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. $t_1<t_2<...t_n$, 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")
}

![enter image description here][1]

I can find no flaw with the above reasoning. So my question is what did Armitage and Dodd calculate here: http://stats.stackexchange.com/questions/145214/i-have-an-epidemiology-question-with-logs ?

[1]: http://i.stack.imgur.com/wfArP.png