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                            176                         Stochastic Processes

                            knowing the limit by showing that, roughly speaking, the terms in the sequence become
                            closer together as one moves down the sequence. See Section A3.2 in Appendix 3 for
                            further details.
                              The same type of approach may be used to establish convergence in mean square of a
                            sequence of random variables. Theorem 6.4 gives such a result; as in the case of convergence
                            of a sequence of real numbers, the advantage of this result is that it may be shown that a
                            limiting random variable exists without specifying the distribution of that random variable.

                            Theorem 6.4. Let Y 0 , Y 1 ,... denote a sequence of real-valued random variables such that
                               2
                            E(Y ) < ∞ for all j = 0, 1,.... Suppose that for every  > 0 there exists an N > 0 such
                               j
                            that
                                                                  2
                                                        E[(Y m − Y n ) ] <
                            for all m, n > N; in this case we say that Y n ,n = 0, 1,..., is Cauchy in mean square.
                              If Y n ,n = 0, 1,..., is Cauchy in mean square, then there exists a random variable Y
                                   2
                            with E(Y ) < ∞ such that
                                                                   2
                                                       lim E[(Y n − Y) ] = 0.
                                                      n→∞
                            Proof. Fix j = 1, 2,.... Since Y 0 , Y 1 ,... is Cauchy in mean square, we can find m, n,
                            m > n, such that
                                                                      1
                                                                 2
                                                       E[(Y m − Y n ) ] ≤  j  .
                                                                     4
                            Thus, we can identify a subsequence n 1 , n 2 ,... such that
                                                                 1
                                                            2
                                                           ) ] ≤  ,  j = 1, 2,....
                                                E[(Y n j+1  − Y n j  j
                                                                4
                            Define
                                                     m

                                                T m =   |Y n j +1 − Y n j  |, m = 1, 2,....
                                                     j=1
                              Let   denote the sample space of the underlying experiment so that, for each m =
                            1, 2,..., T m ≡ T m (ω), ω ∈  . Note that the terms in the sum forming T m are all nonnegative
                            so that, for each ω ∈  , either lim m→∞ T m (ω)exists or the sequence diverges. Define a
                            random variable T by
                                                    T (ω) = lim T m (ω),ω ∈
                                                           m→∞
                            if the limit exists; otherwise set T (ω) =∞.
                                                                                     2
                              Note that for any real-valued random variables Z 1 , Z 2 such that E(Z ) < ∞, j = 1, 2,
                                                                                     j

                                        1        1 2                      1      1
                                      2       2                2        2      2
                                 E Z   2  + E Z  2  − E[(Z 1 + Z 2 ) ] = 2E Z  2  E Z  2  − 2E(Z 1 Z 2 ) ≥ 0,
                                     j        2                        1      2
                            by the Cauchy-Schwarz inequality. Hence,
                                                            1
                                                          2


                                                E[(Z 1 + Z 2 ) ] 2 ≤ E Z 2    1 2  + E Z 2   1 2  .
                                                                   1        2
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