Page 388 - Understanding Machine Learning
P. 388

Technical Lemmas
           370


                 Lemma A.4.   Let X be a random variable and x ∈ R be a scalar and assume that
                                                                                      2

                 there exists a > 0 and b ≥ e such that for all t ≥ 0 we have P[|X − x | > t] ≤ 2be −t /a  2 .


                 Then, E[|X − x |] ≤ a(2 +  log(b)).

                 Proof. For all i = 0,1,2,... denote t i = a (i +  log(b)). Since t i is monotonically
                 increasing we have that
                                                       ∞

                                E[|X − x |] ≤ a  log(b) +  t i P[|X − x | > t i−1 ].


                                                       i=1
                 Using the assumption in the lemma we have
                           ∞                         ∞                   √
                                                                           log(b)) 2
                              t i P[|X − x | > t i−1 ] ≤ 2ab  (i +  log(b))e −(i−1+

                           i=1                      i=1
                                   F
                                     ∞             2
                             ≤ 2ab    √     xe −(x−1)  dx
                                    1+  log(b)
                                   F
                                     ∞
                                                 −y  2
                             = 2ab  √     (y + 1)e  dy
                                     log(b)
                                     ∞        2
                                   F
                             ≤ 4ab  √     ye −y  dy
                                     log(b)

                                   
    2 ∞
                             = 2ab −e −y  √
                                           log(b)
                             = 2ab/b = 2a.
                 Combining the preceding inequalities we conclude our proof.


                 Lemma A.5. Let m,d be two positive integers such that d ≤ m − 2. Then,

                                              d
                                                 m      em  d
                                                     ≤       .
                                                  k      d
                                             k=0
                 Proof. We prove the claim by induction. For d = 1 the left-hand side equals 1 + m
                 while the right-hand side equals em; hence the claim is true. Assume that the claim
                 holds for d and let us prove it for d + 1. By the induction assumption we have

                          d+1
                               m      em        m
                                           d
                                  ≤        +
                               k       d      d + 1
                           k=0

                                                      d
                                      em          d    m(m − 1)(m − 2)···(m − d)
                                           d
                                  =          1 +
                                       d          em           (d + 1)d!

                                                    d
                                      em         d    (m − d)
                                          d
                                  ≤         1 +                .
                                      d          e    (d + 1)d!
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