hw7sol - 36-225 - Homework 7 SOLUTIONS 1. 7 points: (a) 4...

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: 36-225 - Homework 7 SOLUTIONS 1. 7 points: (a) 4 points (2 for each distribution) (b) 3 points (a) Marginal distribution of Y1 0 1 .76 .24 y PY1 (y ) Marginal distribution of Y2 y PY2 (y ) (b) P (Y2 = 0|Y1 = 0) = 0 1 2 .55 .16 .29 P (Y2 = 0, Y1 = 0) .38 = = .5 P (Y1 = 0) .76 P (Y2 = 1|Y1 = 0) = .14 = .18 .76 P (Y2 = 2|Y1 = 0) = .24 = .32 .76 2. 2 points Y1 and Y2 are not independent. For example P (Y1 = 0, Y2 = 0) = P (Y1 = 0)P (Y2 = 0) 3. 8 points: (a) 6 points (1 for the zeros, and 1 for each of the five probabilities) (b) 2 points X can take on the values 0,1,2. Y can take on the values 1,2,3. x\y (a) 1 2 3 0 .1 0 0 1 .2 .4 0 2 0 .2 .1 To find the probabilities, we use basic rules of probability: ∗PXY (0, 1) = P (X = 0, Y = 1) = P (X = 0) · P (Y = 1|X = 0) = 21 1 · ·1= 54 10 ∗PXY (1, 1) = P (X = 1, Y = 1) = P (X = 1) · P (Y = 1|X = 1) = 2 32 2 3 1 +· ·= 54 5 4 3 10 ∗PXY (1, 2) = P (X = 1, Y = 2) = P (X = 1) · P (Y = 2|X = 1) = 4 32 2 3 2 +· ·= 54 5 4 3 10 1 3 5 3 ∗PXY (2, 3) = P (X = 2, Y = 3) = P (X = 2) · P (Y = 3|X = 2) = 5 ∗PXY (2, 2) = P (X = 2, Y = 2) = P (X = 2) · P (Y = 2|X = 2) = 2 4 2 · 4 · 2 2 = 3 10 1 1 ·= 3 10 · (b) X and Y are not independent, because for example, PXY (0, 2) = PX (0) · PY (2) 0 .1×.6 4. 10 points:(a) 2 points (b) 6 points: 3 for each marginal dist (c) 2 points (a) (i) PXY (x, y ) ≥ 0 ∀(x, y ) where x = 0, 1, 2, 3, ... and y = 0, 1, 2, 3, .... (ii) e2λ ∞ ∞ e−3λ λx+y 2y = e−3λ x!y ! y =0 x=0 ∞ y =0 eλ (2λ)y y! ∞ λx = 1. x! x=0 (b) ∞ PX (x) = PXY (x, y ) = y = y =0 ∞ e−3λ λx x! (2λ)y y! y =0 = e−3λ λx+y 2y x!y ! e−λ λx x! x = 0, 1, 2, 3 . . . e2λ =⇒X ∼ P(λ) ∞ Py (y ) = = PXY (x, y ) = x −3λ (2λ)y ∞ e y! x=0 x λ x=0 x! = e−3λ λx+y 2y x!y ! e−2λ (2λ)y y! y = 0, 1, 2, 3 . . . eλ =⇒Y ∼ P(2λ) (c) (X, Y ) are independent since e−λ λx e−2λ (2λ)y · x! y! −3λ λx+y 2y e = PXY (x, y ) = x!y ! PX (x) · PY (y ) = 2 5. 6 points: (a) 2 points (c) 4 points (a) 1 1 1 ky1 y2 dy1 dy2 = k 0 0 0 = 1 y2 y2 · 1 dy2 2 0 2 k y2 · 22 1 k =1 4 = 0 ⇒k=4 (c) 3 1 Y1 ≤ , Y2 ≤ 2 4 P 1 2 = 3 4 0 4y1 y2 dy2 dy1 0 A 1 2 =4 0 y2 y1 · 2 2 3 4 dy1 0 1 2 2 9 y1 · =4· 32 2 =4· 0 91 · 32 8 9 = 64 6. 