2.3.1 - Distribution function

The function that relates a given value of a random variable to its probability is known as the distribution function. As we saw with maximum likelihood estimation, this can also be viewed as the likelihood function with respect to the parameters \(\pi_k\).

The distribution function

The probability that \(X = \left(X_1, \dots, X_k \right)\) takes a particular value \(x = \left(x_1, \dots , x_k \right)\) is

\(f(x)=\dfrac{n!}{x_1!x_2!\cdots x_k!}\pi_1^{x_1} \pi_2^{x_2} \cdots \pi_k^{x_k}\)

The possible values of X are the set of x-vectors such that each \(x_j ∈ {0, 1, . . . , n}\) and \(x_1 + \dots + x_k = n\).

Example: Jury Selection Section

Suppose that the racial/ethnic distribution in a large city is given by the table that follows. Consider these three options as the parameters of a multinomial distribution.

Black Hispanic Other
20% 15% 65%

Suppose that a jury of twelve members is chosen from this city in such a way that each resident has an equal probability of being selected independently of every other resident. There are a number of questions that we can ask of this type of distribution.

Let's find the probability that the jury contains:

  • three Black, two Hispanic, and seven Other members;
  • four Black and eight Other members;
  • at most one Black member.

To solve this problem, let \(X = \left(X_1, X_2, X_3\right)\) where \(X_1 =\) number of Black members, \(X_2 =\) number of Hispanic members, and \(X_3 =\) number of Other members. Then \(X\) has a multinomial distribution with parameters \(n = 12\) and \(\pi = \left(.20, .15, .65\right)\). The answer to the first part is

\begin{align} P(X_1=3,X_2=2,X_3=7) &= \dfrac{n!}{x_1!x_2!x_3!} \pi_1^{x_1}\pi_2^{x_2}\pi_3^{x_3}\\ &= \dfrac{12!}{3!2!7!}(0.20)^3(0.15)^2(0.65)^7\\ &= 0.0699\\ \end{align}

The answer to the second part is

\begin{align} P(X_1=4,X_2=0,X_3=8) &= \dfrac{12!}{4!0!8!}(0.20)^4(0.15)^0(0.65)^8\\ &= 0.0252\\ \end{align}

For the last part, note that "at most one Black member" means \(X_1 = 0\) or \(X_1 = 1\). \(X_1\) is a binomial random variable with \(n = 12\) and \(\pi_1 = .2\). Using the binomial probability distribution,

\(P(X_1=0) = \dfrac{12!}{0!12!}(0.20)^0(0.8)^{12}= 0.0687\)


\(P(X_1=1) = \dfrac{12!}{1!11!}(0.20)^1(0.8)^{11}= 0.2061\)

Therefore, the answer is:

\(P\left(X_1 = 0\right) + P\left(X_1 = 1\right) = 0.0687 + 0.2061 =\) \(0.2748\)