) is a fixed constant

\(L\) is exponentially distributed with parameter \(\lambda\)


Potential customers arrive at a service kiosk in a bank as a Poisson process of rate \(\lambda\). Being impatient, the customers leave immediately unless the assistant is free. Customers are served independently, with mean service time \(\mu\).

Find the mean time between the starts of two successive service periods.
 (Answer) \(\mu + 1/\lambda\)

Find the long run rate at which customers are served
 (Answer) \(\frac{1}{\mu + 1/\lambda}\)

What proportion of customers who arrive at the bank actually get served?
 (Answer) \(\frac{1}{\lambda\mu + 1}\)


Customers arrive at a 24 hour restaurant (which has only one table) according to a poisson process with rate 1 party per hour. If the table is occupied, they leave immediately. If the table is unoccupied, they stay and eat. Customers spend an average of $20 each. The length of time a party stays is uniformly distributed in \([0, N/2 \mathrm{\ hours}]\) where \(N\) is the number of people in the party. Answer the following questions for the cases when \(N\) is uniform\(\{2, 3, 4\}\) and when \(N\) is geometric(\(1/2\)).

Let \(\tau_k\) be the time between when the \((k1)\)st party finishes eating and the \((k)\)th party finishes eating. Compute \(\mathbb{E}[\tau_k]\)
 (Answer) \(\mathbb{E}[N/4]+1\)

Let \(X_k\) be the amount the \((k)\)th party spends. Compute \(\mathbb{E}[X_k]\).
 (Answer) \(20\mathbb{E}[N]\)

Let \(S(t)\) the amount total amount paid by all parties by time \(t\). Compute \(\lim_{t\to \infty}\frac{S(t)}{t}\), and justify the validity of your computation.
 (Answer) \(\frac{20\mathbb{E}[N]}{\mathbb{E}[N/4]+1}\). Justify by SLLN



Useful Formulas

For a discrete random variable \(X\) taking values in \(\{0, 1, 2, \dots\}\), we have \(\mathbb{E}[X] = \sum_{k=0}^\infty P(X > k)\)

For a continuous random variable \(T\) taking values in \([0, \infty)\), we have \(\mathbb{E}[T] = \int_0^\infty P(T > s) ds\)
