3

我正在尝试在 JAGS 中实现为 WinBUGS 编写的以下模型：

model {
  for (i in 1:N) {                         
    wtp[i] ~ dweib(r[G[i]], mu[i])I(lower[i], upper[i])
    mu[i] <- exp(beta[G[i]])
    G[i] ~ dcat(P[])       
  }                                       
  P[1] ~ dunif(0.01, 0.99)
  P[2] <- 1 - P[1]
  r[1] ~ dunif(1, 10)
  r[2] ~ dunif(0.1, 10)
  beta[1] ~ dunif(0, 1000)      
  beta[2] ~ dunif(-1000, 0)                
  weibmed[1] <- pow(log(2) * exp(-beta[1]), 1 / r[1]) 
  weibmed[2] <- pow(log(2) * exp(-beta[2]), 1 / r[2])
  weibmed[3] <- pow(log(1 / (1 - 0.5 + P[1])) * exp(-beta[2]), 1 / r[2])
  weibmean[1] <- pow(exp(-beta[1]), 1 / r[1]) * exp(loggam((1 + r[1]) / r[1]))
  weibmean[2] <- pow(exp(-beta[2]), 1 / r[2]) * exp(loggam((1 + r[2]) / r[2]))
  weibmean[3] <- P[1] * weibmean[1] + P[2] * weibmean[2]
}

我认为在 JAGS 中使用它会很简单：

library(rjags)

txt <- 'model {
  for (i in 1:N) {                         
    wtp[i] ~ dweib(r[G[i]], mu[i])T(lower[i], upper[i])
    mu[i] <- exp(beta[G[i]])
    G[i] ~ dcat(P[])       
  }                                       
  P[1] ~ dunif(0.01, 0.99)
  P[2] <- 1 - P[1]
  r[1] ~ dunif(1, 10)
  r[2] ~ dunif(0.1, 10)
  beta[1] ~ dunif(0, 1000)      
  beta[2] ~ dunif(-1000, 0)                
  weibmed[1] <- pow(log(2) * exp(-beta[1]), 1 / r[1]) 
  weibmed[2] <- pow(log(2) * exp(-beta[2]), 1 / r[2])
  weibmed[3] <- pow(log(1 / (1 - 0.5 + P[1])) * exp(-beta[2]), 1 / r[2])
  weibmean[1] <- pow(exp(-beta[1]), 1 / r[1]) * exp(loggam((1 + r[1]) / r[1]))
  weibmean[2] <- pow(exp(-beta[2]), 1 / r[2]) * exp(loggam((1 + r[2]) / r[2]))
  weibmean[3] <- P[1] * weibmean[1] + P[2] * weibmean[2]
}'

set.seed(3.14159)
dat <- list(N = 1000, lower = rep(0, 1000), upper = runif(1000, 5, 200000))
ini <- list(P = c(0.4, NA), r = c(8.2, 1.2), beta = c(3.8, -6.5))

mod <- jags.model(
  file = textConnection(txt), 
  data = dat,
  inits = c(ini, .RNG.name = 'base::Mersenne-Twister', .RNG.seed = 314159),
  n.chains = 1,
  n.adapt  = 100
)

sam.jags <- coda.samples(
  model = mod,
  variable.names = c('P', 'r', 'beta', 'weibmed', 'weibmean'),
  n.iter = 400,
  n.thin = 1
)

只需将 替换I()为T(). 这会产生coda.samples()错误：

Error: Error in node weibmed[3]
Invalid parent values

如果我忽略监控weibmed然后工作weibmean但coda.samples()参数估计：

                Mean          SD    Naive SE Time-series SE
P[1]       0.4840704   0.2769491  0.01384746     0.01384746
P[2]       0.5159296   0.2769491  0.01384746     0.01384746
beta[1]  509.3614647 295.0860473 14.75430237    14.75430237
beta[2] -487.5362940 285.4126899 14.27063449    14.27063449
r[1]       5.2054730   2.6330434  0.13165217     0.13165217
r[2]       5.0478143   2.9480476  0.14740238     0.14740238

