S631_Quiz1_Solution

S631_Quiz1_Solution - STAT431/631 Quiz 1(Solution 1 Let A=...

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Unformatted text preview: STAT431/631: Quiz # 1 (Solution) 1. Let A= 4 1 1 2 , B= 2 1 1 2 3 4 (a). Find both AB and B T AT . (5 points) 9 4 6 5 16 11 9 TT 6 BA = 16 4 5 11 AB = , (b). Find A−1 . (5 points) A− 1 = (c). For all vectors y y= show that yT Ay ≥ 0. (5 points) 2 2 2 2 yT Ay = 4y1 + 2y1 y2 + 2y2 = 3y1 + (y1 + y2 )2 + y2 ≥ 0 1 7 2 −1 −1 4 y1 y2 2. Let y be an n × 1 column vector y= (a.) Calculate both yT y and yyT . (5 points) n 2 y1 y2 y1 . . . yn y1 y1 y2 . . . yn yT y = i=1 2 yi , yyT = y1 y2 2 y2 . . . yn y2 ... ... .. . ... y1 yn y2 yn . . . 2 yn (b.) Show that yyT is symmetric. (5 points) (yyT )T = (yT )T (y)T = yyT (c.) Calculate tr(yyT ). (5 points) n tr(yyT ) = i=1 1 3. Show that I − n J is idempotent. (5 points) 2 yi (I − 1 1 1 1 11 1 1 1 1 J )(I − J ) = I · I − I J − J · I + J J = I − J − J + J = I − J n n n n nn n n n n 1 therefore, I − n J is idempotent. 4. (STAT631) Show that any variance-covariance matrix V is always positive semi-definite. (2 points) 1 The variance-covariance matrix V = Var(y) for any vector a, by Theorem 1, Var(aT y) = aT V a Since variance cannot be negative, that is, Var(aT y) ≥ 0. Therefore, for any vector a, aT V a ≥ 0, that is, V is always positive semi-definite. 2 ...
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This note was uploaded on 11/10/2009 for the course S 631 taught by Professor Ku during the Spring '09 term at Indiana.

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S631_Quiz1_Solution - STAT431/631 Quiz 1(Solution 1 Let A=...

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