I want to compute the following probability.
Assume that we have a Poisson distribution $P(k,t)=\frac{(a(1-s)^t)^k}{k!}\exp(-a(1-s)^t)$ with mean $a(1-s)^t$.
While $0 \le a \le0.5$ and $0\le s \le 1$ are fixed variables the mean declines with time t. In the following I assume that time is discrete, t=1,2,3,4.... So events could only occur in t=1,2,3,4,...!
Now I want to compute the probability of having k>0 events in an arbitrary time step, while no events (k=0) have occured in earlier times steps. (Phrased differently: I want to know the probability for k events in the first time step an event occur (k $\neq 0$).
Initially I thought the solution would be $P(k)=\sum_{t=1}^\infty P(k,t) \prod_{t'=1}^{t-1} P(0,t')$ but this probability is wrong and not normalized, but should be: $\sum_{k=0}^\infty P(k)=1$. Can you help me? Thanks in advance :)
Assuming that a "time step" is the interval between one integer time mark and its subsequent, and that $\mathsf P(k,t)$ is the probability for $k$ events occurring within the time step starting at $t$, (the count of which is Poisson distributed with rate $a\,(1-s)^t$).
Then the probability that no events occur prior to integer time mark $t$ and $k$ events occur in the step starting at $t$ is $\mathsf P(k,t)\prod_{\tau=0}^{t-1}\mathsf P(0,\tau)$.
So you seek $\mathsf P(k,0)+\mathsf P(k,1)\mathsf P(0,0)+\cdots$, or $$\begin{align}&\sum_{t=0}^\infty\mathsf P(k,t)\prod_{\tau=0}^{t-1}\mathsf P(0,\tau)\\[3ex]=& \dfrac {a^k}{k!}\sum_{t=0}^\infty{(1-s)^{kt}\exp(-a\,(1-s)^t)}\prod_{\tau=0}^{t-1}\exp(-a\,(1-s)^\tau)\\[3ex]=& \dfrac {a^k}{k!}\sum_{t=0}^\infty{(1-s)^{kt}\exp(-a\,(1-s)^t)} \exp(\tfrac as\,(1-s)^t-\tfrac as) \\[3ex]\vdots\end{align}$$