[問題] 獨立性質的一題證明
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Dear all,
有一題證明想要請問大家:
X ~ N(μ,Σ)
Y = a'X, Z = b'X 要證明 Y, Z是獨立的若且唯若(iff) a'Σb = 0
其中, a, b, X, Y, Z 都是colume vector
X1
X2
X = [ ︰ ]
Xm
a'代表a的轉置=>行向量變成列向量;Σ是Cov(Xi,Xj)=E(XiXj)-E(Xi)E(Xj)
以下是我的想法:
(1) Y, Z are independent "=>" a'Σb = 0
(利用Y, Z 獨立 => Cov(Y,Z)=0的特性去證明"=>")
Var(X) = E(XX')-E(X)E(X') = Σ "Var(r) = E(r^2)-[E(r)]^2"
Cov(Y,Z)
= Cov(a'X,b'X)
= E{[Y-E(Y)][Z-E(Z)]} "Covariance的定義"
= E[Y．Z'-Y．E(Z')-E(Y)．Z'+E(Y)．E(Z')] "行向量要相乘必須把後面轉置成列向量"
= E(Y．Z')-E(Y)．E(Z') "E[Y．E(Z')] = E(Y)．E(Z')"
= E[a'X．(b'X)']-E(a'X)E[(b'X)']
= E(a'X．X'b)-E(a'X)E(X'b)
= a'[E(XX')-E(X)E(X')]b = a'Σb "if Y, Z are independent"
= a'[E(X)E(X')-E(X)E(X')]b
= a'0b = 0 ∴if Y, Z are independent, we can prove that a'Σb = 0
(2) Y, Z are independent "<=" a'Σb = 0
a'Σb = a'[E(XX')-E(X)E(X')]b = a'E(XX')b - a'E(X)E(X')b
= E(a'XX'b) - E(a'X)(X'b)
= E(YZ')-E(Y)E(Z')
if a'Σb = 0 then E(YZ')-E(Y)E(Z') must be the form of "E(Y)E(Z')-E(Y)E(Z')"
so that it can equal to 0. thus if a'Σb = 0 "=>" Y, Z are independent.
以上兩個方向的証明是我的想法,請大家能幫我看看有沒有錯誤,或是有正確的解法請
答覆我,謝謝大家
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