hw8sol - STAT 225 Homework 8 Solutions 1 10 points 5 points...

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Unformatted text preview: STAT 225 - Homework 8 - Solutions 1. 10 points: 5 points for E(Y), 5 points for V(Y) Since Λ is random, the Poisson assumption applies to the conditional distribution of Y for a fixed λ. Thus, PY |Λ (y |λ) = λy e−λ y = 0, 1, 2, ... y! So, E (Y |Λ = λ) = λ and, V (Y |Λ = λ) = λ Since Λ has a Gamma distribution with parameters α and β , and has an expected value of αβ and variance αβ 2 , E (Y ) = E [E (Y |Λ)] = E (Λ) = αβ V (Y ) = E [V (Y |Λ)] + V [E (Y |Λ)] = E (Λ) + V (Λ) = αβ + αβ 2 =⇒ αβ + αβ 2 σY = 2. 8 points Note that X Y =y ∼ Beta(1, y ) ⇒ E (X |Y = y ) = 1 1+y . Therefore: 1 1+Y 1 1 · 1dy = (Y ∼ U (0, 1)) = 0 1+y E (X ) = E [E (X |Y )] = E = ln(1 + y ) 1 0 = ln2 − ln1 = ln2 0 3. 20 points: (a) 6 points (b) 6 points (c) 8 points (a) ∞ fY (y ) = x=0 30y 4 (1 − y )2 xe−xy dx ∞ = 30y 4 (1 − y )2 xe−xy x=0 1 almost Ga(2, y ) ∞ 1 = 30y 4 (1 − y )2 ( )2 y dx x=0 xe−xy dx 1 ( y )2 Γ(2) 1 Ga(2, y ) = 30y 2 (1 − y )2 , where 0 < y < 1 ⇒Y ∼ Beta(3, 3) (b) fX |Y (x|y ) = fXY (x, y ) fY (y ) 30y 4 (1 − y )2 xe−xy 30y 2 (1 − y )2 = = y 2 xe−xy ,where x > 0 and 0 < y < 1 1 ⇒ X |Y =y ∼ Ga(2, ) y (c) E (X ) = E [E (X |Y )] = E ( 2 ) since Y ∼ Beta(3, 3) Y 1 2· Y 1 = y =0 2 30y 2 (1 − y )2 dy y 1 y (1 − y )2 = 60 y =0 = 60 12 almost Beta(2,3) 1 y =0 Γ(5) y (1 − y )2 dy Γ(2)Γ(3) 12 = 60 =5 12 dy 4. 12 points: 4 for each part (a) x3 1 1 −x e dy = e−x y 6 6 fX (x) = y =0 x3 0 1 = x3 e−x where x > 0 6 ⇒ X ∼ Ga(4, 1) (b) fY |X (y |x) = 1 −x 6e 1 3 −x 6x e fXY (x, y ) = fX (x) 1 where 0 < y < x3 and x > 0 x3 = ⇒ Y |X =x ∼ U nif orm(0, x3 ) (c) E (Y ) = E [E (Y |X )] = ∗ Recall that if U ∼ U nif orm(a, b) E (U ) = a+b . 2 Therefore, since Y |X =x ∼ U nif orm(0, x3 ) , E (Y |X ) = ∗ E( X3 )= 2 = = ∞ x=0 1 12 X3 . 2 x3 1 3 −x x e dx since X ∼ Ga(4, 1) 26 ∞ x=0 1 Γ(7) 12 x6 e−x almost Ga(7,1) ∞ x6 e−x x=0 17 Γ(7) dx dx Ga(7,1) = 1 6! = 60 12 5. 28 points: (i) 4 points (ii) 4 points (iii) 6 points (iv) 6 points [-5 if E(1/x)=1/E(x)] (v) 8 points [3 for E(T), 5 for V(T)] fXY (x, y ) = 16xe−2(x+2y) X > 0, Y > 0 In HW7 we found: △ X ∼ Ga(α = 2, β = 1 ) =⇒ E (X ) = 1, 2 † Y ∼ exp( 1 ) =⇒ E (Y ) = 1/4, 4 ‡X Y (independent). V (X ) = 1/2 V (Y ) = 1/16 ‡ (i) E (XY ) = E (X )E (Y ) = (2 · 1/2) · 1/4 = 1/4 ‡ (ii) Cov (X, Y ) = 0 ‡ (iii) E (eX +2Y )= E (eX e2Y ⇒ E (eX +2Y ) = (iv) E Y X ) = = E (eX )E (e2Y ) = mX (t = 1) · mY (t = 2) 2 1 1 =4·2=8 1 − (1/2 · 1) 1 − (1/4 · 2) ‡ = E (Y )·E 1 X † = 1/4·E 1 X △ ∞ = 1/4 0 1/2 1 · 4xe−2x dx = x ∞ e−2x dx = 0 (v) T = 2 + 2X − 4Y , E (T ) = E (2 + 2X − 4Y ) = E (2) + 2E (X ) − 4E (Y ) = 2 + 2 · 1 − 4 · 1/4 = 3 V (T ) = V (2 + 2X − 4Y ) = V (2X − 4Y ) = V (2X ) + V (−4Y ) + 2Cov (2X, −4Y ) 0 ⇒ V (T ) = 4V (X ) + 16V (Y ) + 2 · 2 · (−4) C ov (X, Y ) = 4 · 1/2 + 16 · 1/16 = 3 6. 16 points: (a) 4 points (b) 6 points (c) 6 points (3 for cov, 3 for corr) fXY = 60x2 y 0 x > 0, y > 0, x + y ≤ 1 else (a) E (X ) = 1 3·3 1 3 = , V (X ) = = 2 (3 + 3 + 1) 3+3 2 (3 + 3) 28 E (Y ) = 2 1 2·4 2 = , V (X ) = = 2 (2 + 4 + 1) 2+4 3 (2 + 4) 63 (b) 1 1−x E (XY ) = x=0 y =0 1 xy · 60x2 ydydx x3 = 60 x=0 1−x 1 1−x y3 x· = 60 3 x=0 3 1 x=0 = 20 140 dx 0 x3 (1 − x)3 dx = 20 = y 2 dydx y =0 almost Beta(4,4) 1 x=0 140 x3 (1 − x)3 dx B eta(4,4) 20 1 ·1= 140 7 (c) cov (X, Y ) = E (XY ) − E (X )E (Y ) 1 11 −· = 7 23 1 =− 42 cov (X, Y ) ⇒ corr (X, Y ) = V (X ) V (Y ) = = −1 42 1 28 2 63 = (algebra) −1 √ = −.707 2 7. 6 points Cov (αX + a, βY + b) = E [(αX + a)(βY + b)] − E (αX + a)E (βY + b) = E [αβXY + αbX + βaY + ab] − [αE (X ) + a][βE (Y ) + b] = αβE (XY ) + αbE (X ) + βaE (Y ) + E (ab) ab −(αβE (X )E (Y ) + αbE (X ) + βaE (Y ) + ab) = αβE (XY ) − αβE (X )E (Y ) = αβ [E (XY ) − E (X )E (Y )] = αβCov (X, Y ) ...
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This note was uploaded on 07/15/2011 for the course STAT 36225 taught by Professor Tom during the Summer '09 term at Carnegie Mellon.

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