# hw2sol - APPM 4/5520 Solutions to Problem Set Two 1 Let y 1...

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Unformatted text preview: APPM 4/5520 Solutions to Problem Set Two 1. Let y 1 = g 1 ( x 1 ,x 2 ) = x 1 / ( x 1 + x 2 ) and let y 2 = g 2 ( x 1 ,x 2 ) = x 1 + x 2 . Then x 1 = g- 1 1 ( y 1 ,y 2 ) = y 1 y 2 and x 2 = g- 1 2 ( y 1 ,y 2 ) = y 2- y 1 y 2 = y 2 (1- y 1 ) . X 1 ,X 2 iid ∼ exp(rate = λ ) ⇒ f X 1 ,X 2 ( x 1 ,x 2 ) = λ 2 e- λ ( x 1 + x 2 ) I (0 , ∞ ) ( x 1 ) I (0 , ∞ ) ( x 2 ) The Jacobian of the transformation J = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ∂x 1 ∂y 1 ∂x 1 ∂y 2 ∂x 2 ∂y 1 ∂x 1 ∂y 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle y 2 y 1- y 2 1- y 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = y 2 So, f Y 1 ,Y 2 ( y 1 ,y 2 ) = f X 1 ,X 2 ( g- 1 1 ( y 1 ,y 2 ) ,g- 1 2 ( y 1 ,y 2 )) · | J | = λ 2 e- λ ( y 1 y 2 + y 2 (1- y 1 )) I (0 , ∞ ) ( y 1 y 2 ) I (0 , ∞ ) ( y 2 (1- y 1 )) · | y 2 | = λ 2 y 2 e- λy 2 I (0 , 1) ( y 1 ) I (0 , ∞ ) ( y 2 ) . Now we integrate out (or, see alternative to follow) the “unwanted” y 2 : f Y 1 ( y 2 ) = integraltext f Y 1 ,Y 2 ( y 1 ,y 2 ) dy 2 = integraltext ∞ λ 2 y 2 e- λy 2 I (0 , 1) ( y 1 ) dy 2 = I (0 , 1) ( y 1 ) · Γ(2) integraltext ∞ 1 Γ(2) λ 2 y 2 e- λy 2 dy 2 = I (0 , 1) ( y 1 ) · Γ(2) · 1 = I (0 , 1) ( y 1 ) · (2- 1)! = I (0 , 1) ( y 1 ) . (The integral was 1 since it was the integral of the Γ(2 ,λ ) pdf.) Therefore Y = Y 1 ∼ unif (0 , 1) . Note: We did not have to do the above integration since the joint pdf for Y 1 and Y 2 factors into a “ y 1 part” and a “ y 2 part”. This means that they are independent and, as long as we can collect the constants in the right way, we will see a unif (0 , 1) pdf (in y 1 ) times a Γ(2 ,λ ) pdf (in y 2 ). Therefore, we can conclude directly that Y = Y 1 ∼ unif (0 , 1). 2. (a) f X 1 ( x 1 ) = integraltext ∞-∞ f X 1 ,X 2 ( x 1 ,x 2 ) dx 2 = integraltext ∞ x 1 2 e- x 1- x 2 dx 2 = 2 e- 2 x 1 x 1 was below x 2 , but when “marginalizing out” x 2 , we ran it over all values from 0 to ∞ and so there was no upper bound on x 1 . The final answer for the marginal pdf of X 1 is f X 1 ( x ) = 2 e- 2 x I (0 , ∞ ) ( x ) . That is, X 1 ∼ exp ( rate = 2). f X 2 ( x 2 ) = integraltext ∞-∞ f X 1 ,X 2 ( x 1 ,x 2 ) dx 1 = integraltext x 2 2 e- x 1- x 2 dx 1 = 2 e- x 2- 2 e- 2 x 2 x 2 was above x 1 , but since there was no lower bound (other than 0) on x 1 when “marginalizing out” x 1 , x 2 ends up going all the way from 0 to ∞ . The final answer for the marginal pdf of X 2 is f X 2 ( x ) = 2 e- x 2 (1- e- x 2 ) I (0 , ∞ ) ( x ) ....
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hw2sol - APPM 4/5520 Solutions to Problem Set Two 1 Let y 1...

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