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7.4 Expected Value and Variance  487

                                     EXAMPLE 13      Let X and Y be random variables that count the number of heads and the number of tails when
                                                     a coin is flipped twice. Because p(X = 2) = 1/4, p(X = 1) = 1/2, and p(X = 0) = 1/4, by
                                                     Theorem 1 we have
                                                                   1     1      1
                                                        E(X) = 2 ·  + 1 ·  + 0 ·  = 1.
                                                                   4     2      4
                                                    A similar computation shows that E(Y) = 1. We note that XY = 0 when either two heads and
                                                     no tails or two tails and no heads come up and that XY = 1 when one head and one tail come
                                                     up. Hence,
                                                                    1      1   1
                                                        E(XY) = 1 ·   + 0 ·  =  .
                                                                    2      2   2
                                                     It follows that

                                                        E(XY)  = E(X)E(Y).
                                                     This does not contradict Theorem 5 because X and Y are not independent, as the reader should
                                                     verify (see Exercise 16).                                                      ▲

                                                     Variance

                                                     The expected value of a random variable tells us its average value, but nothing about how
                                                     widely its values are distributed. For example, if X and Y are the random variables on the set
                                                     S ={1, 2, 3, 4, 5, 6}, with X(s) = 0 for all s ∈ S and Y(s) =−1if s ∈{1, 2, 3} and Y(s) = 1
                                                     if s ∈{4, 5, 6}, then the expected values of X and Y are both zero. However, the random
                                                     variable X never varies from 0, while the random variable Y always differs from 0 by 1. The
                                                     variance of a random variable helps us characterize how widely a random variable is distributed.
                                                     In particular, it provides a measure of how widely X is distributed about its expected value.


                                   DEFINITION 4       Let X be a random variable on a sample space S. The variance of X, denoted by V(X),is

                                                                                 2
                                                         V(X) =     (X(s) − E(X)) p(s).
                                                                 s∈S
                                                      That is, V(X) is the weighted average of the square of the deviation of X. The standard
                                                                                             √
                                                      deviation of X, denoted σ(X), is defined to be  V(X).

                                                        Theorem 6 provides a useful simple expression for the variance of a random variable.

                                                                                                                      2
                                                                                                             2
                                     THEOREM 6        If X is a random variable on a sample space S, then V(X) = E(X ) − E(X) .

                                                     Proof: Note that

                                                                                2
                                                        V(X) =     (X(s) − E(X)) p(s)
                                                                s∈S

                                                                       2
                                                              =    X(s) p(s) − 2E(X)    X(s)p(s) + E(X) 2   p(s)
                                                                s∈S                  s∈S                 s∈S
                                                                    2
                                                              = E(X ) − 2E(X)E(X) + E(X)  2
                                                                             2
                                                                    2
                                                              = E(X ) − E(X) .

                                                     We have used the fact that  s∈S  p(s) = 1 in the next-to-last step.
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