10 points: (a) 4 points (b) 6 points The support of this joint density is the following: y 1 T =x y E x 1 (a) 1 1 y = k y =0 x=0 14 y 02 ⇒ k = 10 1 kxy 2 dxdy = k = dy = k 3 0 y5 1 10 0 = k 10 y2 x2 y dy 20 (b) y T 1 d y=x d A d d d 1/2 d d d d x+y =1 d E 1 1 y 10xy 2 dxdy = 10 P (X + Y > 1) = y =1/2 A 1 y =1/2 y4 y3 2 − 3 1 y2 1/2 x=1−y y 2 [y 2 − (1 − y )2 ]dy = 5 =5 = 5[ x 1 y =1/2 4 1 1/2 ≈ 0.885 x2 2 y 1−y dy y 2 (2y − 1)dy 7. 11 points: (a) 3 points (c) 4 points (d) 4 points Figure 1: Problem 8(a) (a) See figure above. 2−y1 fY1 (y1 ) = = y2 =y1 2 6y1 2 6y1 y2 dy2 y2 ·2 2 2−y1 y1 2 2 = 3y1 (2 − y1 )2 − y1 2 2 = 3y1 (4 − 4y1 ) = 12y1 (1 − y1 ) 0 < y1 < 1 ⇒Y1 ∼ Beta(3, 2) (c) fY1 ,Y2 (y1 , y2 ) fY1 (y1 ) 2 6y1 y2 = 2 12y1 (1 − y1 ) y2 2(1 − y ) 0 < y1 < 1 1 y1 < y2 < 2 − y1 = 0 else fY2 |Y1 (y2 |y1 ) = 5 (d) fY2 |Y1 (y2 |.6) = 5y2 y2 4 = 2(1 − .6) 0 1.1 ⇒ P (Y2 < 1.1|Y1 = .6) = .6 .6 < y2 < 1.4 else 5y2 dy2 4 1.1 5y 2 =2 8 .6 5 1.12 − .62 = .53125 = 8 8. 10 points: (a) 3 points (b) 3 points (c) 4 points Note: In the solutions to this problem, we’re using X and Y and instead of Y1 and Y2 . (a) Y given X = x has a U (0 , x) distribution. fXY (x, y ) (b) Recall that fY |X (y |x) = fX (x) This implies that: fXY (x, y ) = fY |X (y |x) · fX (x) Therefore in our case: fXY (x, y ) = 1 x 1 ·1= x 0 0<x<1 0<y<x otherwise 1 fXY (x, y )dx = (c) fy (y ) = x =y x y 1 dx = ln x x 1 y = − ln y 0 < y < 1 0 otherwise T 1 y=x E 1 6 x 9. 8 points: (a) 6 points, 3 for each marginal (1 for identifying); (b) 2 points (a) fX (x) = 1−x 3 y =0 120xy 4 dy = 120x y4 1−x = 30x(1 − x)4 , 0 < x < 1 0 ⇒ X ∼ Beta(α = 2, β = 5) fY (y ) = 1−y x=0 2 120xy 3 dx = 120y 3 x 2 1−y = 60y 3 (1 − y )2 , 0 < y < 1 0 ⇒ Y ∼ Beta(α = 4, β = 3) (b) X and Y are not independent since fX (x)fY (y ) = 120x(1 − x)4 y 3 (1 − y )2 = fX,Y (x, y ) 10. 8 points: (a) 6 points (b) 2 points (a) fX (x) = ∞ −2(x+2y ) dy = 16xe−2x ∞ e−4y dy y =0 16xe y =0 −4y ∞ = 4xe−2x (0 − (−1)) = 4xe−2x , 16xe−2x e−4 0 ⇒ X ∼ Gamma α = 2, β = 1 2 ∞ fY (y ) = 16xe−2(x+2y) dx x=0 ∞ = 16e−4y x=0 = 16e−4y · 1 · 4 xe−2x dx ” “ 1 almost Ga 2, 2 ∞ 0 4xe−2x dx ” “ 1 Ga 2, 2 = 4e−4y , y > 0 ⇒ Y ∼ exp 1 4 (b) X and Y are independent since fX (x)fY (y ) = 4xe−2x 4e−4y = 16xe−2(x+2y) = fXY (x, y ) . 