CS代考计算机代写 algorithm Econometrics (M.Sc.) 1 Exercises (with Solutions) 359134

Econometrics (M.Sc.) 1 Exercises (with Solutions) 359134
1. Problem
Program the Newton Raphson algorithm for a numerical computation of the ML estimate θˆ of the parameter θ = P(Coin = HEAD) in our coin toss example of Chapter 6 of our script. Replicate the results in Table 6.1. of our script.
Solution
## Below I use the same data that was used to produce the
## results in Table 6.1 of our script. However, you
## can produce new data by setting another seed-value
theta_true <- 0.2 # unknown true theta value n <- 5 # sample size ## Use a common Random Number Generator: RNGkind(sample.kind = "Rounding") ## Warning in RNGkind(sample.kind = "Rounding"): non-uniform 'Rounding' sampler used set.seed(1) # simulate data: n many (unfair) coin tosses x <- sample(x = c(0,1), size =n, replace = TRUE, prob = c(1-theta_true, theta_true)) ## number of heads (i.e., the number of "1"s in x) h <- sum(x) ## First derivative of the log-likelihood function Lp_fct <- function(theta, h = h, n = n){ (h/theta) - (n - h)/(1 - theta) } ## Second derivative of the log-likelihood function Lpp_fct <- function(theta, h = h, n = n){ - (h/theta^2) - (n - h)/(1 - theta)^2 } t <- 1e-10 check <- TRUE i <- 0 theta <- 0.4 Lp <- Lp_fct( theta, h=h, n=n) Lpp <- Lpp_fct(theta, h=h, n=n) # convergence criterion # for stopping the while-loop # count iterations # starting value Econometrics (M.Sc.) 2 while(check){ i <- i + 1 ## theta_new <- theta[i] - (Lp_fct(theta[i], h=h, n=n) / Lpp_fct(theta[i], h=h, n=n)) Lp_new Lpp_new ## theta Lp <- Lp_fct( theta_new, h = h, n = n) <- Lpp_fct(theta_new, h = h, n = n) <- c(theta, theta_new) <- c(Lp, Lp_new) <- c(Lpp, Lpp_new) Lpp ## if( abs(Lp_fct(theta_new, h=h, n=n)) < t ){check <- FALSE} } cbind(theta, Lp, Lp/Lpp) ## theta Lp ## [1,] 0.4000000 -4.166667e+00 2.400000e-01 ## [2,] 0.1600000 1.488095e+00 -3.326733e-02 ## [3,] 0.1932673 2.159084e-01 -6.558924e-03 ## [4,] 0.1998263 5.433195e-03 -1.736356e-04 ## [5,] 0.1999999 3.539786e-06 -1.132731e-07 ## [6,] 0.2000000 1.504574e-12 -4.814638e-14 2. Problem Assume an i.i.d. random sample X1, . . . , Xn from an exponential distribution, i.e. the un- derlying density of Xi is given by f(x|θ) = θexp(−θx). We then have μ := E(Xi) = 1 as wellasVar(X)= 1 . i θ2 (a) What is the log-likelihood function for the i.i.d. random sample X1 , . . . , Xn ? (b) Derive the maximum likelihood (ML) estimator θˆn of θ. (c) From maximum likelihood theory we know that ˆ􏱣1􏱤 (θn−θ)→dN 0,nJ(θ) θ Derive the expression for the Fischer information I(θ) = nJ (θ). Use the Fisher infor- mation to give the explizit formula for the asymptotic distribution of θˆn. Solution (a) The log-likelihood function is given by n l(θ) = 􏱛 ln(θ exp(−θXi))) i=1 n = 􏱛(ln θ − θXi) i=1 n = n ln θ − 􏱛 θXi i=1 Econometrics (M.Sc.) 3 (b) The ML estimator is defined as θˆn = arg max l(θ). Deriving the ML estimator θˆn: ′ 1􏱛n ln(θ) = nθ − ′ˆ 1􏱛n ln(θn)=0⇔0=nθˆ− Xi n i=1 1 􏱛n ⇔nθˆ= Xi n i=1 ˆ11 ⇔ θ n = 1 􏱯 n X = X ̄ n i=1 i (c) The Fisher information is given by I(θ) = nJ (θ), where J (θ) = − 1 E(l′′(θ)). The n second derivative of l(θ) is given by l′′(θ) = −n 1 i=1 Xi θ2 So, the expression for l′′(θ) is here deterministic as it doesn’t depend on the random variables Xi. 1 ′′ 1􏱣 1􏱤 1 J(θ) = −nE(l (θ)) = −n −nθ2 = θ2 That is, the Fisher information is I(θ) = nJ(θ) = n/θ2. Therefore, the asymptotic distribution of θˆn is ˆ 􏱣 θ2􏱤 (θn−θ)→dN 0,n ⇔ √n(θˆn−θ)→dN􏱡0,θ2􏱢