Page 70 - 4660

Random vectors

The multinomial distribution. The binomial distribution describes the probability of obtaining
x ’successes’ from n independent trials, where each trial has only two possible outcomes. This may
be generalized to the case where each trial has k possible outcomes with respective probabilities
p 1 ,p 2 ,. . . ,p k .IfweconsidertherandomvariablesX i ,i = 1,2,. . . ,n,tobethenumberofoutcomes
of type i in n trials then we may compute their joint probability function

f(x 1 , x 2 , . . . , x k ) = P(X 1 = x 1 , X 2 = x 2 , . . . , X k = x k ),
∑ k
where we must have x i = n. In n trials the probability of obtaining x 1 outcomes of type 1,
i=1
followed by x 2 outcomes of type 2 etc. is given by p p · · · p . However, the number of
x 1 x 2
x k
1 2 k
distinguishable permutations of this result is
n!
x 1 !x 2 ! · · · x k !

and thus
n!
x k
x 1 x 2
f(x 1 , x 2 , . . . , x k ) = p p · · · p . (7.14)
2
1
k
x 1 !x 2 ! · · · x k !
This is the multinomial probability distribution.
If k = 2 then the multinomial distribution reduces to the familiar binomial distribution.
Although in this form the binomial distribution appears to be a function of two random variables,
it must be remembered that, in fact, since p 2 = 1 − p 1 and x 2 = n − x 1 , the distribution of X 1 is
entirely determined by the parameters p and n. That X 1 has a binomial distribution is shown by
remembering that it represents the number of objects of a particular type obtained from
sampling with replacement, which led to the original definition of the binomial distribution. In
fact, any of the random variables X i has a binomial distribution, i.e. the marginal distribution of
each X i is binomial with parameters n and p i . It immediately follows that

2
E(X i ) = np i and Var(X i ) = np i (1 − p i ). (7.15)

Example 7.5. At a village fête patrons were invited, for a 10 p entry fee, to
pick without looking six tickets from a drum containing equal large numbers of
red, blue and green tickets. If five or more of the tickets were of the same
colour a prize of 100 p was awarded. A consolation award of 40 p was made if two
tickets of each colour were picked. Was a good time had by all? ,

Solution. In this case, all types of outcome (red, blue and green) have the same probabilities.
The probability of obtaining any given combination of tickets is given by the multinomial
1
distribution with n = 6, k = 3 and p i = , i = 1, 2, 3.
3
1. The probability of picking six tickets of the same colour is given
0
6
( ) ( ) ( ) 0
6! 1 1 1 1
P(six of the same colour) = 3 × = = .
6!0!0! 3 3 3 243
The factor of 3 is present because there are three different colours.
2. The probability of picking five tickets of one colour and one ticket of another colour is

1
5
( ) ( ) ( ) 0
6! 1 1 1 4
P(five of one colour; one of another) = 3 × 2 × = .
5!1!0! 3 3 3 81
The factors of 3 and 2 are included because there are three ways to choose the colour of
the five matching tickets, and then two ways to choose the colour of the remaining ticket.

70