Is there a nice way to present solution to E(S₄) for Poisson process assuming I am correct

Question

Let S4 the the time of the 4th event in a poisson process with rate $\lambda$ My intuition tells me the expected amount of time to 4th event is $(4/λ)$ But that is not satisfactory $$E(S_4)=E(E(S_4|S_3))=E(E(E(S_4|S_3|S_2)))=E(E(E(E(S_4|S_3|S_2|S_1)))) =E(E(E((1/λ)+S_3|S_2|S_1)))=E(E((1/λ)+(1/λ)+S_2|S_1))= \text{etc}...$$

Is this usual notation or am I missing something?

Answer 1

We can simplify the question if we write $S_n$ as a sum of interarrival times. If we let $T_1, T_2, \cdots $ be the interarrival times of each event, then $T_i\sim \text{Exp}(\lambda)$. Hence \begin{align*} S_4&=T_1+T_2+T_3+T_4\\ \mathbb E(S_4)&=\mathbb E(T_1+T_2+T_3+T_4)=\mathbb E(T_1)+\mathbb E(T_2)+\mathbb E(T_3)+\mathbb E(T_4)=4\cdot \frac1\lambda =\frac4\lambda \end{align*}