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2006solutions1

# 2006solutions1 - Statistics 351 Fall 2006(Kozdron Midterm#1...

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Statistics 351 Fall 2006 (Kozdron) Midterm #1 — Solutions 1. (a) By definition, f X | Y = y ( x ) is given by f X | Y = y ( x ) = f X,Y ( x, y ) f Y ( y ) . We being by calculating f Y ( y ) = 4 - y 2 0 1 π dx = 1 π 4 - y 2 for 0 < y < 2. Therefore, f X | Y = y ( x ) = 1 π 1 π 4 - y 2 = 1 4 - y 2 , 0 < x < 4 - y 2 . 1. (b) We find E ( X | Y = y ) = -∞ x · f X | Y = y ( x ) dx = 4 - y 2 0 x · 1 4 - y 2 dx = 1 2 4 - y 2 2 4 - y 2 = 4 - y 2 2 . 1. (c) Solving for X and Y we find X = U cos V and Y = U sin V. The Jacobian of this transformation is J = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v = cos v - u sin v sin v u cos v = u cos 2 v + u sin 2 v = u. The joint density of ( U, V ) is therefore given by f U,V ( u, v ) = f X,Y ( u cos v, u sin v ) · | J | = u π provided that 0 < u < 2 and 0 < v < π 2 . 1. (d) From (c) , we see that U and V are independent since we can write f U,V ( u, v ) = f U ( u ) · f V ( v ) where f U ( u ) = u 2 , 0 < u < 2, and f V ( v ) = 2 π , 0 < v < π 2 . 2. By the law of total probability, f X ( x ) = -∞ f X | M = m ( x ) f M ( m ) dm. Since X | M = m U (0 , m ), we know that f X | M = m ( x ) = 1 m , 0 < x < m . We also find that f M ( m ) = F M ( m ) = 12 m 2 - 12 m 3 = 12 m 2 (1 - m ) , 0 < m < 1 . Therefore, f X ( x ) = 1 x 1 m · 12 m 2 (1 - m ) dm = (6 m 2 - 4 m 3 ) 1 x = 2 - 6 x 2 + 4 x 3 , 0 < x < 1 . 1

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3. Notice that P ( X (1) = X 1 ) = P ( X 1 < X 2 ). Therefore, by the law of total probability, P ( X 1 < X 2 ) = 0 P ( X 2 > x | X 1 = x ) · f X 1 ( x ) dx. Since X 1 and X 2 are independent, we find P ( X 2 > x | X 1 = x ) = P ( X 2 > x ) = x 2 e - 2 y dy = e - 2 x . Thus, P ( X 1 < X 2 ) = 0 e - 2 x · e - x dx = - 1 3 e - 3 x 0 = 1 3 . An alternative solution can be given by conditioning on the value of
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2006solutions1 - Statistics 351 Fall 2006(Kozdron Midterm#1...

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