318 On Random Occurrences in Space and Time 
^1, ... tn. Let /(/9) be the chance of no occurrence happening during the interval 
/3 before the start of the period T. Then the chance of the observed event is 
m m ni m ^ ^ ' 
where q (tn) is the chance of no event occurring between tn and T. If ^„ + /3 < T, 
this is e ™ , but if tn + P> T, this is certainty, for the time between tn and T 
will be closed. Accordingly there is discontinuity in the form of the function q (tn) 
and we leave it for the moment as q (tn). We have therefore for the chance 
6 
BO=f(0) dt,dt, ... dtnq(tn)e 
tn 
We have now to integrate this expression for all possible values of the t's. The 
limits of tn are clearly tn-i + ^ to T and our first integral 
dtnqitn) = q (tn) e '"dtn +\ e '"Xlxdtn 
= 1 e dtn + m(e - e ) 
i 
Having thus freed ourselves from t^ our chance now becomes 
SC" =f(^) dt,dt,... dtn-, \(T- tn-, - 2/3) + m {^1 - e 
Now tn-i must be integrated from tn-^ + /S up to /3 of the final position of tn ; 
this was (T — /3) for the first part of the integral of tn, i.e. (T — tn-i — 2^8), and T 
for the second part, i.e. m 
Thus our htn~i integral is . 
{T- 1,,., - 2/3) dtn-, + ni{l-e dtn-. 
= 2", (T-tn-, - 3/3)"- + m (l-e (T-t,-, - W). 
Continuing this process we reach finally 
11 1 \ III J \ / (?i — 1 ) ! V m 
(X). 
Now it is clear that this is a very complicated expression for the frequency of 
n occurrences and will be hard to deal with if m and /3 are as a rule unknown and 
