Proof: E(C*zi)=c^2/2, where c is a constant, Z~N(0,1), i=1 2 3...

这样简化是不对的……
显然由于期望函数的线性性质，E(CZi)=C*E(Zi)=0(因为EZ=0)
其实原题的左边就是随机变量的矩母函数（moment generating function）的定义啊，正态的矩母函数超重要的。
证明：
E(exp(CZi))=将{exp(-0.5x^2+cx)/根号(2pai)}dx从负无穷到无穷
=将{exp(-0.5(x-c)^2+0.5c^2)/根号(2pai)}dx从负无穷到无穷
=将{exp(-0.5(x-c)^2+0.5c^2)/根号(2pai)}d(x-c)从负无穷到无穷(变量代换)
=将{exp(-0.5y^2+0.5c^2)/根号(2pai)}dy从负无穷到无穷
=[将{exp(-0.5y^2)/根号(2pai)}dy从负无穷到无穷]*exp(0.5c^2)(积分里面是标准正态的密度函数)
=exp(0.5c^2)

where 2【(5)is the algebra of divided power polynomiMs of 5 variables.To abbreviate the
notations,in the foHowing we shall write D(f)simply as f,and we have(cf.[1】)
2 2
If,g]=(Ds,)(9一∑(Dz+t9) z+t)-(Ds9)(,一∑(Dz",) z+t).
2
江 (1.1)
,一一,
+∑(( ,)( +29)一(Dig)(Di+2,)), ,,g∈ (5).
':1
Now we introduce the G2 and its variations.
Let L=G2 or G for i=3,4,5,6,7 with gradation L=
L一1=(Xl,x2,x3, 4),L0=(hi,h2,el,e2)(ore3,e4),L1=
satisfying
2 0 Li,where L一2=(1),
i=一2