in Poisson counting process calculate $P(N_{k + w + 1, k + 1}\ge t | N_{k + w, k} < t)$ in essence it's a question of calculating probability if moving average will pass certain value in new window knowing that in previous window it didn't pass this value
having "Poisson counting process" $N$ with intensity $\lambda$, and for given arbitrary $w \in \mathbb{N}$ ($w$ is length of moving sampling window), notation $N_{a, b} = N(a) - N(b)$
I would like to calculate $$P(N_{k + w + 1, k + 1}\ge t | N_{k + w, k} < t)$$
I am not sure if this approach is correct:
$$P(N_{k + w + 1, k + 1}\ge t | N_{k + w, k} < t) = 
\frac{\sum_{i=0}^{t-1} P(N_{k + w + 1,k + w} \ge t-i) * P(N_{k + w,k + 1} = i) * P(N_{k + 1,k} \le t - 1 - i)}{P(N_{k + w,k} \le t-1)}$$ and this should be computed with for loop
any remarks and links to reading material would be very much appreciated. Thanks!
 A: $$
P(N_{k+w+1,k+1} \geq t | N_{k+w,k} < t) = \frac{P(N_{k+w,k}<t, N_{k+w+1,k+1} \geq t)}{P(N_{k+w,k}< t)}
$$
let $X = N_{k+1,k}, Y = N_{k+w,k+1}, Z = N_{k+w+1,k+w}$. Then $X,Y,Z$ are independent, and note that $N_{k+w,k} = X+Y$, $N_{k+w+1,k+1} = Y+Z$ in distribution. The denominator is easily calculated, it's the numerator that is the problem. For that, we write :
$$
P(X+Y < t, Y+Z \geq t) = \sum_{n \in \mathbb N \cup \{0\}} P(Y=n,X < t-n, Z \geq t-n)
$$
using the fact that the event $X+Y < t, Y+Z \geq t$ breaks into separate events upon fixing $Y$. We use the fact that $X,Y,Z$ are independent to get $$
\sum_{n \in \mathbb N \cup \{0\}} P(Y=n,X < t-n, Z \geq t-n) = \sum_{n \in \mathbb N \cup \{0\}} P(Y=n)P(X < t-n) P(Z \geq t-n)
$$
Now we remember the distributions : suppose we have a homogenous point process of rate $\lambda$, then $X,Z \sim Poi(\lambda)$ and $Y \sim Poi(\lambda(w-1))$. Thus, the numerator equals
$$
\sum_{n \in \mathbb N \cup \{0\}} \frac{e^{-\lambda(w-1)}\lambda^n(w-1)^n}{n!} \left(\sum_{z < t - n} \frac{\lambda^z}{z!}\right)\left(\sum_{y \geq t - n} \frac{\lambda^y}{y!}\right)
$$
since the term doesn't exist for $t \leq n$, we get only the finite sum $$
\sum_{n =0}^{t-1} \frac{e^{-\lambda(w-1)}\lambda^n(w-1)^n}{n!} \left(\sum_{z < t - n} \frac{\lambda^z}{z!}\right)\left(\sum_{y \geq t - n} \frac{\lambda^y}{y!}\right)
$$
using $e^{\lambda} - \sum_{y < t - n}\frac{\lambda^y}{y!} = \sum_{y \geq t - n} \frac{\lambda^y}{y!}$ gives us
$$
\sum_{n =0}^{t-1} \frac{e^{-\lambda(w-1)}\lambda^n(w-1)^n}{n!} \left(\sum_{z < t - n} \frac{\lambda^z}{z!}\right)\left(e^{\lambda} - \sum_{y < t - n}\frac{\lambda^y}{y!}\right)
$$
and the final answer (after seeing that the denominator is $Poi(\lambda w)$) is:
$$
\frac{\sum_{n =0}^{t-1} \left[\frac{e^{-\lambda(w-1)}\lambda^n(w-1)^n}{n!} \left(\sum_{z =0}^{t - n-1} \frac{\lambda^z}{z!}\right)\left(e^{\lambda} - \sum_{y=0}^{ t - n-1}\frac{\lambda^y}{y!}\right)\right]}{\sum_{u =0}^{t-1}\frac{e^{-\lambda w}(\lambda w)^u}{u!}}
$$
which doesn't really admit to much simplification, possibly some $\exp$ terms may get cancelled but that's barely anything. It still demonstrates that the moving average is quite difficult to study conditionally. I don't expect this to simplify, unfortunately.
