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solutions2008

# solutions2008 - Statistics 351–Intermediate Probability...

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Unformatted text preview: Statistics 351–Intermediate Probability Fall 2008 (200830) Final Exam Solutions Instructor: Michael Kozdron 1. (a) We see that f X,Y ( x,y ) ≥ 0 for all 0 < x,y < 1, and that Z ∞-∞ Z ∞-∞ f X,Y ( x,y ) d x d y = Z 1 Z y 10 xy 2 d x d y = Z 1 5 y 4 d y = y 5 1 = 1 . Thus, f X,Y is a legitimate density. 1. (b) We compute f X ( x ) = Z ∞-∞ f X,Y ( x,y ) d y = Z 1 x 10 xy 2 d y = 10 x (1- x 3 ) 3 , < x < 1 . 1. (c) We compute E ( X ) = Z ∞-∞ xf X ( x ) d x = Z 1 x · 10 x (1- x 3 ) 3 d x = 10 9- 10 18 = 5 9 . 1. (d) We compute f Y | X = x ( y ) = f X,Y ( x,y ) f X ( x ) = 10 xy 2 10 x (1- x 3 ) 3 = 3 y 2 1- x 3 , x < y < 1 . 1. (e) We compute E ( Y | X = x ) = Z ∞-∞ yf Y | X = x ( y ) d y = Z 1 x y · 3 y 2 1- x 3 d y = 3(1- x 4 ) 4(1- x 3 ) . 1. (f) Using properties of conditional expectation (Theorem II.2.1), we compute E ( Y ) = E ( E ( Y | X ) ) = E 3(1- X 4 ) 4(1- X 3 ) = Z 1 3(1- x 4 ) 4(1- x 3 ) · 10 x (1- x 3 ) 3 d x = 5 2 Z 1 x- x 5 d x = 5 6 . 1. (g) If U = XY and V = Y , then solving for X and Y gives X = U/V and Y = V , so that the Jacobian of this transformation is J = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = 1 /v- u/v 2 1 = 1 v . By Theorem I.2.1, the joint density of ( U,V ) is therefore given by f U,V ( u,v ) = f X,Y ( u/v,v ) · | J | = 10 uv- 1 v 2 · v- 1 = 10 u provided that 0 < u < v 2 < 1 and v > 0. It then follows that the density function of U is given by f U ( u ) = Z ∞-∞ f U,V ( u,v ) d v = Z 1 √ u 10 u d v = 10 u (1- √ u ) provided that 0 < u < 1. 2. (a) The density function of X is given by f X ( x ) = Z ∞-∞ f X,Y ( x,y ) d x d y = Z ∞-∞ f X | Y = y ( x ) · f Y ( y ) d y. Substituting in gives f X ( x ) = Z ∞ θyx θ- 1 exp {- yx θ } · a p Γ( p ) y p- 1 e- ay d y = a p Γ( p ) θx θ- 1 Z ∞ y p exp n- y ( a + x θ ) o d y. Properties of the gamma function imply that Z ∞ y p exp n- y ( a + x θ ) o d y = Γ( p + 1) · ( a + x θ )- ( p +1) so that f X ( x ) = a p Γ( p ) θx θ- 1 · Γ( p + 1) · ( a + x θ )- ( p +1) = pa p θ x θ- 1 ( a + x θ ) p +1 provided that x > 0 (and using the fact that Γ( p + 1) = p Γ( p ).) 2. (b) If Z = X θ , then the density function of Z is f Z ( z ) = f X ( z 1 /θ ) · d d z z 1 /θ = pa p θ z 1- 1 /θ ( a + z ) p +1 · 1 θ z 1 /θ- 1 = pa p ( a + z ) p +1 provided that z > 0. 3. There are two ways to compute the resulting integral, depending on the order of integration. One way is much easier than the other. Solution 1 : ( d y d x ) We begin by noticing that P { X < Y } = ZZ { x<y } f X,Y ( x,y ) d y d x = Z ∞ Z ∞ x x 27 exp- x + y 3 d y d x = Z ∞ x 27 e- x/ 3 Z ∞ x e- y/ 3 d y d x....
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