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Review of Probability and Random Variables  3.23

                      Equation (3.30) can be simplified using the following two well-known linear
                      algebra identities:

                      1.

                                               AB                 −1
                                           det       = det[A − BD C]det[D]               (3.31)
                                               CD
                      2.
                                	    
 −1  	  −1  
  	    −1
                                 AB         A   0     −A B             −1  −1      −1
                                         =         +           [D − CA B]     −CA    I
                                 CD          00          I
                                            00            I            −1  −1       −1

                                         =      −1  +     −1    A − BD C] [ I −BD        (3.32)
                                            0 D        −D C
                      where A, B, C, and D are arbitrary matrices. Using Eq. (3.31) the ratio of the
                      determinants in Eq. (3.30) is given as

                                     det[C Z ]                −1
                                             = det C X − C XY C C YX = det[C X|Y ]       (3.33)
                                                              Y
                                     det[C Y ]
                      Using Eq. (3.32) to reformulate C Z in Eq. (3.30) gives
                                 1        T   −1                  1         T  −1

                                                                    y
                           exp − [ z − m Z ] C [ z − m Z ] = exp −  [  − m Y ] C [  y − m Y ]
                                                                               Y
                                              Z
                                 2                                2

                                                                   1  T  −1
                                                           × exp − F G F                 (3.34)
                                                                   2
                      where
                                                      −1
                                   F = − m X − C XY C [  y − m Y ] = x − E[X|Y = y ]
                                       x

                                                      Y
                      and
                                                           −1
                                            G = C X − C XY C C T XY  = C X|Y
                                                           Y
                      Using Eq. (3.33) and Eq. (3.34) in Eq. (3.30) and cancelling the common terms
                      gives the desired result.
          3.4 Homework Problems
                      Problem 3.1. This problem exercises axiomatic definitions of probability. A com-
                      munications system used in a home health care system needs to communicate
                      four patient services, two digital and two analog. The two digital services are
                      a 911 request and a doctor appointment request. The two analog services are
                      the transmission of an electrocardiogram (EKG) and the transmission of audio
                      output from a stethoscope. The patient chooses each of the services randomly
                      depending on the doctor’s prior advice and the patient’s current symptoms.
                      Assume only one service can be requested at a time.
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