Stat 544 – Spring 2005 Homework assignment 4 answer key Exercise 1 µ+ µ2 + 4 2 20 05 20 05 K (α) = 0 ⇔ α2 − αµ − 1 = 0 ⇔ α = since α > 0. ii. No choice α = µ the notation: p(y) is our truncated normal (our target distribution), and g(y) is probability of obtaining one observation ≥ a. So p = 1 − Φ(a). Note that Range(X) the shifted exponential that we use as the instrumental density. M is the smallest i. Draw a candidate value X from g(x) b.1 - Upper bound Exercise 2 Thus, h(x) attains its maximum when x = α. Hence, the upper bound ignoring (a) Let θ = (α, β, σ 2 ), θ−α = (β, σ 2 ), and so on. at truncation is 1 α2 /2−αµ e α • Likelihood St Now, if we introduce the truncation we only need to consider values of α such that α ≥ µ. So, we consider two cases for the bound: 2 /2−αµ ii. α = µ: Upper bound is α1 e−α 2 /2 • Priors := K(α) = µ1 e−µ iv. Set Y = X with probability p/gM v. Go back to step (i). ∂ 2 2 h(x) = αe−x /2 eα −αµ ∂x i. α > µ: Upper bound is α1 eα iii. Draw a uniform U ∼ U(0, 1) - 1 α(x−µ) −x2 /2 e e α 4 ⇒ h(x) = 54 54 4 - ii. Compute the ratio p/M g evaluated at X at 2 D Io epa w r a tm St e at n t e U of S n i v ta S er tis p s i t i ri ty c s n straightforward: (b) Note that a has been replaced by µ for the sake of clarity f (x) = e−x /2 g(x) = αe−α(x−µ) value of the upper bound of the ratio p/g. So the steps in the algorithm are St D Io epa w r a tm St e at n t e U of S n i v ta S er tis p s i t i ri ty c s n = {1,2,...}. Hence, the average number of simulations is E(X) = [1 − Φ(a)]−1 g truncated normal. So X ∼ Geometric(p), where p is the acceptance rate, i.e. the g (a) Let X be the number of simulations needed to produce one observation from the b.3 - You have already found the M for each of the two cases (α > µ or α = µ). We use 2 /2 p(y|θ) ∝ σ −n exp − n 1 (yi − α − βxi )2 2 2σ i (α − α0 )2 p(α) ∝ σa−1 exp − 2 2σa (β − β0 )2 p(β) ∝ σb−1 exp − 2 2σb b.2 - Now need to find α that minimizes upper bound in each case i. 2 ∂ 1 2 1 2 eα /2−αµ 2 K(α) = K (α) = − 2 eα /2−αµ + eα /2−αµ (α−µ) = (α −αµ−1) ∂α α α α2 1 p(σ 2 ) ∝ (σ 2 )−(ν0 /2+1) exp 2 ν0 s20 2σ 2 p(θ|y) ∝ σ −(ν0 +2+n) exp − 12 σ12 ( ni (yi − α − βxi )2 + ν0 s20 ) + σ 2 |y, θ−σ2 ∼ Inv-χ2 (ν0 + n, n y[i] ~ dnorm(mu[i],tau.y) } alpha ~ dnorm(mu.a,tau.a) beta (yi − α − βxi )2 + ν0 s20 ) tau.y <- nu0*s20/tau.y0 σβ = σb2 σ2σ2 2b xi + σ 2 4 where 54 µα = Data α|y, θ−α ∼ N(µα , σα2 ) σa2 n(ȳ − β x̄) + σ 2 α0 nσa2 + σ 2 σα = list(y = c(....),x = c(....), N= , σ 2 σa2 nσa2 + σ 2 (c) To obtain a sample of size m using Gibbs sampling do: 2 - Set t = 1. 4 - Sample αt from a N(µα , σα2 ) i (yi − αt−1 − βt−1 xi )2 + ν0 s20 ) distribution. distribution with σ 2 n(ȳ − βt−1 x̄) + σt2 α0 µα = a nσa2 + σt2 5 - Sample βt from a N(µβ , σβ2 ) distribution with σ 2 ( xi yi − nαt x̄) + σt2 β0 µβ = b σb2 x2i + σt2 list(alpha = , beta = , tau.y0 = ) at at parameters, say β0 and α0 . n mu.a = , tau.a = , mu.b = , tau.b = , nu0 = , s20 = , ) Inits 1 - Choose two of the three parameters, say, β and α. Assign initial values for those 3 - Sample σt2 from a Inv-χ2 (ν0 + n, that this part was not required. - - • α. Below you will find how the data and inital values specification should look like. Note 4 σb2 ( xi yi − nαx̄) + σ 2 β0 σb2 x2i + σ 2 D Io epa w r a tm St e at n t e U of S n i v ta S er tis p s i t i ri ty c s n µβ = } σα = σβ = σt2 σa2 nσa2 + σt2 σb2 St D Io epa w r a tm St e at n t e U of S n i v ta S er tis p s i t i ri ty c s n sigma2 <- 1/tau.y β|y, θ−β ∼ N(µβ , σβ2 ) where ~ dnorm(mu.b,tau.b) tau.y0 ~ dchisq(nu0) g i • β. St model{ for (i in 1:N){ mu[i] <- alpha + beta*x[i] g • σ2 (d) WinBugs code 54 (b) Conditional posterior distributions 20 05 + (β 2 −2ββ0 ) σb2 (α2 −2αα0 ) 2 σa 20 05 • Joint posterior σ2σ2 t 2b xi + σt2 6 - Set t = t+1. Go back to step (3) and repeat m times. 3 4