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

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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 ...
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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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