Question:
N(t) is a Poisson process with parameter $\lambda$, let $S_n$ denotes the time when the $n$-th event happens. Try to calculate $E(S_4|N(1)=2)$.
Solution 1:
Let $F(x|N(1)=2)$ be the CDF. $$F(x|N(1)=2)=Pr(S_4 \leq x|N(1)=2)=\frac{Pr(N(x)-N(1) \geq 2,N(1)=2)}{Pr(N(1)=2)}=Pr(N(x)-N(1) \geq 2)$$ It's easy to get: $$Pr(N(x)-N(1) \geq 2)=1-e^{\lambda(x-1)}-\lambda(x-1)e^{-\lambda(x-1)}$$ So $$ F(x|N(1)=2)=\left\{ \begin{aligned} &0 &x \leq 1 \\ &1-e^{\lambda(x-1)}-\lambda(x-1)e^{-\lambda(x-1)} &x>1 \\ \end{aligned} \right. $$ Thus: $$E(S_4|N(1)=2)=\int_{0}^{\infty}1-F(x|N(1)=2)dx=1+\frac{2}{\lambda} $$
Solution 1 is correct and easy to understand. But there is a solution 2 which i can not understand.
Solution 2:
What we need to calculate is $E(S_4|N(1)=2)$, it's a conditional expectation. The time that the first and second events need equals 1. As we know, the time interval between events is exponential distribution with parameter $\lambda$, whose mean queals $\frac{1}{\lambda}$. After the first and second events, There are two events need to happen, thus the total expected time is $1+\frac{2}{\lambda}$
My question:
I am wondering if the logic of solution 2 is right? Because the time that the first and second events need is less than or equals 1.
The result is right but i am not sure if it's a conincidence.
The condition $N(1)=2$ means that exactly two events have occurred before time $1$.
Thus this condition specifies that a third event has not occurred by time $1$. (And, of course, neither the forth.)
The memoryless property of exponential distributions that the expected wait duration until the next event after any given time point will be $1/\lambda$.
(It does not matter when the second event has occurred before the given time point, only that it has occurred while the third has not occurred by then.)
Therefore the expected time for the occurrence of the third and forth event when given that the third does not occur before time $1$, will be the $1$ plus the expected wait duration for two events.$$\mathsf E(S_4\mid N(1)=2)=1+\mathsf E(S_2)$$
In general, where $m>n$ are integer event counts and $t$ a positive real time mark, $$\mathsf E(S_n\mid N(m)=t)~{=t+\mathsf E(S_{n-m})\\=t+(n-m)/\lambda}$$