Give the pdf of $T_1|N_1=2$ where $T_1$ is the time of first arrival

Question

Give the pdf of $T_1|N_1=2$ where $T_1$ is the time of first arrival, and $(N_t, t\geq 0)$ is a Poisson process.

Attempt:

First find the cdf:

For $0<t<1$: \begin{align} P(T_1\le t\ |\ N_1=2) &={P(T_1\le t,N_1=2)\over P(N_1=2)}\\ &={P(N_t= 1,N_1-N_t=1)\over P(N_1=2)}\\ &={P(N_t= 1)P(N_1-N_t=1)\over P(N_1=2)}\\ &={P(N[0,t]= 1,N(t,1]=1)\over P(N[0,1]=2)}\\ &={P(N[0,t]= 1)P(N(t,1]=0)\over P(N[0,1]=2)}\\ &={2\lambda te^{-\lambda t}\cdot \lambda(1-t)e^{-\lambda (1-t)}\over \lambda^2 e^{-\lambda}} \\ &=2t(1-t) \end{align} (then derive to find pdf)

This is wrong since the cdf must be non-decreasing? What has gone wrong? My concern is that we can also have $0<T_1<T_2<t$

Answer 1

I believe your concern has correctly diagnosed what went wrong. To compute the cdf, try writing $\ P\left(T_1\le t,N_1=2\right)\ $ as $\ \int_\limits{0}^{\,t}P\left(N(s,1]=1\left\vert \,T_1=s\right.\right)\lambda e^{-\lambda s}ds \ $.

Answer 2

Following up on @lonza leggiera's hint, we have \begin{align} \mathbb P(T_1\leqslant t\mid N_1=2) &= \frac{\mathbb P(T_1\leqslant t,N_1=2)}{\mathbb P(N_1=2)}\\ &= \frac{\int_0^t \mathbb P(N(1)-N(s)=1\mid T_1=s)\lambda e^{-\lambda s}\ \mathsf ds}{e^{-\lambda}\lambda^2/2!}\\ &= \frac{\int_0^t \lambda(1-s)e^{-\lambda(1-s)}\lambda e^{-\lambda s}\ \mathsf ds}{e^{-\lambda}\lambda^2/2!}\\ &= \frac{2e^\lambda}{\lambda^2}\lambda^2e^{-\lambda}\int_0^t(1-s)\ \mathsf ds\\ &= 2\left(t - \frac{t^2}2\right)\\ &= t(2-t). \end{align}