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# What is the probability that a sequence of events occurs within a given time interval?

If an event has probability p of occurring in some time interval, then the probability the event does not occur q is: $$q=1-p$$

The probability of the event not occurring by time t will be:

$$P(q_1 \cap q_2 \cap...q_t) =q_1q_2...q_t = q^t$$

So the probability it did occur by is one minus this value ($$q^t$$), which is equal to the cumulative sum of p times the probability it had not occurred up to that point ($$q^{t-1}$$): $$p_t=1-q^t=\sum\limits_{i=1}^t pq^{t-1}$$

If we are concerned with n independent events occurring with probabilities $$p_1, p_2, ... p_n$$ by time t, then: $$P(p_{1t} \cap p_{2t} \cap...p_{nt}) =p_{1t}p_{2t}...p_{nt}$$

If $$p_1=p_2= ... p_n$$ then the above will be simply $$p_t^n$$. So the probability that all n events have occurred by time t will be:

$$P(t_{Allevents}\leq t)=(1-q^t)^n=\sum\limits_{i=1}^t (pq^{t-1})^n$$

If there is only one sequence of these events that results in the outcome of interest (e.g. $$t_1, where $$t_i$$ refers to time of occurrence), the probability it is the observed sequence will be one over the total number of permutations ($$1/n!$$). So:

$$P(t_{Sequence}\leq t)=\frac{(1-q^t)^n}{n!}=\sum\limits_{i=1}^t \frac{(pq^{t-1})^n}{n!}$$

That gives us the CDF. To get the PDF we take the first derivative which is: $$P(t\geq t_{Sequence}\leq t+1)=\frac{-nq^tln(q)(1-q^t)^{n-1}}{n!}$$

Given the above assumptions, we would expect the probability that the sequence events occurs at any given time interval to follow the PDF, shown in the lower row of plots:

t=1:100; p=.025; q=1-p
par(mfrow=c(2,4))
for(n in c(1,2,4,6)){
plot(t,(cumsum((p*q^(t-1)))^n)/factorial(n), xlab="Time",
ylab="P(t.Seq <= t)",main=paste(n, "Events"))
lines(t,((1-q^(t))^n)/factorial(n))
}

for(n in c(1,2,4,6)){
plot(t,(-n*(q^t)*log(q)*(1-q^t)^(n-1))/factorial(n), log="xy", xlab="Time",
ylab="P(t<= t.Seq<= t+1)",main=paste(n, "Events"))
lines(t[-1]-.5,diff(((1-q^(t))^n)/factorial(n)), col="Red")
}


I can find no flaw with the above reasoning. So my question is what did Armitage and Dodd calculate here: I have an epidemiology question with logs ?