Let $N(t)$ is a poisson process with rate $\lambda$, $T^* \sim \operatorname{Exp}(\lambda^*)$, find the expectation of $N(\min(t, T^*))$

Question

Currently, my approach is to split $N(\min(t, T^*))$ like the following by the law of total expectation.

\begin{align*} &E(N(\min(t,T^*))) = E(N(t \wedge T^*)) \\ = {}&E(N(t\wedge T^*) \mid t \le T^*) \cdot P(t \le T^*) + E(N(t\wedge T^*) \mid t > T^*) \cdot P(t > T^*). \end{align*}

To make it less confusing and more concrete, one could think of the target expectation as $E(N(\min(3, T^*)))$ when $t=3$. Later, generalize the solution to suit any given $t$.

The first and second parts become

\begin{align*} &E(N(t\wedge T^*)\mid t \le T^* )\\ ={}& \sum_{x = 0}^\infty x P(N(t\wedge T^*) = x\mid t \le T^*)\\ ={}& \sum_{x = 0}^\infty x \frac{P(N(t\wedge T^*) = x, t \le T^*)}{P(t \le T^*)}\\ ={}& \sum_{x = 0}^\infty x \frac{P(N(t) = x, t \le T^*)}{P(t \le T^*)}\\ ={}& \sum_{x = 0}^\infty x \frac{P(N(t) = x) P(t \le T^*)}{P(t \le T^*)}\\ ={}& \sum_{x = 0}^\infty x P(N(t) = x)\\ ={}& E(N(t)) = \lambda t; \end{align*}

\begin{align*} &E(N(t\wedge T^*) \mid t > T^*)\\ ={}& \sum_{x = 0}^\infty x P(N(t\wedge T^*) = x\mid t > T^*) \hspace{1cm}\color{red}{T^* < t}\\ ={}& \sum_{x = 0}^\infty x \frac{P(N(t\wedge T^*) = x \mid t > T^*)}{P(t > T^*)} \\ ={}& \sum_{x = 0}^\infty x \frac{P(N(t\wedge T^*) = x ,t > T^*)}{P(t > T^*)}\\ ={}& \sum_{x = 0}^\infty x \frac{P(N(T^*) = x ,T^* < t)}{P(T^* < t)} \end{align*}

by the conditional expectation.

So, the problem is $N(T^*)$ and $T^*$ are less likely to be independent; how to find the probability $P(N(T^*) = x ,T^* < t)$?

Answer 1

We can use the law of total expectations to simplify the problem as follows:

$$\mathbb{E} N(\min(t,T^*)) = \lambda \mathbb{E}\min(t,T^*)$$

Finding $\mathbb{E} \min(t,T^*)$ is straightforward:

$$\mathbb{E} \min(t,T^*) = \mathbb{E} (T^*|T^*\leq t)P(T^*\leq t) + tP(T^*> t) \tag 1$$

Using the memoryless property of the exponential distribution allows us to simplify further, as:

$$\mathbb{E} T^* = \mathbb{E}(T^*|T^*\leq t)P(T^*\leq t) + \mathbb{E}(T^*|T^*> t)P(T^*> t)$$

and $\mathbb{E} (T^*|T^*> t) = \lambda^*+t$ while $\mathbb{E} T^* = \lambda^*$. Therefore, rearranging terms,

$$\mathbb{E}(T^*|T^*\leq t)P(T^*\leq t) = \lambda^* - (\lambda^*+t)P(T^*>t)$$

Substituting into the right hand side of eq. (1) gives us:

$$\mathbb{E}\min(t,T^*) = \lambda^* - (\lambda^*+t)P(T^*>t) + tP(T^*> t) = \lambda^*(1-e^{-t/\lambda^* })$$

One more substitution gets us to:

$$\mathbb{E} N (\min(t,T^*)) = \lambda \lambda^*(1-e^{-t/\lambda^* })$$

Now for a check via simulation. We will set $\lambda = 1.138$, $\lambda^* = 0.842$, and $t = 1.234$:

lambda <- 1.138
lambdastar <- 0.842
tc <- 1.234
 
tstar <- rexp(100000, 1/lambdastar)
x <- rpois(100000, lambda*pmin(tstar, tc))
 
lambda*lambdastar*(1-exp(-tc/lambdastar))
[1] 0.7369016
 
mean(x)
[1] 0.73378
sd(x)/sqrt(length(x))
[1] 0.003133204

which seems to match as well as we would hope.