91Ó°ÊÓ

The interarrival times of the renewal process are given to be uniformly distributed over the interval (0,1). A uniform distribution has a probability density function (PDF) given by: \[f(t) = \begin{cases} 1, & 0 \le t \le 1 \\ 0, & \text{otherwise} \end{cases}\]

Now we want to find the expected remaining time from \(t=1\) until the next renewal. In other words, let \(R\) be the remaining time, and we want to find \(E[R | R>t]\) for \(t=1\). The Conditional Probability Density Function (CPDF) of \(R\) given \(R>t\) can be calculated as: \[f(r|R>t) = \frac{f(r)}{P(R>t)}\] Where: - \(f(r)\) is the PDF of the interarrival times - \(P(R>t)\) is the probability that the remaining time is greater than \(t\)

We will calculate \(P(R>t)\) for \(t=1\) using the interarrival time distribution. Since the interarrival times are uniformly distributed over (0,1), any \(t\) in the interval (0,1) will have a \(1-t\) probability of having the next interarrival time greater than \(t\). Thus, \(P(R>1) = 1 - 1 = 0\).

The conditional probability density function for t=1 was calculated in step 2: \[f(r|R>1) = \frac{f(r)}{P(R>1)}.\] Since \(P(R>1) = 0\), we will have an undefined CPDF. However, we can adjust our approach slightly. Instead of considering \(t=1\) exactly, let's consider an interval \((1-\epsilon, 1+\epsilon)\) and then take the limit as \(\epsilon\) approaches 0.

Now, we will calculate the expected value of the remaining time \(R\) starting at \(t=1\), or \(E[R | R>t]\) for \(t=1\). Given the CPDF of step 4, we will compute this with the following integral: \[E[R | R\in(1-\epsilon,1+\epsilon)] = \lim_{\epsilon \to 0}\frac{\int_{1-\epsilon}^{1+\epsilon} r f(r)\, dr}{\int_{1-\epsilon}^{1+\epsilon} f(r)\, dr}\] Now integrate, using the PDF given in step 1: \[E[R | R\in(1-\epsilon,1+\epsilon)] = \lim_{\epsilon \to 0}\frac{\int_{1-\epsilon}^{1+\epsilon} 1\cdot r\, dr}{\int_{1-\epsilon}^{1+\epsilon} 1\cdot dr}\] \[= \lim_{\epsilon \to 0}\frac{\left[\frac{1}{2}r^2\right]_{1-\epsilon}^{1+\epsilon}}{[r]_{1-\epsilon}^{1+\epsilon}}\] Now apply the limits of integration and take limit as \(\epsilon\) approaches 0: \[= \lim_{\epsilon \to 0}\frac{\frac{1}{2}[(1+\epsilon)^2 - (1-\epsilon)^2]}{2\epsilon}\] \[= \lim_{\epsilon \to 0}\frac{1}{2}\cdot\frac{4\epsilon}{2\epsilon}\] \[= \frac{1}{2}\]

The expected remaining time from t=1 until the next renewal for the given renewal process with interarrival times uniformly distributed over (0,1) is 1/2.