91Ó°ÊÓ

To compute the conditional expectation \(E[X \mid N(t)]\), we will first consider the total waiting time of one person who arrives at the bus stop at time \(s \in [0, t]\). Since the waiting time is uniformly distributed on the interval \([0, t]\), the expected waiting time for one person is \(t/2\). Now, given that there are \(N(t)\) people who arrived at the bus stop, their total waiting time is \(N(t)\) times the waiting time of a single person, which is \((N(t))(t/2)\). Therefore, the conditional expected value of \(X\) is: \[E[X \mid N(t)] = N(t) \cdot \frac{t}{2}\]

To derive the conditional variance of \(X\) given \(N(t)\), we first consider the variance of the waiting time of a single person who arrives at time \(s \in [0, t]\). Since it is uniformly distributed on the interval \([0, t]\), the variance for a single person is \(t^2 / 12\). Now, given that there are \(N(t)\) people who arrived at the bus stop, and since their waiting times are independent, the total variance of their waiting times is the sum of the variances of their individual waiting times. Thus, the conditional variance of \(X\) is: \[\operatorname{Var}[X \mid N(t)] = N(t) \cdot \frac{t^2}{12}\]

To find the overall variance of \(X\), we will use the law of total variance which states that: \[\operatorname{Var}(X) = E[\operatorname{Var}[X \mid N(t)]] + \operatorname{Var}(E[X \mid N(t)])\] We already computed the conditional variance in part (b) and the conditional expectation in part (a): \(\operatorname{Var}[X \mid N(t)] = N(t) \cdot \frac{t^2}{12}\) \(E[X \mid N(t)] = N(t) \cdot \frac{t}{2}\) Now, we need to compute \(E[\operatorname{Var}[X \mid N(t)]]\) and \(\operatorname{Var}(E[X \mid N(t)])\). Since the arrivals follow a Poisson process with rate \(\lambda\), we know that: \(E[N(t)] = \lambda t\) Therefore, \[E[\operatorname{Var}[X \mid N(t)]] = E\left[N(t) \cdot \frac{t^2}{12}\right] = \frac{t^2}{12} \cdot E[N(t)] = \frac{t^2}{12} \cdot (\lambda t)\] Next, we need to compute \(\operatorname{Var}(E[X \mid N(t)])\): \(\operatorname{Var}(E[X \mid N(t)]) = \operatorname{Var}\left(N(t) \cdot \frac{t}{2}\right) = \left(\frac{t}{2}\right)^2 \cdot \operatorname{Var}(N(t))\) Since \(N(t)\) follows a Poisson distribution with mean \(\lambda t\), its variance is also \(\lambda t\): \(\operatorname{Var}(E[X \mid N(t)]) = \left(\frac{t}{2}\right)^2 \cdot \lambda t\) Now, we can plug in the values we found for \(E[\operatorname{Var}[X \mid N(t)]]\) and \(\operatorname{Var}(E[X \mid N(t)])\) in the equation for the law of total variance and calculate the overall variance of \(X\): \[\operatorname{Var}(X) = E[\operatorname{Var}[X \mid N(t)]] + \operatorname{Var}(E[X \mid N(t)]) = \frac{t^2}{12} \cdot (\lambda t) + \left(\frac{t}{2}\right)^2 \cdot \lambda t\] Therefore, the variance of the total waiting time of people boarding the bus at time \(t\) is: \[\operatorname{Var}(X) = \lambda t^3\left(\frac{1}{12} + \frac{1}{4}\right)\]