Page 21 - Elements of Distribution Theory

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1.3 Random Variables 7

d
The minimal support of the distribution is the smallest closed set X 0 ⊂ R such that
Pr(X ∈ X 0 ) = 1.
That is, the minimal support of X is a closed set X 0 that is a support of X, and if X 1 is
another closed set that is a support of X, then X 0 ⊂ X 1 .
The distribution of a real-valued random variable X is said to be degenerate if there
exists a constant c such that

Pr(X = c) = 1.
Fora random vector X, with dimension greater than 1, the distribution of X is said to be
T
degenerate if there exists a vector a = 0, with the same dimension as X, such that a X
is equal to a constant with probability 1. For example, a two-dimensional random vector
X = (X 1 , X 2 )hasadegeneratedistributionif,asinthecaseofareal-valuedrandomvariable,
it is equal to a constant with probability 1. However, it also has a degenerate distribution if

Pr(a 1 X 1 + a 2 X 2 = c) = 1
for some constants a 1 , a 2 , c.In this case, one of the components of X is redundant, in the
sense that it can be expressed in terms of the other component (with probability 1).

Example 1.6 (Polytomous random variable). Let X denote a random variable with range

X ={x 1 ,..., x m }
where x 1 ,..., x n are distinct elements of R. Assume that Pr(X = x j ) > 0 for each j =
1,..., m.Any set containing X is a support of X; since X is closed in R,it follows that the
minimal support of X is simply X.If m = 1 the distribution of X is degenerate; otherwise
it is nondegenerate.

Example 1.7 (Uniform distribution on the unit cube). Let X denote the random variable
3
defined in Example 1.5. Recall that for any A ⊂ R ,

Pr(X ∈ A) = dt 1 dt 2 dt 3 .
A∩(0,1) 3
3
The minimal support of X is [0, 1] .
Example 1.8 (Degenerate random vector). Consider the experiment considered in Exam-
ple 1.2 and used in Example 1.4 to define the binomial distribution. Recall that an element
ω of is of the form (x 1 ,..., x n ) where each x j is either 0 or 1. Define Y to be the
two-dimensional random vector given by
n n

2
Y(ω) = x j , 2 x j .
j=1 j=1
Then
T
Pr((2, −1) Y = 0) = 1.
Hence, Y has a degenerate distribution.