Derivation of the Poisson distribution
- From Bob Deserio’s Lab handout
A better way of describing  is as a probability per unit time that an event will occur.
That is
dP = dt
(3)
where dP is the differential probability that an event will occur in the infinitesimal time
interval dt. Of course, some care must be taken when translating a rate to a probability
per unit time. For example, if  = 10/s, it is obviously not true that the probability is 10
that an event will occur in any particular second. However, if that same rate is expressed
 = 0.01/ms it is roughly true that the probability is 0.01 that an event will happen in any
particular millisecond. Eq. 3 only becomes exact in the limit of infinitesimal dt. It is
approximately correct for any finite t.
P=t
(4)
to the extent that P << 1.
There are several possible derivations of the Poisson probability distribution. It is
often derived as a limiting case of the binomial probability distribution. The derivation to
follow relies on Eq. 3 and begins by determining the probability P(0; t) that there will be
no events in some finite interval t. The first step is to break the interval from 0 to t into N
intervals of equal length t = t/N. (The limit as N   will be performed at the end.) The
probability of an event in a small enough but finite interval t will be given by Eq. 4 and
thus the probability of no event in this same interval is given by 1 - t For the full
interval from 0 to t to have no events, each and every one of the N subintervals must have
no events. Consequently, the probability of no events in a time t is the product of the N
probabilities for no events in the subintervals.
P(0; t) = (1 - t)N (5)
Substituting t = t/N
P(0; t) = (1 - t/N)N (6)
Taking the limit N   and recognizing that
N
lim 
x
1    e x
(7)
N   
N
gives
P(0;t) = e-t
(8)
Next, a relation is derived for the probability, denoted P(n + 1; t), for there to be n + 1
events in a time t. It will be a recursion relation because it will be based on the
probability P(n; t) of one less event. For there to be n + 1 events in t, three independent
events must happen in the following order (their probabilities given in parentheses).
• There must be n events up to some point t’ in the interval from 0 to t (P(n, t’) by
definition)
• An event must occur in the infinitesimal interval from t’ to t’ + dt’ (dt’ by Eq. 3).
• There must be no events in the interval from t’ to t (P(0, t - t’) by definition).
The probability of n + 1 events in the interval from 0 to t would be the product of the
three probabilities above integrated over all t’ from 0 to t to take into account that the last
event may occur at any time in the interval. That is,
t
P  n  1; t    P  n; t '  dt ' P  0; t  t ' 
(9)
0
From Eq. 8 we already have P(0; t -t’) = e-t(t-t’) and with the following definition
P  n; t   et P  n; t  (10)
substituted, Eq. 9 becomes (after canceling e-t from both sides):
t
P  n  1; t     P  n; t ' dt '
(11)
0
From Eqs. 8 and 10, P  0; t   1 and then P 1; t  can be found from an application of Eq.
11
t
P 1, t     P  0, t  dt   t
(12)
0
Applying Eq. 11 for the next few terms … the pattern clearly emerges that
P  n; t  
 t 
n
(15)
n!
Thus with Eq. 10, the Poisson probabilities result
P  n; t   e
 t
 t 
n
(16)
n!
BobC - truncating and shortening BobD Using  = t as the “expected” number of
events
P  n   e 
n
(17)
n!
BobD also derives that

<n> =
 nP  n  = 
(21)
n 
And

n2   n2 P  n    2  
(23)
n 0
Leading to
 2  n   
2
 n2   2  
(22)