4.2 Interarrival and Waiting Time Distributions
Consider a Poisson process \(\{N(t), \, t\geq 0\}\) of rate \(\lambda \).
Interarrival time: Time between two consecutive arrivals.
Waiting time of the \(n^{\text {th}}\) event \(W_n\) is the waiting time of the \(n^{\text {th}}\) event i.e time from \(t = 0\) to the instant when \(n^{\text {th}}\) event occurred.
\[P(T_1 > t) = P(N(t) = 0) = e^{-\lambda t}.\]
Hence \(P(T_1 \leq t) = 1 - e^{-\lambda t}\,\) (Distribution function of \(T_1\))
\[f_{T_1}(t) = \lambda \, e^{-\lambda t}.\]
Hence \(T_1\) follows exponential distribution with parameter \(\lambda \).
Distribution of \(T_2\). \begin {align*} P\left (T_2 > t\, |\, T_1 = s\right ) & = P\left (N(s, s + t) = 0\, |\, N(s) = 1\right )\\ & =P\left (N(s, s+ t) = 0\right )\\ & = e^{-\lambda t}. \end {align*}
Distribution of \(T_2\) is also exponential.
\(\{T_1, \, T_2, \, \cdots \, T_n\}-\) each is an exponential.
Let \(F_{S_n}(t)\) be the distribution function of random variable \(S_n\) (waiting time) \begin {align*} F_{S_n}(t) & = P\left (S_n \leq t\right )\\ & = P\left (N(t) \geq n\right )\\ & = \sum ^{\infty }_{i = n} \frac {e^{-\lambda \, t}\, (\lambda t)^i}{i!}. \end {align*}
Density function \[f_{S_n}(t) = \frac {d}{dt}\, F_{S_n}(t)\] where \(f_{S_n}(t)\) is the pdf of \(S_n\) \begin {align*} f_{S_n}(t) & = \sum ^{\infty }_{i = n} \frac {1}{i!} \left (-\lambda \, e^{-\lambda t}\, (\lambda t)^i + i\, \lambda \, (\lambda t)^{i -1}\, e^{-\lambda t}\right )\\ & = -\lambda \, \sum ^{\infty }_{i = n} \frac {e^{-\lambda t}\, (\lambda t)^i}{i!} \, + \, \sum ^{\infty }_{i = n} \frac {\lambda \, e^{-\lambda t}\, (\lambda t)^{i - 1}}{(i - 1)!}\\ & = -\lambda \, e^{-\lambda t}\, \sum ^{\infty }_{i = n} \frac {(\lambda t)^i}{i!} + \frac {\lambda \, e^{-\lambda t}\, (\lambda t)^{n - 1}}{(n - 1)!} + \sum ^{\infty }_{i = n+ 1} \lambda \, e^{-\lambda t}\, \frac {(\lambda t)^{i - 1}}{(i - 1)!}\\ & = \frac {\lambda \, e^{-\lambda t}\, (\lambda t)^{n - 1}}{(n - 1)!}\hspace {0.2cm} , \hspace {0.5cm} t >0. \end {align*}
\(S_n\) is Gamma random variable.
- 1.
- Suppose that people immigrate into a territory at a Poisson rate \(\lambda = 1\) per day.
- (a)
- What is the expected time until the tenth immigrant arrives?
Solution. \(E\left (S_{10}\right ) = \dfrac {10}{1} = 10\) days. □
- (b)
- what is the probability that the elapsed time between the tenth and eleventh arrival
exceeds two days?
Solution. \(\begin {aligned}[t] P(T_{11} > 2) & = \int ^{\infty }_2 e^{-x}\, dx \\ & = \lim _{a\rightarrow \infty } \int ^a_2 e^{-x}\, dx\\ & = \lim _{a\rightarrow \infty } \left [-e^{-x}\right ]^a_2\\ & = \lim _{a\rightarrow \infty } \left [-e^{-a} + e^{-2}\right ]\\ & = e^{-2}. \end {aligned}\)
□
- 2.
- The time required to repair a machine is an exponential distribution random variable with
mean \(\frac {1}{2}\) hrs. What is the probability that repair time exceeds \(\frac {1}{2}\) hours.
Solution. \(\dfrac {1}{\lambda } = \dfrac {1}{2}\, \implies \, \lambda = 2.\) \[Pdf. \, = 2\, e^{-2x}\hspace {0.3cm} , \hspace {0.4cm} x \geq 0\] \begin {align*} P\left (X\geq \frac {1}{2}\right ) & = \int ^{\infty }_{\frac {1}{2}} 2\, e^{-2\,x }\, dx\\ & = \lim _{k \rightarrow \infty } \int ^k_{\frac {1}{2}} 2\, e^{-2\, x}\, dx\\ & = e^{-1}. \end {align*} □
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