7 = x>0 11. 12 points: (a) 6 points; (b) 2 points; (c) 4 points (a) fY (y ) = ∞ −y (x+1) dx = ye−y ∞ e−yx dx = x=0 ye x=0 ∞ −y (0 − (−1)) = e−y , y > −y e−yx ye −y 0 = e 0 ⇒ Y ∼ exp(1). This implies that fX,Y (x,y ) fY (y ) fX |Y (x|y ) = = ye−y (x+1) e −y = ye−yx , x > 0, y > 0 ⇒ X | Y = y ∼ exp(1/y ). Comment: So, for example: X | Y = 2 ∼ exp(1/2) X | Y = 8 ∼ exp(1/8), etc. (b) Clearly, X and Y are dependent since fX |Y (x|y ) depends on the value of y . (c) ∞ ye−y(x+1) fX (x) = y =0 = dy ” “ 1 almost Ga 2, x+1 Γ(2) (x + 1)2 ∞ y =0 ye−y(x+1) Γ(2) 1 x+1 ” “ 1 Ga 2, x+1 = 1 (x+1)2 0 x>0 else 8 2 dx = [Γ(2) = 1] Y= kX 0.8 1.0 12. 8 points: 4 for correct setup, 4 for work 0.0 0.2 0.4 Y 0.6 A 0.0 0.2 0.4 0.6 0.8 1.0 X X ∼ exp(λ1 ) , Y ∼ exp(λ2 ) , X ⊥ Y ⊥ ⇒ fXY (x, y ) = fX (x)fY (y ) = 1 −x/λ1 1 −y/λ2 e e ,x>0y>0 λ1 λ2 ∞ ∞ x=0 y =x P (Y > X ) = A ∞ = x=0 ∞ = x=0 ∞ = x=0 = = = = 1 −x/λ1 1 −y/λ2 e e dydx λ1 λ2 1 −x/λ1 ∞ 1 −y/λ2 e e dydx λ1 y =x λ2 ∞ 1 −x/λ1 dx −e−y/λ2 e λ1 x 1 −x/λ1 −x/λ2 e e dx λ1 ∞ 1 −x ( 1 + 1 ) e λ1 λ2 dx λ1 x=0 1 1 1 −x ( 1 + 1 ) −( + )−1 e λ1 λ2 λ1 λ1 λ2 11 1 ( + )−1 λ1 λ1 λ2 λ2 λ2 + λ1 9 ∞ 0 Extra Credit Assign one of : 0,2,5,8,10 =X Y 0.5 Y 1.0 1.5 Y =X +1 2.0 Note that the support of fXY (x, y ) is the following region: 0.5 1.0 1.5 2.0 X fXY (x, y )dx = (a) fY (y ) = x y 3x2 · dx = y 3 0<y<1 x=0 1 3x2 · dx = 1 − (y − 1)3 1 < y < 2 x =y −1 (b) fX |Y (x|y ) = fXY (x, y ) fY (y ) For 0 < y < 1: fX |Y (x|y ) = 3x2 , y3 0<x<y For 1 < y < 2: fX |Y (x|y ) = 3x2 , 1 − (y − 1)3 y−1<x<1 Important comments: Note that fY (y ) is composed of 2 parts, which together make up the marginal density. However for fX |Y (x|y ) we have 2 cases, depending on what value of y we are conditioning on. It’s important to understand that each by itself is a density. For example: 1 ⋆fX |Y (x| 2 ) = 5 ⋆fX |Y (x| 4 ) = 3x2 13 2 = 24x2 , 0<x< 64x2 3x2 , = 1 − (5/4 − 1)3 21 1 2 1 <x<1 4 10 (c) Using part (e) for y = 1.5 fX |Y (x|1.5) = 24x2 3x2 = , 1 − (1.5 − 1)3 7 .5 < x < 1 Therefore, .75 P (X < .75 | Y = 1.5) = x =. 5 24x2 8 dx = x3 7 7 11 .75 ≈ .34 .5 ...
View Full Document

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.

Ask a homework question - tutors are online