Random Signals for Engineers using MATLAB and Mathcad
Copyright  1999 Springer Verlag NY
Example 5.2 Poisson Processes
The Poisson Random Variable T is formed by a summation of IID exponential Random Variables.
We have shown chapter 5.2 that the Density function for T is Erlang or
syms k t lam
kfac=sym('k!');
f= k/t*(lam*t)^k/subs(kfac,k,k)*exp(-lam*t);
pretty(f)
k
k (lam t) exp(-lam t)
---------------------t k!
If we define a new random variable , which is bounded by t and t k+1 or t <  < w . We have defined the
IID Random Variable tk+1 as w for convenience in notation. The use of a new random variable,  now
means that this new random variable can fall anywhere after t k and before tk+1. The range of the random
variable t is 0 < t <  and w is  < w < . The Random Variable T and W are independent and the joint
density function, f(t,w) becomes the product of the Erlang and the exponential as
k   t 
1
f TW t , w  
 e  t   e  w
t
k!

k
The Probability that there will be k events in  is just the integral of f(t,w) evaluated over the range of t and
w in relation to The integration range is shown below on the t-w plane. The w integral is taken along the
line u and is bounded by  - t and  as shown
2
t( w )
( u
1)
0
0
1
2
3
w u
4
5
6
The integral becomes
k   t   t

 e    e  w  dw  dt
t
k
!
t
 
PK  k in    
0


Matlab can perform the symbolic integration and the substitution at the limits for w   and t 0. The
assumptions that allow this to proceed are the conditions that  > 0 and k > 0. This must be imposed on the
Maple computations
gt=sym('lam>0');
maple('assume',gt);
gtt=sym('k>0');
maple('assume',gtt);
syms w tau
ap=int(int(f*lam*exp(-lam*w),w,tau-t,inf),t,0,tau);
pretty(ap)
k~
k~
tau
lam~
exp(-lam~ tau)
--------------------------k~!
We will now show that in the range {0 to } each ti for i = 1 .. k will be uniformly distributed in this range.
Let us first take the case for k =1. This is a joint event where there is one event in {0 t i} and no events in
{ti }. This probability can be written as
Pone event in 0    PT     Pone event in t one event in  
The expression on the right hand side is the conditional probability and it can be evaluated using Bayes'
rule
Pone event in 0   
PK  1 in t  and K  1 in  
PK  1 in  
The Joint probability may be evaluated below since the two events are independent
PK  1 in t  and K  1 in    PK  1 in t  PK  0 in   t 
Using the Poisson distribution for each of the probability evaluations in the right hand side of the equation
above, we have
PT    
  t  e   t  e   t  t


    e  
The probability distribution can be recognized as the cumulative distribution function for a uniform
distribution in the range { 0  }. The same reasoning can be used to show that the i th event will also b e
uniformly distributed for each i = 1 .. k and all the events are independently located in the range { 0  }.
The mean and variance for a Poisson are computed next. For simplicity we let  = t. The gamma
function has replaced k! since the two are mathematically equivalent.
syms alp
app=subs(ap,lam*tau,alp);
app=subs(app,tau,alp/lam);
app=simplify(app);
pretty(app)
k~
alp
exp(-alp)
--------------gamma(k~ + 1)
The mean of E[K] is
EK=symsum(k*app,0,inf)
simplify(EK)
EK =
alp*exp(-alp)*exp(alp)
ans =
alp
The second moment and then the variance become
EK2=symsum(k^2*app,0,inf);
simplify(EK2)
ans =
alp*(alp+1)
and
'VAR[K] = '
VAR=EK2-EK^2;
simplify(VAR)
ans =
VAR[K] =
ans =
alp
