# Random Process ```Week of September 20, 2005
Multinomial
 Extension of Binomial when more
than two outcomes are possible.
 Setting: Consider a sequence of
trials for which:
- # of trials (n) is fixed
- k possible outcomes ( E1 , E2 ,
are possible per trial
, Ek )
- the probabilities of outcomes
E1 , E2 , , Ek stay constant across trials
(on a given trial, P ( Ei )  pi )
- the trials are independent
Multinomial probability function gives
P ( E1 occurs x1 times,
E2 occurs x2 times,
Ek occurs xk times )
where x1  x2 
 xk  n
Multinomial Probability Function:
f ( x1 , x2 ,
, xk ; p1, p2 ,
, pk , n) 
n

 x1 x2
 x , x , , x  p1 p2
k
 1 2
where
x1  x2 
 xk  n
p1  p2 
 pk  1
and
n


n!
 x , x , , x   x !x ! x !
k
 1 2
1 2
k
pkxk
Notes:
(1) Binomial is special case of k  2
(2) pk  1  p1 
 pk 1
i.e. the probabilities are completely specified
by pi , i  1, , k  1
(in binomial case, we only had to specify p )
Note:
Binomial and Multinomial
Sampling is WITH Replacement
- when a “trial” consists of sampling from
a population, the binomial is applicable
to sampling WITH replacement
(this gives independent outcomes
-- probabilities don’t change)
- binomial is a good approximation for
sampling without replacement when
# of trials (observations) is small
with respect to population size from
which the sample is taken
Hypergeometric
Setting:
1. Random sample of size n is selected without
replacement from N items
2. Two types of items among the N
-- k of one type (“success”)
-- N  k are of the other type (“failure”)
3. We want to know the probability of getting x
successes out of the n objects selected.
Hypergeometric Probability Function:
 k  N  k 
 x  n  x 
,
h( x; N , n, k )   
N
n
 
x  0,1,
,n
Mean and Variance of Hypergeometric:
nk

N
N n k
k
 
n (1  )
N 1 N
N
2
Relationship between Hypergeometric
and Binomial
- if N is large with respect to n, then these
formulas become very similar to those for
the binomial
We will not cover:
- Multivariate Hypergeometric (page 130)
- Negative Binomial and Geometric (sec 5.5)
Poisson Process
Assume that events occur at random
throughout an interval such that:
1. The # of outcomes in any one interval is
independent of the # occurring in any
disjoint interval
2. The probability of a single occurrence in a
short interval is proportional to the length
of the interval
3. The probability that &gt; 1 outcome will occur
in a short interval is negligible
Poisson Distribution
X = # of occurrences in an interval of
length t during a Poisson experiment.
- the probability function for X is
e   t ( t ) x
p( x; t ) 
, x  0,1,2,
x!
where  (&gt;0) is the mean number
of outcomes in a unit interval
Mean and Variance of Poisson:
  t
 2  t
Poisson Tables (Table A.2 – pp. 667-)
Tabled values are Poisson probability sums
r
P ( X  r )   p ( x; t )
x 0