如何找到给定分布的区间可行性？

Question

假设我有一些数据并且我将它们拟合到gamma分布，如何找到Pr(1 < x <= 1.5)的间隔概率，其中x是样本外数据点？

require(fitdistrplus)

a <- c(2.44121289,1.70292449,0.30550832,0.04332383,1.0553436,0.26912546,0.43590885,0.84514809,
0.36762336,0.94935435,1.30887437,1.08761895,0.66581035,0.83108270,1.7567334,1.00241339,
0.96263021,1.67488277,0.87400413,0.34639636,1.16804671,1.4182144,1.7378907,1.7462686,
1.7427784,0.8377457,0.1428738,0.71473956,0.8458882,0.2140742,0.9663167,0.7933085,
0.0475603,1.8657773,0.18307362,1.13519144)

fit <- fitdist(a, "gamma",lower = c(0, 0))

Answer 1

这里是一个使用MCMC技术和贝叶斯推理模式来估计新观察值落入区间（1：1.5）的后验概率的例子。这是一个无条件估计，与通过将伽马分布与最大似然参数估计相结合而获得的条件估计相反。

此代码要求在您的计算机上安装JAGS（免费且易于安装）。

library(rjags)

a <- c(2.44121289,1.70292449,0.30550832,0.04332383,1.0553436,0.26912546,0.43590885,0.84514809,
       0.36762336,0.94935435,1.30887437,1.08761895,0.66581035,0.83108270,1.7567334,1.00241339,
       0.96263021,1.67488277,0.87400413,0.34639636,1.16804671,1.4182144,1.7378907,1.7462686,
       1.7427784,0.8377457,0.1428738,0.71473956,0.8458882,0.2140742,0.9663167,0.7933085,
       0.0475603,1.8657773,0.18307362,1.13519144)

# Specify the model in JAGS language using diffuse priors for shape and scale
sink("GammaModel.txt")
cat("model{

    # Priors
    shape ~ dgamma(.001,.001)
    rate ~ dgamma(.001,.001)

    # Model structure
    for(i in 1:n){
    a[i] ~ dgamma(shape, rate)
    }
    }
    ", fill=TRUE)
sink()

jags.data <- list(a=a, n=length(a))

# Give overdispersed initial values (not important for this simple model, but very important if running complicated models where you need to check convergence by monitoring multiple chains)
inits <- function(){list(shape=runif(1,0,10), rate=runif(1,0,10))}

# Specify which parameters to monitor
params <- c("shape", "rate")

# Set-up for MCMC run
nc <- 1   # number of chains
n.adapt <-1000   # number of adaptation steps
n.burn <- 1000    # number of burn-in steps
n.iter <- 500000  # number of posterior samples
thin <- 10   # thinning of posterior samples

# Running the model
gamma_mod <- jags.model('GammaModel.txt', data = jags.data, inits=inits, n.chains=nc, n.adapt=n.adapt)
update(gamma_mod, n.burn)
gamma_samples <- coda.samples(gamma_mod,params,n.iter=n.iter, thin=thin)
# Summarize the result
summary(gamma_samples)

# Compute improper (non-normalized) probability distribution for x
x <- rep(NA, 50000)
for(i in 1:50000){
  x[i] <- rgamma(1, gamma_samples[[1]][i,1], rate = gamma_samples[[1]][i,2])
}

# Find which values of x fall in the desired range and normalize.
length(which(x>1 & x < 1.5))/length(x)

答案：      Pr(1 < x <= 1.5) = 0.194 非常接近条件估计，但通常情况并非如此。

Answer 2

有人不喜欢我的上述方法，这是以MLE为条件的;现在让我们看看无条件的东西。如果我们采用直接集成，我们需要三重集成：一个用于shape，一个用于rate，最后一个用于x。这不具吸引力。我将改为制作蒙特卡洛估计。

在中心极限定理下，MLE通常是分布式的。 fitdistrplus::fitdist没有给出标准错误，但我们可以使用MASS::fitdistr来执行精确推理。

fit <- fitdistr(a, "gamma", lower = c(0,0))

b <- fit$estimate
#   shape     rate 
#1.739737 1.816134 

V <- fit$vcov  ## covariance
          shape      rate
shape 0.1423679 0.1486193
rate  0.1486193 0.2078086

现在我们想从参数分布中抽样并获得目标概率的样本。

set.seed(0)
## sample from bivariate normal with mean `b` and covariance `V`
## Cholesky method is used here
X <- matrix(rnorm(1000 * 2), 1000)  ## 1000 `N(0, 1)` normal samples
R <- chol(V)  ## upper triangular Cholesky factor of `V`
X <- X %*% R  ## transform X under desired covariance
X <- X + b  ## shift to desired mean
## you can use `cov(X)` to check it is very close to `V`

## now samples for `Pr(1 < x < 1.5)`
p <- pgamma(1.5, X[,1], X[,2]) - pgamma(1, X[,1], X[,2])

我们可以制作p的直方图（如果需要，也可以进行密度估算）：

hist(p, prob = TRUE)

现在，我们经常需要预测器的样本均值：

mean(p)
# [1] 0.1906975

Answer 3

您可以在pgamma中使用fit估算参数。

b <- fit$estimate
#   shape     rate 
#1.739679 1.815995 

pgamma(1.5, b[1], b[2]) - pgamma(1, b[1], b[2])
# [1] 0.1896032

  

感谢。但P(x > 2)怎么样？

查看lower.tail参数：

pgamma(q, shape, rate = 1, scale = 1/rate, lower.tail = TRUE, log.p = FALSE)

默认情况下，pgamma(q)会评估Pr(x <= q)。设置lower.tail = FALSE会产生Pr(x > q)。所以你可以这样做：

pgamma(2, b[1], b[2], lower.tail = FALSE)
# [1] 0.08935687

或者您也可以使用

1 - pgamma(2, b[1], b[2])
# [1] 0.08935687