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3. Multivariate Random Variables  159

                           even though we may not know the exact distribution of ,    , k = 1, 2, ... .
                                                                                     2ξ
                           The following result actually gives an upper bound for E[|      – µ | ] with
                           ξ ≥ 1/2.
                              Theorem 3.9.11 (Central Absolute Moment Inequality) Suppose that
                           we have iid real valued random variables X , X , ... having the common mean
                                                               1
                                                                  2
                                                        2ξ
                           µ. Let us also assume that E[| X  | ] < ∞ for some ξ ≥ 1/2. Then, we have
                                                      1
                           gwhere k does not depend on n, and  T = 2ξ – 1 or ξ according as 1/2 ≤ ξ < 1
                           or ξ ≥ 1 respectively.
                              The methods for exact computations of central moments for the sample
                           mean      were systematically developed by Fisher (1928). The classic text-
                           book of Cramér (1946a) pursued analogous techniques extensively. The par-
                           ticular inequality stated here is the special case of a more general large devia-
                           tion inequality obtained by Grams and Serfling (1973) and Sen and Ghosh
                           (1981) in the case of Hoeffding’s (1948) U-statistics. A sample mean      turns
                           out to be one of the simplest U-statistics.


                           3.10   Exercises and Complements

                              3.2.1 (Example 3.2.2 Continued) Evaluate f (i) for i = –1, 0, 1.
                                                                   2
                              3.2.2 (Example 3.2.4 Continued) Evaluate E[X  | X  = x ] where x  = 1, 2.
                                                                     1  2   2        2
                              3.2.3 (Example 3.2.5 Continued) Check that     =  16.21. Also
                           evaluate
                              3.2.4 Suppose that the random variables X  and X  have the following joint
                           distribution.                         1    2


                                     X  values               X  values
                                      2                        1
                                                     0           1           2


                                        0            0         3/15        3/15

                                        1           2/15       6/15          0


                                        2           1/15         0           0

                           Using this joint distribution.
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