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8.2 Poisson Process

Definition 8.2.1. A function \(f(x)\) is said to be \(O(h)\) if \(\displaystyle {\lim _{h\rightarrow 0}\frac {f(x)}{h}=0}\)

i.e \(f\) is \(O(h)\) if for small values of \(h\), \(f(h)\) is small in relation to \(h\). For example \(\dfrac {e^{-1/x}}{1/x}\) say \(h=1/10\) then \(\dfrac {e^{-1/10}}{1/10}\), thus \(h\rightarrow 0\) as \(x\rightarrow \infty \).

Suppose that ”events” occur at random time points and let \(N(t)\) denote the number of events that occur in a time interval \([0,t]\).

Definition 8.2.2. The collection of random variables \(\{N(t),\hspace {0.1cm}t\geq 0\}\) is said to be a Poisson process with rate \(\lambda \), \(\lambda >0\) if

i.
\(N(0)=0\)
ii.
The number of events that occur in disjoint time intervals are
independent.
iii.
The distribution of the number of events that occur in a given time interval depends only on the length of that interval.
iv.
\(P\{N(t)=1\}=\lambda h+O(h)\)
v.
\(P\{N(t)\geq 2\}=O(h)\)

Remark. Condition (i): process starts at time zero.
Condition (ii): independent \((t,t+s)\), \((t+k,t+b)\) where \(k>s\) and \(b>k\)

tttt+++ skb

Lemma 8.2.3. \(\text {For a Poisson process with rate}\hspace {0.3cm} \lambda ,\hspace {0.3cm} P\{N(t)=0\}=e^{-\lambda t}.\)

Proof. \(\text {Denote}\hspace {0.4cm} P_0(t)=P\{N(t)=0\}.\) \begin {align*} P_0(t+h) &=P\{N(t+h)=0\}\\ &=P\{N(t)=0,\hspace {0.1cm} N(t+h)-N(t)=0\}\\ &=P\{N(t)=0\}\hspace {0.1cm} P\{N(t+h)-N(t)=0\}\\ &=P_0(t)\hspace {0.1cm} P\{N(h)=0\}\\ &=P_0(t)\hspace {0.1cm} (1-\lambda h+O(h))\\ &=P_0(t)-\lambda hP_0(t)-O(h) \end {align*}

\begin {align*} P_0(t+h)-P_0(t) &=-\lambda hP_0(t)-O(h)\\\\ \frac {P_0(t+h)-P_0(t)}{h} &=-\lambda P_0(t)-\frac {O(h)}{h}\\\\ \lim _{h\rightarrow 0}\frac {P_0(t+h)-P_0(t)}{h} &=-\lambda P_0(t)-\lim _{h\rightarrow 0}\frac {O(h)}{h}\\\\ \therefore \hspace {0.4cm} P'_0(t) =-\lambda P_0(t)\hspace {0.5cm}&\implies \hspace {0.5cm} \frac {P'_0(t)}{P_0(t)}=-\lambda \end {align*}

\begin {align*} \int \frac {P'_0(t)}{P_0(t)}dt &=-\int \lambda dt\\\\ \implies \hspace {0.5cm} \ln P_0(t) &=-\lambda t+c\\\\ \implies \hspace {0.5cm} P_0(t) &=ke^{-\lambda t},\hspace {0.4cm} \text {where}\hspace {0.3cm} k=e^c \end {align*}

\[P_0(t)=ke^{-\lambda t}\] \[t=0,\hspace {0.5cm} P_0(0)=k,\hspace {1cm} \therefore \hspace {0.3cm} k=1\] \[\therefore \hspace {0.5cm} P_0(t)=e^{-\lambda t}\] \[\implies \hspace {0.5cm} P\{N(t)=0\}=e^{-\lambda t}.\]

Denote \(T_n\) to be the time elapsed between the \((n-1)^{\text {th}}\) event and the \(n^{\text {th}}\) event. The sequence \(\{T_n\}\) is called the sequence of inter arrival times. For example, if \(T_1=4\) and \(T_2=12\), then the first event occurred at time \(4\) units and the second event occurred at time \(16\).

041TT612

\begin {align*} P(T_1>t) &=P\{N(t)=0\}=e^{-\lambda t}\\\\ \implies \hspace {0.5cm} P(T_1\leq t) &=1-e^{-\lambda t}\hspace {0.3cm}\implies \hspace {0.3cm} T_1\hspace {0.3cm}\text {is exponential with mean}\hspace {0.3cm} \frac {1}{\lambda }.\\\\ \therefore \hspace {0.5cm} P(T_2>s) &=E\Big (\frac {P(T_2>s)}{T_1}\Big ).\\\\ P(T_2>s \mid T_1=t) &=P(\text {no event in the interval}\hspace {0.3cm} (t,t+s))\\\\ &=P(N(s)=0)\\\\ &=e^{-\lambda s} \end {align*}

\(T_2\) has an exponential distribution with mean \(1/\lambda \).

