Question

Suppose that the interarrival distribution for a renewal process is Poisson distributed with mean \(\mu .\) That is, suppose $$ P\left\\{X_{n}=k\right\\}=e^{-\mu} \frac{\mu^{k}}{k !}, \quad k=0,1, \ldots $$ (a) Find the distribution of \(S_{n}\). (b) Calculate \(P\\{N(t)=n\\}\).

Short answer

(a) The distribution of \(S_n\) is a Poisson distribution with mean \(n\mu\): $$P\{S_n=k\} = e^{-n\mu} \frac{(n\mu)^k}{k!}, \quad k=0,1,\ldots$$ (b) The probability \(P\{N(t)=n\}\) is given by: $$P\{N(t)=n\} = \left(\sum_{k=0}^{\lfloor t \rfloor} e^{-n\mu} \frac{(n\mu)^k}{k!}\right) - \left(\sum_{k=0}^{\lfloor t \rfloor} e^{-(n+1)\mu} \frac{((n+1)\mu)^k}{k!}\right)$$

Step-by-step solution

(a) Finding the distribution of \(S_n\)

Since the interarrival times are each distributed according to a Poisson distribution with mean \(\mu\), we have \(n\) independent Poisson-distributed random variables. Let \(X_1, X_2, \ldots, X_n\) represent these interarrival times, then the sum of the first \(n\) interarrival times can be denoted as: $$S_n = X_1 + X_2 + \ldots + X_n$$ Given that the interarrival times are independent, we can use the property of independent Poisson distributions which states that the sum of independent Poisson random variables is also a Poisson variable, with its mean equal to the sum of the means of the independent variables: $$E(S_n) = E(X_1) + E(X_2) + \ldots + E(X_n) = n\mu$$ Therefore, the distribution of \(S_n\) is a Poisson distribution with mean \(n\mu\). We can write its probability mass function as: $$P\{S_n=k\} = e^{-n\mu} \frac{(n\mu)^k}{k!}, \quad k=0,1,\ldots$$

(b) Calculating \(P\{N(t)=n\}\)

To calculate the probability \(P\{N(t)=n\}\), we need to determine the probability that there are \(n\) arrivals by time \(t\). In other words, we want to find the probability that \(S_n \le t\) and \(S_{n+1} > t\). We can use the cumulative distribution function (CDF) of the Poisson distribution for \(S_n\) to find this probability: $$P\{S_n \le t\} = \sum_{k=0}^{\lfloor t \rfloor} P\{S_n=k\} = \sum_{k=0}^{\lfloor t \rfloor} e^{-n\mu} \frac{(n\mu)^k}{k!}$$ $$P\{S_{n+1} > t\} = 1 - P\{S_{n+1} \le t\} = 1 - \sum_{k=0}^{\lfloor t \rfloor} e^{-(n+1)\mu} \frac{((n+1)\mu)^k}{k!}$$ Now, we find the probability \(P\{N(t)=n\}\) by taking the difference of the CDFs: $$P\{N(t)=n\} = P\{S_n \le t\} - P\{S_{n+1} \le t\}$$ $$= \left(\sum_{k=0}^{\lfloor t \rfloor} e^{-n\mu} \frac{(n\mu)^k}{k!}\right) - \left(\sum_{k=0}^{\lfloor t \rfloor} e^{-(n+1)\mu} \frac{((n+1)\mu)^k}{k!}\right)$$