Discrete Random Variables 521
The term random variable is unfortunate, because X is not a variable in the mathe-
matical sense and also is not random. Each co E 2 is sent by X to a uniquely defined real
number X (w). In other words, X is a function. Nevertheless, we shall use the term random
variable, because it is standard terminology.
Because the domain of X is a sample space endowed with a probability density func-
tion, the range of X-namely,
{x : Xt(w) = x for some w e Q2}
which is a subset of ]R-can easily be made into a sample space QX and endowed with its
own probability density. To see this, note that the preimage of any element x in the range
of X is the event
{o e 2Q: X(wo) = x} C Q*
Notation. The event [0 e 2 : X((o) = X} g2 will be denoted by (X = x). The proba-
bility of this event,
P({o I X(co) = x})
will be denoted by P(X = x).
The next theorem shows that the assignment of the probability P(X = x) to each x
defines a probability density function on the range of X.
Theorem 1. (Probability Density Defined by a Random Variable) Let £Žx be the
range of a random variable X defined on a sample space Q2 endowed with probabil-
ity density function p. Then, px(x) = P(X = x) defines a probability density function
on Q^2 x.
Proof. Since each px(x) is the probability of an event in Q, we have 0 < px(x) < 1. To
see that E px (x) = 1, we observe that the sets
{[o E Q : X(wo) = x}
where x ranges over i2x form a partition of Q2. Hence, by the Additive Principle of Disjoint
Events, the probabilities of these events sum to 1. 0
Definition 2. The probability density function defined on the range of random variable
X by px (x) = P(X = x) is called the distribution of X.
Example 2. Let the random variable X be defined on the value of the top face after the
roll of a fair die as follows: X(1) = 1, X(2) = 2, X(3) = 2, X(4) = 3, X(5) = 2, and
X (6) = 3. Determine the distribution of X.
Solution. The range of the random variable X is {1, 2,31. Probability px(1) =
P(X = 1) = P({1}) = 1/6; px(^2 ) = P(X = 2) = P({2, 3, 5}) = 1/2; px(^3 ) = P(X =
- = p({4, 6}) = 1/3. 0