Discrete Random Variables 525
spots is, respectively,
1 1 1 1 1 1
6' 2' 12' 12' 12' 12
These probabilities define a random variable X with domain
0 = {1, 2, 3, 4, 5, 6}
and range
O^2 x = ($1,-$1, -$3, -$51
Calculate the distribution pk induced by X on its range Q^2 j, and then determine E(X).
Solution. The outcomes in Q that have a $1 payoff are 2, 4, and 6, so
1 1 1 2
px($1) = p(2) + p(4) + p(6) = I +
I -^2
Similarly,
1
px(-$1) = p(M) = -
6
and
1
px(-$^3 ) = px(-$5) = p(^3 ) = p(5)= 12
Now that we know px, we can compute E(X):
E(X) = ($1) • px($1) + (-$1) • px(-$1) + (-$3) -px(-$^3 ) + (-$5) • px(-$5)
= ($ + ($1) + (-$3)(1) + (-$5) =1
Note that the expectation of a random variable does not tell us what to "expect" the
value of X to be. In the above example, we can never owe -$1/6 as a result of rolling the
die once. We can only get one of the four values in QX. A more extreme case would be to
flip a fair coin that pays $100 for heads and charges $100 for tails. The expected value of
the payoff is zero, which is far from either of the two actual payoffs.
Since the probability px (x) assigned to an element x E £x is the probability of the
event (X = x) in ý2, we can also compute the expectation of X by returning to the original
sample space Q, as the next theorem shows.
Theorem 2. (Basic Property of Expectation) The expectation E(X) of a random vari-
able X defined on a sample space Q2 satisfies
E(X) --- X-(0)p()
Proof. By definition,
E(X) X -•x
(X px(x)
xFE2x