与我在使用 WinBUGS 时得到的那些没有可比性：

library(R2WinBUGS)

sam.bugs <- bugs(
  model.file = 'model.bug',
  data = dat,
  inits = list(ini),
  parameters.to.save = c('P', 'r', 'beta'), #, 'weibmed', 'weibmean'),
  n.chains = 1,
  n.burnin = 100,
  n.iter = 500,
  n.thin = 1,
  debug = F,
  DIC = F,
  bugs.seed = 314159
)

 Inference for Bugs model at "3mixout2.bug", fit using WinBUGS,
 1 chains, each with 500 iterations (first 100 discarded)
 n.sims = 400 iterations saved
        mean  sd 2.5%  25%  50%  75% 97.5%
P[1]     0.4 0.0  0.3  0.4  0.4  0.4   0.4
P[2]     0.6 0.0  0.6  0.6  0.6  0.6   0.7
r[1]     7.2 0.8  6.2  6.6  6.9  7.7   9.3
r[2]     1.5 0.0  1.4  1.4  1.5  1.5   1.6
beta[1]  5.5 0.5  4.6  5.0  5.4  5.8   6.6
beta[2] -7.2 0.2 -7.5 -7.3 -7.2 -7.1  -6.9

有什么想法或建议吗？

4

2 回答 2

1

看来这是dinterval应该使用的情况，而不是T()：

txt2 <- '
data {
  x <- rep(1, N)
}
model {
  for (i in 1:N) {     
    x[i] ~ dinterval(wtp[i], B[i, ])                 
    wtp[i] ~ dweib(r[G[i]], mu[i])
    mu[i] <- exp(beta[G[i]])
    G[i] ~ dcat(P[])
  }                                     
  P[1] ~ dunif(0, 1)
  P[2] <- 1 - P[1]
  r[1] ~ dunif(0, 10)
  r[2] ~ dunif(0, 10)
  beta[1] ~ dunif(1, 1000)      
  beta[2] ~ dunif(-1000, 0)                
  weibmed[1] <- pow(log(2) * exp(-beta[1]), 1 / r[1]) 
  weibmed[2] <- pow(log(2) * exp(-beta[2]), 1 / r[2])
  # weibmed[3] <- pow(log(1 / (1 - 0.5 + P[1])) * exp(-beta[2]), 1 / r[2])
  weibmean[1] <- pow(exp(-beta[1]), 1 / r[1]) * exp(loggam((1 + r[1]) / r[1]))
  weibmean[2] <- pow(exp(-beta[2]), 1 / r[2]) * exp(loggam((1 + r[2]) / r[2]))
  # weibmean[3] <- P[1] * weibmean[1] + P[2] * weibmean[2]
}'

mod <- jags.model(
  file = textConnection(txt2), 
  data = list(N = dat[[1]], B = cbind(dat$lower, dat$upper)),
  inits = c(ini, .RNG.name = 'base::Mersenne-Twister', .RNG.seed = 314159),
  n.chains = 1,
  n.adapt  = 100
)
于 2016-11-28T02:14:10.040 回答
1

JAGS 手册说；

战俘（x，z）|| 幂函数 || 真实 || 如果 x < 0 那么 z 是整数

当0.5 < P[1], log(1 / (1 - 0.5 + P[1])) * exp(-beta[2]) < 0, 所以weibmed[3],pow(negative, non_integer)变成NaN. 据我所知，coda.samples()不允许监控变量取NaN.

如果您通过更改名称使用P[1] ~ dunif(0.01, 0.5)或weibmed[3]从受监控变量列表中排除，例如weibmed3 <- pow(log(1 / (1 - 0.5 + P[1])) * exp(-beta[2]), 1 / r[2])，您的代码将运行。

于 2016-11-25T07:39:38.033 回答