Question

Let \(D\) denote the time between successive departures in a stationary \(M / M / 1\) queue with \(\lambda<\mu .\) Show, by conditioning on whether or not a departure has left the system empty, that \(D\) is exponential with rate \(\lambda\). Hint: By conditioning on whether or not the departure has left the system empty we see that $$ D=\left\\{\begin{array}{ll} \text { Exponential }(\mu), & \text { with probability } \lambda / \mu \\ \text { Exponential }(\lambda) * \text { Exponential }(\mu), & \text { with probability } 1-\lambda / \mu \end{array}\right. $$ where Exponential \((\lambda) *\) Exponential \((\mu)\) represents the sum of two independent exponential random variables having rates \(\mu\) and \(\lambda\). Now use moment-generating functions to show that \(D\) has the required distribution. Note that the preceding does not prove that the departure process is Poisson. To prove this we need show not only that the interdeparture times are all exponential with rate \(\lambda\), but also that they are independent.

Short answer

In an M/M/1 queue, the interdeparture time D can be characterized by two cases: 1. With probability \(\frac{\lambda}{\mu}\), the departure leaves the system empty, and D is exponentially distributed with rate μ. 2. With probability \(1 - \frac{\lambda}{\mu}\), the departure doesn't leave the system empty, and D is the sum of two independent exponential random variables with rates λ and μ. We determine the overall moment-generating function M(t) for D by combining the moment-generating functions for the two cases and their probabilities: M(t) = \(\frac{\lambda(\mu-t) + (\lambda\mu-\lambda^2)}{(\mu-t)(\lambda-t)}\) By simplifying this expression, we obtain that M(t) = \(\frac{\lambda\mu - \lambda t}{(\mu-t)(\lambda-t)}\), which is the moment-generating function of an exponential distribution with rate λ. Therefore, D is exponentially distributed with rate λ, conditioned on whether or not a departure has left the system empty.

Step-by-step solution

Identify the events

There are two possibilities: 1. The departure leaves the system empty with probability λ/μ, and the interdeparture time, D, is Exponential(μ). 2. The departure does not leave the system empty with probability 1 - λ/μ, and D is the sum of two independent exponential random variables having rates λ and μ.

Determine the moment-generating function for the two cases

For an exponential distribution with rate r, the moment-generating function is given by M(t) = \(\frac{r}{r-t}\), for t < r. In Case 1, D is Exponential(μ), so the moment-generating function of D is \(M_1(t) = \frac{\mu}{\mu-t}\). In Case 2, D is the sum of two independent exponential random variables with rates λ and μ. Let X ~ Exponential(λ) and Y ~ Exponential(μ) be the two independent random variables. The moment-generating functions of X and Y are \(M_X(t) = \frac{\lambda}{\lambda-t}\) and \(M_Y(t) = \frac{\mu}{\mu-t}\), respectively. Since X and Y are independent, the moment-generating function of D, the sum of X and Y, is given by \(M_2(t) = M_X(t) * M_Y(t) = \frac{\lambda}{\lambda-t} * \frac{\mu}{\mu-t}\).

Determine the overall moment-generating function of D

We can find the overall moment-generating function of D, M(t), by using the probabilities and individual moment-generating functions: M(t) = (Probability of Case 1) * \(M_1(t)\) + (Probability of Case 2) * \(M_2(t)\) M(t) = \((\frac{\lambda}{\mu}) * \frac{\mu}{\mu-t} + (1 - \frac{\lambda}{\mu}) * \frac{\lambda\mu}{(\lambda-t)(\mu-t)}\)

Simplify the expression for the moment-generating function M(t)

Now, we simplify the expression obtained above: M(t) = \(\frac{\lambda(\mu-t) + (\lambda\mu-\lambda^2)}{(\mu-t)(\lambda-t)}\) M(t) = \(\frac{\lambda\mu - \lambda t}{(\mu-t)(\lambda-t)}\) This is the moment-generating function of an exponential distribution with rate λ, since M(t) = \(\frac{\lambda}{\lambda-t}\) for t < λ. Thus, we have shown that the random variable D representing the interdeparture time in an M/M/1 queue is exponentially distributed with rate λ, conditioned on whether or not a departure has left the system empty.