Econometric take home APPS_Part_45

Econometric take home APPS_Part_45 - For part (c), we just...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
For part (c), we just note that γ = θ /( β + θ ). For a sample of observations on x , the log-likelihood would be ln L = n ln γ + ln(1- γ ) x i i n = 1 ln L /d γ = n/ γ - /(1- γ ). x i i n = 1 A solution is obtained by first noting that at the solution, (1- γ )/ γ = x = 1/ γ - 1. The solution for γ is, thus, γ = 1 / (1 + x ).Of course, this is what we found in part b., which makes sense. For part (d) f ( y | x ) = fxy fx (,) () = θ β βθ βθ θβ βθ ey xx yx x −+ ++ ( ) ( ) ! . Cancelling terms and gathering the remaining like terms leaves f ( y | x ) = so the density has the required form with λ = ( β + θ ). The integral is {} . This integral is a Gamma integral which equals Γ ( x +1)/ λ x +1 , which is the reciprocal of the leading scalar, so the product is 1. The log-likelihood function is [ ] / βθβθ ye x xy ! y x y [] / ! λ λ x xe y d +− 1 0 l n L = n ln λ - λ + ln λ - y i i n = 1 x i i n = 1 ln ! x i i n = 1 ln L / ∂λ = ( + n )/ λ - . x i i n = 1 y i i n = 1 2 ln L / ∂λ 2 = -( + n )/ λ 2 . x i i n = 1 Therefore, the maximum likelihood estimator of λ is (1 + x )/ y and the asymptotic variance, conditional on the x s is Asy.Var. = ( λ 2 / n )/(1 + λ x ) Part (e.) We can obtain f ( y ) by summing over x in the joint density. First, we write the joint density as . The sum is, therefore, . The sum is that of the probabilities for a Poisson distribution, so it equals 1. This produces the required result. The maximum likelihood estimator of θ and its asymptotic variance are derived from e e y x yy x ( ) / ! = −− fy e e y x yy x x = = 0 l n L = n ln θ - θ y i i n = 1 ln L / ∂θ = n / θ - y i i n = 1 2 ln L / ∂θ 2 = - n / θ 2 . Therefore, the maximum likelihood estimator is 1/ y and its asymptotic variance is θ 2 / n . Since we found f ( y ) by factoring f ( x , y ) into f ( y ) f ( x | y ) (apparently, given our result), the answer follows immediately. Just divide the expression used in part e. by f ( y ). This is a Poisson distribution with parameter β y . The log-likelihood function and its first derivative are l n L = - β + ln y i i n = 1 x i i n = 1 + ii i n ln = 1 - ln ! x i i n = 1 ln L / ∂β = - + / β , y i i n = 1 x i i n = 1 from which it follows that β = /.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

Econometric take home APPS_Part_45 - For part (c), we just...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online