As the title suggests, I have a question on the find the expectation of a pure birth process.
Let $\{X(t), t \geq0 \}$ be a pure birth process with $X(0)=1$ and birth rates $\lambda _{k}=k $ for $k=1,2, \dots$. Let $T=\inf \{t \geq0: X(t)=4\}$. Find $E(\int_{0}^{T}X \big(t)dt \big)$.
I am sorry if this question seem rather trivial as it is simply some computations. I have recently learnt this in class but have yet to gain enough exposure to know how to 'start' doing such question. Could the question be solved by simply just applying the formula: $E(X)= \int_{\mathbb{R}} xf(x) dx$ for a continuous random variable?
Since $X(t)$ is a pure birth process, it will behave like a step function on random intervals. For example, let $T_1$ be the first jump time, so that $X(0) = 1$ on $[0, T_1)$ until the first jump to $2$. And so on. Specifically, on the interval $[0, T]$,
$$ X(t) = \begin{cases}1 & t \in [0,T_1) \\ 2 & t \in [T_1,T_2) \\ 3 & t \in [T_2,T) \\ 4 & t = T\end{cases} $$
So you can calculate the integral like you would a step function,
$$ \int_0^TX(t)dt = T_1 + 2(T_2-T_1) + 3(T-T_2)$$
Finally, what you do know about the "inter-arrival" times, $T_i - T_{i-1}$?