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158   Artificial Intelligence for the Internet of Everything


          we have  d  EðVjbÞ¼ PðL > bÞ  b EðCފ   f L ðbÞ, where the term f L ðbÞ is
                  db                   ½
          the density function f(l)of L evaluated at l ¼ b. The optimal b o in the non-
          trivial region solves the integral equation:
                                                        ∞
                                                   1
                                                      Z
                                PðL > bÞ
                      b ¼ EðCÞ +        ¼ EðCÞ +          f L ðlÞdl
                                                       b
                                  f L ðbÞ        f L ðbÞ
                                                                      2
                                                            ð PðL > b o ÞÞ
                                                        b
                            and in the nontrivial region dhVie ¼
                                                               f L ðb o Þ
          9.5 CLAIRVOYANCE ABOUT C

          Let us go back to Eq. (9.4), but now let us assume that the company knows
          the cost C. In this case the company will never bid less than the cost or it will
          lose money! Note that our results in this section differ from Howard’s results
          on clairvoyance.
             Assumption 9.3 The company has knowledge of the cost.
             We must modify Theorem 9.1 so that:

                                         b c,   if c   b   l
                                                                      (9.17)
                                         0,     otherwise
                            EðVjb,c,lÞ¼
          We have that:
                          ZZ
                  EðVjbÞ¼        EðVjb,c,lÞ  f ðcÞf ðlÞdc dl ðas aboveÞ
                                2
                          Z  b        Z  ∞                            (9.18)
                                         f ðlÞdl f ðcÞdc
                        ¼     ðb cÞ
                             ∞         b
                                    Z  b
                                        ðb cÞf ðcÞdc                  (9.19)
                        ¼ PðL > bÞ
                                      ∞

                                        Z  b
                                                                      (9.20)
                        ¼ b   PðC   bÞ      cf ðcÞdc   PðL > bÞ
                                          ∞
             We will go through an example similar to what we did before. Previ-
          ously, we followed Howard and modeled C so that EðCÞ ¼ 1=2 and
          L¼ U½0,2Š. Note that, as before, the distribution of C did not matter, only
          its mean. We see from the above that this is no longer true. Let us try some
          examples.
             Example 9.1 [L¼ U½0,2Š and PðC ¼ 1=2Þ¼ 1] So we have that
          f(c) ¼ δ(c   1/2), and Eq. (9.20) becomes (Fig. 9.5):
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