Theorem: 1. P and I P are idempotent. 2. rank I P trI P n p 3. I PX 0 4. E mean residual sum of square Y Yˆ t Y Yˆ E s2 E n p [proof:] PP X X tX 1. 1 XtX XtX 1 2 Xt X XtX 1 Xt P and I PI P I P P P P I P P P I P . 2. Since P is idempotent, rank P trP . Thus, rank P trP tr X X t X 1 X t tr X t X 1 X t X trI p p p Similarly, rank I P trI P trI trP tr A B tr A trB n p 3. 4. I P X X PX X X X t X 1 XtX X X 0 t RSS model p et e Y Yˆ Y Yˆ Y Xb Y Xb t Y PY Y PY t Y t I P I P Y t Y t I P Y I P is idempotent 25 Thus, E RSS model p E Y t I P Y E Z , V Y t t trI P V Y X I P X E Z AZ t tr A A tr I P 2 I 0 I P X 0 2trI P n p 2 Therefore, RSS model p E mean residual sum of square E n p 2 Supplement: Multivariate Normal Distribution 1. Definition Intuition: Let Y ~ N , 2 . Then, the density function is 1 2 y 2 1 f y exp 2 2 2 2 1 2 1 2 1 1 1 1 exp y y 2 Var Y 2 Var Y Definition (Multivariate Normal Random Variable): A random vector 26 Y1 Y Y 2 ~ N , Yn with EY , V Y has the density function 1 2 n 2 1 1 1 t f y f y1 , y2 , , yn exp y 1 y 2 det 2 Then, the moment generating function for Y is M Y t M Y t1 , t2 ,, tn E exp t tY E exp t1Y1 t2Y2 tnYn 1 exp t t t t t 2 2. Important results: 1.(a) If Y ~ N , and C is a pn matrix of rank p, then CY ~ N C , CC t 2 (b) If Y ~ N , I . then TY ~ N T , 2 I , where T is an orthogonal matrix. (c) If Y ~ N , , then the marginal distribution of subset of the elements of Y is also multivariate normal. 27 Y1 Yi1 Y Y 2 Y ~ N , Y i 2 ~ N , , then , where Yn Yim i21i1 i21i2 i1 2 2 i i i i 2 2i2 m n, i1 , i2 , , im 1,2, , n , , 2 1 2 2 im imi1 imi2 2. i21im i22im i2mim Q Y 1 Y ~ n2 . t 2 3. If Y ~ N , I and let P be an n n symmetric matrix of rank r. Then, t Y PY Q 2 2 is distributed as r if and only if P 2 P (i.e., P is idempotent). 2 4. If Y ~ N , I Q1 and let t Y P1 Y ,Q 2 2 t Y P2 Y 2 2 2 If Q1 ~ r1 , Q2 ~ r2 , Q1 Q2 0 , then Q1 Q2 and Q2 2 are independent and Q1 Q2 ~ r1 r2 . 5.Let Y ~ N 0, I and let Q1 Y t P1Y and Q2 Y t P2Y be both distributed as chi-square. Then, Q1 and Q2 are independent if and only if P1P2 0 . 28 [proof of result 3]: : Suppose P 2 P and rank P r . Then, P has r eigenvalues equal to 1 and n r eigenvalues equal to 0. Thus, without loss generalization, 1 0 t P TT T 0 0 0 0 0 1 0 0 0 0 0 0 t T 0 0 where T is an orthogonal matrix. Then, t t Y PY Y TT t Y Q 2 Z t Z 2 2 Z T Y Z t 1 Zn t Z2 Z1 Z 1 2 Z1 Z 2 Z n 2 Z n Z12 Z 22 Z r2 2 Since Z T t Y and Y ~ N 0, 2 I , thus Z T t Y ~ N T t 0, T tT 2 N 0, 2 I . Z1 , Z 2 ,, Z n