《数理统计》课程教学资源（参考资料）Hoeffding's Indequality 证明中的一个不等式的证明 complementary

Proof of Inequality Eesz0, Ees2≤e26-a2/8 Proof.By the convexity of the exponential function,(i.e.,the definition of convexity function:f(ta+(1-t)y)≤tf(x)+(1-t)f(y),0≤t≤1） ≤6+0ass6 thus, Ee2≤E-e+E 一zea b-a b-a -a esb+ b-a sa b-a =(1-0+0es6-a)e-8s(6-a), where 0=-a Now let u =s(b-a)and define o(u)=-Ou+log(1-6+0e"),then we have EesZ≤e(u). By Taylor expansion l=ol0+gou+pre)u2,for some0≤u≤u and Deu (u=-9+1-0+e,→0(0)=0 Deu 12 0"(四）=1-日+0e“ 1-0+0eu Deu beu =1-9+9e1-1-9+9e） 1 =p1-p)≤ where p=1-0+0e" 0eu Hence, Eesz eu2/8=es(-a)218 口

Proof of Inequality 퐸푒푠푍 ≤ 푒 푠 2 (푏−푎) 2/8 Theorem 1. if 푍 is a random variable with 퐸[푍] = 0 and 푎 ≤ 푍 ≤ 푏, then for any real number 푠 > 0, 퐸푒푠푍 ≤ 푒 푠 2 (푏−푎) 2 /8 Proof. By the convexity of the exponential function, (i.e., the definition of convexity function: 푓(푡푥 + (1 − 푡)푦) ≤ 푡푓(푥) + (1 − 푡)푓(푦), 0 ≤ 푡 ≤ 1 ) 푒 푠푧 ≤ 푧 − 푎 푏 − 푎 푒 푠푏 + 푏 − 푧 푏 − 푎 푒 푠푎, 푎 ≤ 푧 ≤ 푏, thus, 퐸푒푠푍 ≤ 퐸 푧 − 푎 푏 − 푎 푒 푠푏 + 퐸 푏 − 푧 푏 − 푎 푒 푠푎 = −푎 푏 − 푎 푒 푠푏 + 푏 푏 − 푎 푒 푠푎 = (1 − 휃 + 휃푒푠(푏−푎) )푒 −휃푠(푏−푎) , where 휃 = − 푎 푏−푎 . Now let 푢 = 푠(푏 − 푎) and define 휙(푢) = −휃푢 + 푙표푔(1 − 휃 + 휃푒푢 ), then we have 퐸푒푠푍 ≤ 푒 휙(푢) . By Taylor expansion 휙(푢) = 휙(0) + 휙 ′ (0)푢 + 1 2 휙 ′′(푣)푢 2 , for some 0 ≤ 푣 ≤ 푢 and 휙 ′ (푢) = −휃 + 휃푒푢 1 − 휃 + 휃푒푢 , =⇒ 휙 ′ (0) = 0 휙 ′′(푢) = 휃푒푢 1 − 휃 + 휃푒푢 − [ 휃푒푢 1 − 휃 + 휃푒푢 ]2 = 휃푒푢 1 − 휃 + 휃푒푢 (1 − 휃푒푢 1 − 휃 + 휃푒푢 ) = 휌(1 − 휌) ≤ 1 4 , where 휌 = 휃푒푢 1−휃+휃푒푢 . Hence, 퐸푒푠푍 ≤ 푒 푢 2/8 = 푒 푠 2 (푏−푎) 2/8