derivation: counts in an interval

[John Denker <jsd@MONMOUTH.COM>]

At 07:11 AM 7/9/00 -0400, I wrote:

> The probability of recording m counts in time t given an average counting
> rate of r is given by
>          g(m,t) = (rt)^m exp(-rt) / m!

For those of you who might be wondering where that came from....  It's
easy.  It's even easier if we do it step by step:

  Say that the interval t is made up of "slots" of size dt.
  number of slots:  t/dt
  average counting rate: r
  probability of a count in a given slot:  r dt
  probability of no count in a given slot (1 - r dt)
  probability of no count in the whole interval:  (1 - r dt)^[t/dt]
  ... which in the limit dt -> 0 is just exp(-rt)
  ... so g(0,T) = exp(-rt)

====

  probability of no count in the first (t/dt-1) slots followed
    by a count in the final slot:        (1 - r dt)^[t/dt - 1] (r dt)
  then apply a shuffle factor to get the
  probability of one count in SOME slot: (1 - r dt)^[t/dt - 1] (r dt) t/dt
  ... which is just rt exp(-rt)

===

  prob. of no count in the first (t/dt-m) slots followed by a count
    in each of the last m slots: (1 - r dt)^[t/dt - m] (r dt)^m
  then apply a shuffle factor to get the
  prob. of m counts somewhere:   (1 - r dt)^[t/dt - m] (r dt)^m (t/dt)^m / m!
  ... note in the limit dt->0 we can ignore the possibility of
    collisions, i.e. multiple counts in the same slot.
  bottom line:  g(m,t) = (rt)^m exp(-rt) / m!