are i.i.d. normal random variables with common variance 2 . Therefore, 29 Q : Z12 Z 22 Z r2 2 Since P is symmetric, 2 2 2 Z Z Z 1 2 r ~ r2 P TT t , where T is an orthogonal matrix and is a diagonal matrix with elements 1 , 2 ,, r . Thus, let Z T t Y . Since Y ~ N 0, 2 I , Z T t Y ~ N T t 0, T t T 2 N 0, 2 I That is, Z1 , Z 2 , , Z r . are independent normal random variable with variance 2 . Then, t t Y P Y Y TT t Y Q 2 Z T Y Z Z t Z 2 t Z2 1 2 Zn t r Z i 1 i 2 i 2 r The moment generating function of Q E exp t r i 1 Z i 1 i 2 2 i is i Z r ti Z i2 i 1 E exp 2 2 i 1 r 2 i ti zi2 zi2 exp exp 2 2 2 2 2 1 30 dzi r 1 2 2 i 1 r 1 1 2i t i 1 r z i2 1 2i t exp 2 dz i 2 z i2 1 2i t 1 2i t exp dz i 2 2 2 2 1 1 2i t i 1 r 1 2i t 1 2 i 1 Also, since Q is distributed as 1 2t r 2 r2 , the moment generating function is also equal to . Thus, for every t, E exp tQ 1 2t r r 2 1 2i t 1 2 i 1 Further, r 1 2t 1 2i t . r i 1 By the uniqueness of polynomial roots, we must have i 1 . Then, P2 P by the following result: a matrix P is symmetric, then P is idempotent and rank r if and only if it has r eigenvalues equal to 1 and n-r eigenvalues equal to 0. ◆ Theorem: If Then, 1. Y ~ N X , 2 I , b ~ N , 2 X t X where X is a n p matrix of rank p . 1 31 2. b t X t X b ~ 2 2 p RSS model p 2 3. n p s 2 2 ~ n p 2 b t X t X b 2 4. is independent of RSS model p 2 n p s 2 2 . [proof:] 1. Z Since for a normal random variable , Z ~ N , CZ ~ N C , CC t thus for Y ~ N X , 2 I , 1 b XtX ~ N X t X 1 t 2. by result 1 (a), X tY X X , X X t t 1 1 t 1 2 1 t 1 t X I X X t X X X X N , X X b ~ N 0, X X . N , X tX 2 t 1 X t t 2 2 Thus, by result 2, b t X X t 1 2 1 t b X t X b b 2 Z ~ N 0, ~ t 1 2 Z Z ~ p . 2 p 32 3. I PI P I P and rank I P n p , thus by result 3 2 for A A , rank A r Y X t I P Y X ~ 2 and Z ~ N , 2 I n p 2 t Z AZ 2 ~ r 2 Since I P X 0, Y t I P X 0, X t I P Y X 0 , RSS model p n p s 2 Y t I P Y 2 2 2 t Y X I P Y X ~ 2 n p 2 4. Let Q1 t Y X Y X 2 t t Y Xb Y Xb Xb X Xb X 2 t Y t I P Y b X t X b 2 2 Q2 Q2 Q1 where Q2 t t Y Xb Y Xb Y PY Y PY 2 2 Y t I P Y 2 33 . and Q1 Q2 t t Xb X Xb X b X t X b 2 Xb X 2 2 2 0 Since Q1 t t Y X Y X Y X I 1 Y X ~ 2 2 2 Z Y X ~ N 0, I Z I 2 t 2 1 Z Q1 ~ n2 and by result 4, Q2 t Y Xb Y Xb RSS model p 2 2 t Y X I P Y X ~ n2 p 2 therefore, Q2 , RSS model p 2 is independent of Q1 Q2 t b X t X b 2 . Q1 ~ r21 , Q2 ~ r22 , Q1 Q2 0, Q1 , Q2 are quadratic form of multivaria te normal Q is independen t of Q Q 2 1 2 34 n
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