jackson23sun 發表于 2015-7-28 07:41
1/nΣ[(Xi-X')2+(X')2]=1/nΣ{[(Xi2-2XiX'+(X')2]+(X')2}=1/n∑[Xi2-2XiX'+2(X')2]=1/n(∑Xi2-2X' ...

哦哦，我以為求和符號里不包含后面那個(X')2，thank you.

1/nΣ[(Xi-X')2+(X')2]=1/nΣ{[(Xi2-2XiX'+(X')2]+(X')2}=1/n∑[Xi2-2XiX'+2(X')2]=1/n(∑Xi2-2X'∑Xi+2∑(X')2)=1/n(∑Xi2-2n(X')2+2n(X')2)=1/n∑Xi2  (X'表示Xi的均值)

來自iPhone客戶端