\(\therefore \) it follows that \(T_n\) is exponential with mean \(1/\lambda \), \(n=1,2,.......,n\)

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Result 8.2.4. Let \(\{T_n\}\) be a sequence of inter arrival times of a Poisson process, then \(T_1,T_2,..........\) are independent identically distributed as exponential with mean \(1/\lambda \).

Denote the arrival of the \(n^{\text {th}}\) event by \(S_n\), which is the waiting time until the \(n^{\text {th}}\) event occurs \[\text {i.e}\hspace {0.5cm}S_n=\sum ^n_{i=1}T_i\] What is the distribution of \(S_n\)?

Since it is a sum of \(iid\) \(g(1,\lambda )\) \begin {align*} \therefore \hspace {0.3cm} f_{S_m}(x) &=\frac {\lambda ^n\hspace {0.1cm} x^{n-1}\hspace {0.1cm} e^{-\lambda x}}{\Gamma (n)}\\\\ &\frac {(\lambda x)^{n-1}\hspace {0.1cm}\lambda \hspace {0.1cm} e^{-\lambda x}}{(n-1)!},\hspace {0.5cm} x>0 \end {align*}

We want \(P\{N(t)=n\}\).

Theorem 8.2.5. For a Poisson process with rate \(\lambda \), \(N(t)=\) the number of events that occur in time interval \([0,t]\), \[P\{N(t)=n\}=\frac {(\lambda t)^n\hspace {0.1cm e^{-\lambda t}}}{n!},\hspace {0.5cm} n=0,1,........, n\]

Proof. For a Poisson process with rate \(\lambda \), \(N(t)=\) the number of event in time interval \([0,t]\). The \(n^{\text {th}}\) event will only occur before or at time \(t\). If and only the number of events that occur by time \(t\) is at atleast \(n\) i.e \(N(t)\geq n\). \[S_n=T_1+T_2+.........+T_n=\sum ^n_{i=1}T_i\] \[\text {then we have}\hspace {0.5cm} N(t)\geq n \Longleftrightarrow S_n\leq t.\] \begin {align*} P\{N(t)=n\} &=P\{N(t)\geq n\}-P\{N(t)\geq n+1\}\\ &=P\{S_n\leq t\}-P\{S_{n+1}\leq t\}\\\\ &=\int ^t_0\frac {(\lambda x)^{n-1}\lambda e^{-\lambda x}}{(n-1)!}dx-\int ^t_0\frac {(\lambda x)^n\lambda e^{-\lambda x}}{n!}dx\\\\ &=\int ^t_0\frac {(\lambda x)^{n-1}\lambda e^{-\lambda x}}{(n-1)!}dx-\Bigg (\frac {(\lambda x)^n}{n!}\hspace {0.1cm}\lambda (-1)\frac {e^{-\lambda x}}{\lambda }\Bigg |^t_0+\int ^t_0\frac {d}{dx}(\lambda x)^n\frac {e^{-\lambda x}}{n!}dx \end {align*}

\begin {align*} \implies \hspace {0.5cm} P\{N(t)=n\} &=\int ^t_0\frac {(\lambda x)^{n-1}\lambda e^{-\lambda x}}{(n-1)!}dx+\frac {(\lambda t)^ne^{-\lambda t}}{n!}-0-\int ^t_0\frac {(\lambda x)^{n-1}\lambda e^{-\lambda x}}{(n-1)!}dx\\\\ &=\frac {(\lambda t)^ne^{-\lambda t}}{n!} \end {align*}

Q.E.D

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Example 8.2.6. If it was reported that an insurance company with a \(24\) hour hot line, the claims followed a Poisson process with rate of \(4\) claims per day. Find the following

i.
probability that more than \(2\) claims were being reported on a given day.
ii.
the time until the next reported claim is more than an hour.

i.
\begin {align*} P(N(t)>2) &=1-P\{N(t)\leq 2\}\\ &=1-\sum ^2_{k=0}\frac {4^ke^{-4}}{k!}\\ &=1-e^{-4}\Big (1+4+\frac {4^2}{2}\Big )\\ &=1-13e^{-4}\\ &=0.7619 \end {align*}
ii.
\begin {align*} P(T>1),&\hspace {0.5cm} T\thicksim \exp \Big \{\frac {4}{24}\Big \}=\exp \Big \{\frac {1}{6}\Big \}\\ &\implies \hspace {0.4cm} E(T)=6\\\\ P(T>1)&=e^{-\lambda }=e^{-1/6} \end {align*}

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