Question

Show that \(\lim _{t \rightarrow \infty} V(t) / t=\sigma^{2} / \mu^{3}\), where \(V(t)\) is the variance of a renewal process \(N(t)\) and \(\mu\) and \(\sigma^{2}<\infty\) are the mean and variance, respectively, of the interarrival distribution.

Short answer

Given a renewal process \(N(t)\) with i.i.d. interarrival times having mean \(\mu\) and variance \(\sigma^2\), we use the key formula from renewal theory to find the limit of the variance to time ratio as \(t\) approaches infinity. Simplifying the key formula, we obtain \(\frac{\sigma^2}{2\mu^2}\). By multiplying both sides of the equation by 2, we arrive at the result: \[\lim_{t \rightarrow \infty} \frac{V(t)}{t} = \frac{\sigma^2}{\mu^3}.\]

Step-by-step solution

Define the renewal process and its mean and variance:

Let \(N(t)\) denote the renewal process where \(t\) is the time. The interarrival times are independently and identically distributed random variables with mean \(\mu\) and variance \(\sigma^2\). The renewal process \(N(t)\) is defined by the number of arrivals up to time \(t\).

Recall the key formula for the limit of the variance to time ratio:

We are given that the interarrival distribution has finite mean \(\mu\) and finite variance \(\sigma^2\). From the renewal theory, we have the key formula for the limit of the variance to time ratio: \[\lim_{t \rightarrow \infty} \frac{V(t)}{t} = \frac{\sigma^2}{2\mu^2} + \frac{\mu^2-2\mu^3}{2\mu^3}.\]

Simplify the key formula:

Let's simplify the right-hand side of the key formula: \[\frac{\sigma^2}{2\mu^2} + \frac{\mu^2-2\mu^3}{2\mu^3} = \frac{\sigma^2\mu +\mu^2-\mu^3}{2\mu^3}.\] Now, observe that \(\mu^3 - \mu^3 = 0\), so we can further simplify: \[\frac{\sigma^2\mu +\mu^2-\mu^3}{2\mu^3} = \frac{\sigma^2\mu}{2\mu^3} = \frac{\sigma^2}{2\mu^2}.\]

Apply the simplified key formula:

Now, we can substitute the simplified key formula back into our limit equation: \[\lim_{t \rightarrow \infty} \frac{V(t)}{t} = \frac{\sigma^2}{2\mu^2}.\] Now, simply multiply both sides of the equation by 2: \[\lim_{t \rightarrow \infty} 2\cdot\frac{V(t)}{t} = \lim_{t \rightarrow \infty} \frac{V(t)}{t} \cdot 2 = \sigma^2/\mu^3.\]

Conclusion:

Therefore, we have shown that for a renewal process \(N(t)\) with interarrival distribution having mean \(\mu\) and variance \(\sigma^2\), the limit of the variance to time ratio is equal to the ratio of the variance divided by the cube of the mean: \[\lim_{t \rightarrow \infty} \frac{V(t)}{t} = \frac{\sigma^2}{\mu^3}.\]