I am trying to do the following problem in order to prepare for my Stochastic Processes exam:
Problem statement: Consider $N(t)$ to be a non homogeneous Poisson process with rate $\lambda(t)$ over the interval [0, a]. Let $T_{1}$ denote the time for the first event to occur. Find the distribution of $T_{1}$ given that $N(a) = 1$. Here, $N(t)$ denotes the number of occurrences up until time t.
->What have i tried so far? I started by investigating the following: $$P(T_{1} > t) = P(N(T_{1})=0)$$
Which, since we have a non homogeneous Poisson process, turns out to be:
$$\frac{\left[\int_{0}^{T_{1}} \lambda(s) d s)\right]^{0}e^{-\int_{0}^{T_{1}} \lambda(s)ds}}{0 {\displaystyle !\,}} = e^{-\int_{0}^{T_{1}} \lambda(s)ds}$$
Finally: $$P\left(N\left(T_{1}\right)=0\right) = e^{-\int_{0}^{T_{1}} \lambda(s) d s}$$
Now, i tried to use the remaining information to finish the problem. However, i got stuck. I don't know how to proceed in regards to conditioning the variables (I dont know if that is indeed the path to solve the problem). I tried doing:
$$P\left(N\left(T_{1}\right)=0 \mid N(a)=1\right) = \frac{P\left(N\left(T_{1}\right)=0, N(a)=1\right)}{P(N(a)=1)}$$
Now, i tried a few things, but i dont know if that is indeed correct. Knowing that $N(T) = 0$ and $N(a) = 1$, one can infer that $T \le a$, otherwise $N(T)$ would not equal zero, as one event would have ocurred. This information can be translated into: $$P(N(a)=1) = e^{-\int_{a}^{T_{1}} \lambda(s) d s}$$
Now, if i could expand the below term: $$P\left(N\left(T_{1}\right)=0, N(a)=1\right)$$ into something useful, i could start manipulating some terms. What do you guys think of my approach? Any help is appreciated.
Thanks in advance!
On your approach
You have done all the right things initially. $P(N(T_1) = 0) = e^{-\int_0^{T_1} \lambda(s)ds}$ is correct. The conditional formula $P(N(T_1) = 0 | N(a) = 1) = \frac{P(N(T_1) = 0, N(a) = 1)}{P(N(a) = 1)}$ is correct (because the denominator of RHS is not zero). Furthermore, this is the way to solve the problem, so you are going in the right direction.
Now, $P(N(a) = 1)$ is incorrectly written. Note that $N(a)$ is a Poisson random variable with parameter $\int_0^a \lambda(s)ds$. Therefore, we have : $$ P(N(a) = 1) = \frac{\left(\int_0^a \lambda(s)ds\right)^1e^{-\int_0^a \lambda(s)ds}}{1!} = \left(e^{-\int_0^a \lambda(s)ds}\right)\int_0^a \lambda(s)ds $$
Answer
Now, how do we handle the numerator? Recall the independence of increments of the Poisson process. This means, in particular, that $N(T_1)$ and $N(a) - N(T_1)$ are independent for any $T_1$.
Now, it is easy to see that $$P(N(T_1) =0, N(a)= 1) = P(N(T_1) = 0,N(a) - N(T_1) =1) = P(N(T_1)= 0)P(N(a) - N(T_1) = 1)$$
thanks to independence of the two random variables. Finally, each of these involves a Poisson random variable, so we should be through here. One part you've already done : $$ P(N(T_1) = 0) = e^{-\int_0^{T_1} \lambda(s)ds} $$
The other part is easy as well : $$ P(N(a) - N(T_1) = 1) = \frac{\left(\int_{T_1}^a \lambda(s)ds\right)^1e^{-\int_{T_1}^a \lambda(s)ds}}{1!} = \left(\int_{T_1}^a \lambda(s)ds\right)e^{-\int_{T_1}^a \lambda(s)ds} $$
Multiplying these two, and combining the integrals sitting on top of the exponentials (basically, $\int_a^c f + \int_c^b f = \int_a^b f$ for any $a,b,c$ and function $f$), gives us : $$ P(N(T_1) = 0 , N(a) = 1) = \left(\int_{T_1}^a \lambda(s)ds\right)e^{-\int_{0}^a \lambda(s)ds} $$
Dividing by $P(N(a) = 1)$ and doing some change of variables to get back to our original situation, gives an answer that feels right, in addition to being right of course!
$$\bbox[yellow, 2pt, border:2px solid red]{ P(T_1 > t | N(a) = 1) = \frac{\int_{t}^a \lambda(s)ds}{\int_{0}^a \lambda(s)ds}} $$