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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ) ...
View Full Document

Ask a homework question - tutors are online