1.令(x1,x2)=X(t-1)月 2.从f(x1E2)中产生候选点X(t): 3.更新x1=X(t): 4.从f(x21)中产生X5(t): 5.令X(t)=(X(),X(t): R代码如下 Cde #initialize constants and parameters N<-5000 #length of chain burn<-1000 #burn-in length X <-matrix(0,N,2) #the chain,a bivariate sample rho<--.75 #correlation mu1<-0 mu2<-2 sigma1 <-1 s1gma2<-.5 s1 <sqrt(1-rho2)*sigmal Previous Next First Last Back Forward 4
1. -(x1, x2) = X(t − 1); 2. lf(x1|x2)•)ˇ¿:X∗ 1 (t). 3. ç#x1 = X∗ 1 (t). 4. lf(x2|x1)•)X∗ 2 (t). 5. -X(t) = (X∗ 1 (t), X∗ 2 (t)). R ìËXe ↑Code #initialize constants and parameters N <- 5000 #length of chain burn<- 1000 #burn-in length X <- matrix(0, N, 2) #the chain, a bivariate sample rho <- -.75 #correlation mu1 <- 0 mu2 <- 2 sigma1 <- 1 sigma2 <- .5 s1 <- sqrt(1-rho^2)*sigma1 Previous Next First Last Back Forward 4
s2 <sqrt(1-rho 2)*sigma2 ####generate the chain##### X[1,]<-c(mu1,mu2) #initialize for (i in 2:N){ x2<-X[1-1,2] m1 <mu1 rho (x2 -mu2)*sigma1/sigma2 X[1,1]<-rnorm(1,m1,s1) x1<-X[1,1] m2 <mu2 rho (x1 mu1)*sigma2/sigma1 X[i,2]<rnorm(1,m2,s2) b <burn +1 x<-X[b:N,] Code 产生的链开始的1000个观测被丢弃掉，剩下的观测存在x中，对此样本计算均 值和协方差矩阵如下。各参数的样本估计离真值很近，散点图也显示出二元正 Previous Next First Last Back Forward 5
s2 <- sqrt(1-rho^2)*sigma2 ###### generate the chain ##### X[1, ] <- c(mu1, mu2) #initialize for (i in 2:N) { x2 <- X[i-1, 2] m1 <- mu1 + rho * (x2 - mu2) * sigma1/sigma2 X[i, 1] <- rnorm(1, m1, s1) x1 <- X[i, 1] m2 <- mu2 + rho * (x1 - mu1) * sigma2/sigma1 X[i, 2] <- rnorm(1, m2, s2) } b <- burn + 1 x <- X[b:N, ] ↓Code )Ûm©1000á*ˇøÔK, êe*ˇ3x•, Èd Oé˛ ä⁄ê › Xe. àÎÍ Ol˝äÈC, —:„èw´— Previous Next First Last Back Forward 5
态所具有的球面对称性和负相关性特征 TCode compare sample statistics to parameters colMeans(x) cov(x) cor(x) plot(x,main="",cex=.5,xlab=bquote(X[1]), ylab=bquote(X[2]),ylim=range(x[,2])) Code 例2（贝叶斯分析例子：身体温度数据）考虑Mackowiak et al. (1992)的数据，该数据记录了130个人的身体温度（华氏），性别和每分钟的 心跳.实验的目的是检验Carl Wunderlich的观点-健康成年人的体温平 均为37C(=98.6°F). Previous Next First Last Back Forward 6
§‰k•°È°5⁄KÉ'5A. ↑Code # compare sample statistics to parameters colMeans(x) cov(x) cor(x) plot(x, main="", cex=.5, xlab=bquote(X[1]), ylab=bquote(X[2]), ylim=range(x[,2])) ↓Code ~ 2 (ìd©¤~f: Nߛ͂) ƒ Mackowiak et al. (1992)Í‚, TÍ‚P¹ 130á< Nß›(uº), 5O⁄z©® %a. ¢8¥uCarl Wunderlich*:– Ëx§c<Nß² ˛è37oC(=98.6oF). Previous Next First Last Back Forward 6
记温度为，i=1,·,n,并假设正态模型 V(4,a2) 以及取先验分布为 N(uo,ao),2~IG(ao,bo) 则此时我们的目标分布为4，σ的后验分布 f(4,o21g)xf(4,a2)π(4)r(a2) 为使用Gibbs抽样算法，我们必须计算f(叫a2,)与f(a2|4,y).经过计算我们 得到 哈 4o,g≈Ng+=o,wnw=2n+品 n ou.y1Ga+号o+∑-P. n 2 i=1 Previous Next First Last Back Forward 7
Pß›èyi, i = 1, · · · , n, øb . yi ∼ N(µ, σ2 ) ±9k©Ÿè µ ∼ N(µ0, σ0), σ2 ∼ IG(a0, b0) Kdû·Ç8I©Ÿèµ, σ2©Ÿ f(µ, σ2 |y) ∝ f(y|µ, σ2 )π(µ)π(σ 2 ) è¶^Gibbsƒé{, ·Ç7LOéf(µ|σ 2 , y)Üf(σ 2 |µ, y). ²LOé·Ç µ|σ, y ∼ N(ωy¯ + (1 − ω)µ0, ω σ 2 n ), ω = σ 2 0 σ2/n + σ 2 0 . σ 2 |µ, y ∼ IG(a0 + n 2 , b0 + 1 2 Xn i=1 (yi − µ) 2 ). Previous Next First Last Back Forward 7
使用此结果，Gibbs抽样算法如下： 对t=1,…,T, 1.令μ=ht-1),g=at-1) 2.计第w=7m=u呵+(1-ω)o和s2=w民 3.从N(m,s2)中产生μ. 4.令u(8=μ 5.计算a=a0+受，b=b0+∑=1(h-4)2 6.从G(a,b)中产生T 7.令g2=1/r以及(0=. 从而R代码如下： 下coda bodytemp<-read.table("bodytemp.txt",header=T) y<-bodytempStemp bary<-mean(y);n<-length(y) Iterations<-3500 mu0<-0;s0<-100;a0<-0.001;b0<-0.001 Previous Next First Last Back Forward 8
¶^d(J, Gibbsƒé{Xe: Èt = 1, · · · , T, 1. -µ = µ (t−1), σ = σ (t−1) . 2. Oéω = σ 2 0 σ2/n+σ2 0 , m = ωy¯ + (1 − ω)µ0 ⁄s 2 = ω σ 2 n . 3. lN(m, s2 )•)µ. 4. -µ (t) = µ. 5. Oéa = a0 + n 2 , b = b0 + 1 2 Pn i=1(yi − µ) 2 . 6. lG(a, b)•)τ. 7. -σ 2 = 1/τ±9σ (t) = σ. l RìËXe: ↑Code bodytemp<-read.table("bodytemp.txt",header=T) y<-bodytemp$temp bary<-mean(y); n<-length(y) Iterations<-3500 mu0<-0; s0<-100; a0<-0.001; b0<-0.001 Previous Next First Last Back Forward 8