Question

Suppose that events occur in accordance with a Poisson process at the rate of five events per hour.

a. Determine the distribution of the waiting time \({{\bf{T}}_{\bf{1}}}\) until the first event occurs.

b. Determine the distribution of the total waiting time \({{\bf{T}}_{\bf{k}}}\) until k events have occurred.

c. Determine the probability that none of the first k events will occur within 20 minutes of one another.

Short answer

a. The distribution is exponential with parameters \(\beta = 5\).

b. The distribution is Gamma with parameters \(\alpha = k\,\,and\,\,\beta = 5\)

c. The required probability is \({e^{ - \frac{{5\left( {k - 1} \right)}}{3}}}\)

Step-by-step solution

Given information

The rate of A random variable following the Poisson process is 5 events per hour.

(a) Distribution for waiting time until the first event occurs

We know that the proofs of the results in the Poison process have established that theexponential distribution models the time to wait for the first event. It is because of the memoryless property of this distribution.

Therefore, the waiting time T₁until the first event occurs follows an exponential distribution with the same parameter as the Poisson process.

Therefore,

\({T_1} \sim \exp \left( {\beta = 5} \right)\)

(b) Distribution for waiting time until the first k event occurs

We know that the proofs of the results in the Poison process have established that the Gamma distribution is the sum of some Exponential distributions and thus is the waiting time distribution forkevents.

Now,

\({T_1} \sim \exp \left( {\beta = 5} \right)\)

And we know that

\(Gamma\left( {\alpha ,\beta = 5} \right) = \sum\limits_\alpha {Exp\left( {\beta = 5} \right)} \)

Since we are waiting for the first k events, \({T_k}\), \(\alpha = k\).

Therefore, the distribution follows Gamma distribution, that is

\({T_k} \sim Gamma\left( {\alpha = k,\beta = 5} \right)\)

(c) Probability that none of the first k events will occur within 20 minutes of one another

20 minutes are to be converted to hours:

\(\begin{array}{c}t = \frac{{20}}{{60}}\\ = \frac{1}{3}\end{array}\)

The parameter value is 5, with the occurrence of first k events not happening.

The parameter value hence changes to k-1.

Therefore, the distribution is,

\(\begin{array}{l} = {e^{ - \lambda t}}\\ = {e^{ - \frac{{5\left( {k - 1} \right)}}{3}}}\end{array}\)