Question

Let \(\\{N(t), t \geqslant 0\\}\) be a conditional Poisson process with a random rate \(L\). (a) Derive an expression for \(E[L \mid N(t)=n]\). (b) Find, for \(s>t, E[N(s) \mid N(t)=n]\). (c) Find, for \(s

Short answer

In summary, we have derived expressions for the following quantities: (a) \(E[L \mid N(t) = n] = \frac{\int_{0}^{\infty} l \cdot f_{N(t)|L}(n|l) \cdot f_{L}(l) dl}{\frac{e^{-Lt}(Lt)^n}{n!}}\) (b) \(E[N(s) \mid N(t) = n] = (s-t)E[L] + n\) for \(s > t\) (c) \(E[N(s) \mid N(t) = n] = \sum_{k=0}^{n} k \cdot e^{-L(t-s)} \frac{(L(t-s))^{n-k}}{(n-k)!}\) for \(s < t\)

Step-by-step solution

Find an expression for E[L | N(t) = n]

Recall the definition of the conditional expectation: \[ E[L | N(t) = n] = \frac{E[L \cdot 1_{\{N(t) = n\}}]}{P(N(t) = n)} \] where 1 is the indicator function, taking the value 1 when the observed N(t) = n and 0 otherwise. Given that N(t) follows a Poisson process with random rate L, we have: \[ P(N(t) = n) = \frac{e^{-Lt}(Lt)^n}{n!} \] We know that the joint probability density function of L and N(t) is: \[ f_{L,N(t)}(l,n) = f_{L}(l) \cdot P(N(t) = n|L=l) \] So, in order to find E[L | N(t) = n], we need to find E[L · 1_{N(t) = n}]. This can be obtained as follows: \[ E[L \cdot 1_{\{N(t) = n\}}] = \int_{0}^{\infty} l \cdot f_{N(t)|L}(n|l) \cdot f_{L}(l) dl \] Plug this back into the formula for the conditional expectation: \[ E[L | N(t) = n] = \frac{\int_{0}^{\infty} l \cdot f_{N(t)|L}(n|l) \cdot f_{L}(l) dl}{\frac{e^{-Lt}(Lt)^n}{n!}} \] We can simplify this further.

Find E[N(s) | N(t) = n] for s > t

To find E[N(s) | N(t) = n] for s > t, we use the fact that increments in a Poisson process are independent. Therefore, given N(t) = n, the expected value of N(s) doesn't depend on the history before time t. So we have: \[ E[N(s) | N(t) = n] = E[N(s) - N(t) + N(t) | N(t) = n ] \] Using the linearity of expectations, we get: \[ E[N(s) | N(t) = n] = E[N(s) - N(t)] + E[N(t) | N(t) = n] \] Now, since increments are independent, the expected value of N(s) - N(t) is simply the expected value of a Poisson process over the interval (t, s] with the rate L: \[ E[N(s) - N(t)] = (s-t)E[L] \] And we have: \[ E[N(t) | N(t) = n] = n \] Putting the pieces together, we have: \[ E[N(s) | N(t) = n] = (s-t)E[L] + n \]

Find E[N(s) | N(t) = n] for s < t

For s < t, we first note that N(s) and (N(t) - N(s)) are independent by the properties of Poisson processes. We have: \[ E[N(s) | N(t) = n] = E[N(s) | N(t) - N(s) = n - N(s)] \] Now, we can use the conditioning and the fact that (N(t) - N(s)) is independent of N(s) to rewrite the conditional expectation as follows: \[ E[N(s) | N(t) = n] = \sum_{k=0}^{n} E[N(s) | N(t) - N(s) = n - k] P(N(t) - N(s) = n - k) \] For each k, (N(t) - N(s)) being Poisson with parameter L(t-s), we have: \[ P(N(t) - N(s) = n - k) = e^{-L(t-s)} \frac{(L(t-s))^{n-k}}{(n-k)!} \] Therefore, \[ E[N(s) | N(t) = n] = \sum_{k=0}^{n} E[N(s) | N(t)-N(s) = n-k] e^{-L(t-s)} \frac{(L(t-s))^{n-k}}{(n-k)!} \] Now, using the properties of Poisson processes, we can find the conditional expectation for each k: \[ E[N(s) | N(t)-N(s) = n-k] = k \] Finally, substitute this back into the formula for E[N(s) | N(t) = n]: \[ E[N(s) | N(t) = n] = \sum_{k=0}^{n} k \cdot e^{-L(t-s)} \frac{(L(t-s))^{n-k}}{(n-k)!} \]