Let $N(t)$ be a Poisson Process of rate λ, let X1, X2, X3, . . . denote the interarrival times, and let S1 = X1, S2 = X1 + X2, . . . be the times of the arrivals (jumps). Compute:
E[X3|N(t) = 2]
My first gut instinct reminds me of the memoryless property of exponential distribution and tells me that the answer is $\frac{1}{\lambda}$ and $E[S3|N(t)=2] = \frac{1}{\lambda} + 2$. However, it is wrong.
I instead used:
X3 = S3 −S2 = (t−S2)+(S3 −t). Conditional on N(t) = 2, we notice: For the first term, that S2 = max{U1,U2}, where U1,U2 are independent, uniform RV’s on [0, t]. For the second term, that S3 − t is Exp(λ) by the no-memory property. Thus, we have $E[X3|N(t)=2]=t−Emax\{U1,U2\}+E[Exp(λ)]=t−\frac{2}{3}t+\frac{1}{\lambda} =\frac{t}{3} +\frac{1}{\lambda}$.
I can't wrap my head around how the result is $> \frac{1}{\lambda}$, which is the average time for $X_i$. I also would think that since it is unknown that the second occurrence happened exactly, or (most likely) before t, the time to wait for the third occurrence would be shorter than average value $\frac{1}{\lambda}$. If anyone can explain, that will be awesome. Thank you!