Ralph Vince - Portfolio Mathematics

ch02 JWBK035-Vince February 12, 2007 6:50 Char Count= 0

Probability Distributions 95

N′(X)=

GAM((M+N)/2)∗(M/N)M/^2

(GAM(M/2)∗GAM(N/2)∗(1+M/N))(M+N)/^2

(2.57)

where: M=The number of degrees of freedom of the first parameter.

N=The number of degrees of freedom of the second

parameter.

GAM( )=The standard gamma function.

The Multinomial Distribution

TheMultinomial Distributionis related to the Binomial, and likewise is

a discrete distribution. Unlike the Binomial, which assumes two possible

outcomes for an event, the Multinomial assumes that there are M different

outcomes for each trial. The probability density function, N′(X), is given as:

N′(X)=

(

N!

/∏M

i= 1

Ni!

)

∗

∏M

i= 1

PNii (2.58)

where: N=The total number of trials.

Ni=The number of times the ith trial occurs.

Pi=The probability that outcome number i will be the result

of any one trial. The summation of all Pi’s equals 1.

M=The number of possible outcomes on each trial.

For example, consider a single die where there are six possible out-

comes on any given roll (M=6). What is the probability of rolling a 1 once,

a 2 twice, and a 3 three times out of 10 rolls of a fair die? The probabilities

of rolling a 1, a 2, or a 3 are each 1/6. We must consider a fourth alternative

to keep the sum of the probabilities equal to 1, and that is the probability

of not rolling a 1, 2, or 3, which is 3/6. Therefore, P 1 =P 2 =P 3 =1/6, and

P 4 =3/6. Also, N 1 =1, N 2 =2, N 3 =3, and N 4 =10–3–2–1=4. Therefore,

Equation (2.58) can be worked through as:

N′(X)=(10!/(1!∗2!∗3!∗4!))∗(1/6)^1 ∗(1/6)^2 ∗(1/6)^3 ∗(3/6)^4

=(3628800/(1∗ 2 ∗ 6 ∗24))∗. 1667 ∗. 0278 ∗. 00463 ∗. 0625

=(3628800/288)∗. 000001341

= 12600 ∗. 000001341

=. 0168966 (2.58)