Prove a weaker version of Theorem 7.1: under the conditions of Theorem 7.1,                 E  |mn(x) − m(x)| 2 µ(dx)                 ≤ E  |m(hˆ) nl (x) − m(x)| 2 µ(dx)+8√ 2L2 :log(2|Qn|) nt ....


Prove a weaker version of Theorem 7.1: under the conditions of Theorem 7.1,


                E  |mn(x) − m(x)| 2 µ(dx)


                ≤ E  |m(hˆ) nl (x) − m(x)| 2 µ(dx)+8√ 2L2 :log(2|Qn|) nt .


Hint:


                               |mn(x) − m(x)| 2 µ(dx) −  |m(hˆ) nl (x) − m(x)| 2 µ(dx)


                              = E (m(H) nl (X) − Y ) 2 Dn  − E  (m(hˆ) nl (X) − Y ) 2 Dn


                              = E  (m(H) nl (X) − Y ) 2 Dn  − 1 nt n l+nt i=nl+1 (m(H) nl (Xi) − Yi)


                              + 1 nt n l+nt i=nl+1 (m(H) nl (Xi) − Yi) 2 − 1 nt n l+nt i=nl+1 (m(hˆ) nl (Xi) − Yi) 2


                              + 1 nt n l+nt i=nl+1 (m(hˆ) nl (Xi) − Yi) 2 − E{(m(hˆ) nl (X) − Y ) 2 Dn}


                              ≤ E  (m(H) nl (X) − Y ) 2 Dn  − 1 nt n l+nt i=nl+1 (m(H) nl (Xi) − Yi) 2


                               + 1 nt n l+nt i=nl+1 (m(hˆ) nl (Xi) − Yi) 2 − E  (m(hˆ) nl (X) − Y ) 2 Dn


                                ≤ 2 max h∈Qn 1 nt n l+nt i=nl+1 (m(h) nl (Xi) − Yi) 2 − E  (m(h) nl (X) − Y ) 2 Dn


                                = 2 max h∈Qn 1 nt n l+nt i=nl+1 (m(h) nl (Xi) − Yi) 2 − E  (m(h) nl (X) − Y ) 2 Dnl .


Use Hoeffding’s inequality (cf. Lemma A.3) to conclude


                                      P




|mn(x) − m(x)| 2 µ(dx) −  |m(hˆ) nl (x) − m(x)| 2 µ(dx) > |Dnl


                                      ≤ 2|Qn|e −nt 2 32L4 .


Compare also Problem 8.2.

May 03, 2022
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