
[英]Using optimize() to find the shortest interval that takes 95% area under a curve in R
[英]Finding an interval that covers 95% area under an asymmetrical curve
我想从称为posterior
的曲线的模式(红色垂直线)移向尾巴,并在覆盖posterior
95%区域时停止。 我的愿望是找到可以做到的最短间隔(以X轴为单位)。 需要这样的间隔的两个极限值吗?
注意:我已经在这里尝试了第一个解决方案。 但是该解决方案不适用于当前问题!
PS请注意,我的posterior
曲线不是对称的。 因此,尽可能短的95%是最佳选择。
这是我的功能:
prior = function(x) dnorm(x)
likelihood = function(x) dt(1.46, 19, x*sqrt(20))
posterior = function(x) prior(x)*likelihood(x)
mode = optimize(posterior, interval = c(-2, 2), maximum = TRUE, tol = 1e-12)[[1]]
curve(posterior, -2, 2, n = 1e4)
abline(v = mode, col = 2)
我相信解决此问题的方法类似于在coda::HPDinterval
(适用于密度)中找到的方法; 从曲线的峰值开始,向下移动一条水平线; 对于每个级别,将曲线的两半反转以找到交点; 测量相交点之间的面积。
设定:
prior = function(x) dnorm(x)
likelihood = function(x) dt(1.46, 19, x*sqrt(20))
posterior = function(x) prior(x)*likelihood(x)
mode = optimize(posterior, interval = c(-2, 2), maximum = TRUE, tol = 1e-12)[[1]]
curve(posterior, -2, 2, n = 1e4)
abline(v = mode, col = 2)
后验分布的函数逆,一次为一侧:
inverse.posterior <- function(x,side="left") {
target <- function(y) posterior(y)-x
ur <- switch(side,
left=uniroot(target,interval=c(-2,mode)),
right=uniroot(target,interval=c(mode,2)))
return(ur$root)
}
i1 <- inverse.posterior(0.07,"left")
i2 <- inverse.posterior(0.07,"right")
abline(h=0.07,col="gray")
abline(v=c(i1,i2),col="gray")
计算与给定的水平边界对应的面积:
areafun <- function(h) {
i1 <- inverse.posterior(h,"left")
i2 <- inverse.posterior(h,"right")
return(integrate(posterior,i1,i2)$value)
}
areafun(0.07)
找到给出特定比例密度的高度:
post.area <- integrate(posterior,-2,2)$value
find.lims <- function(a) {
ur <- uniroot(function(h) areafun(h)/post.area-a,
c(0.01,posterior(mode)-0.01))
return(ur$root)
}
试试看:
f <- find.lims(0.95)
## critical height = 0.02129
lwr <- inverse.posterior(f,"left") ## -0.124
upr <- inverse.posterior(f,"right") ## 0.753
integrate(posterior,lwr,upr)$value/post.area ## 0.9499
对于第二个问题(Cauchy),我决定将解决方案封装到一个函数中。 tl; dr如果您将限制设置得足够宽,它将起作用。
get.HPDinterval <- function(posterior,lwr=-2,upr=2,level=0.95,eps=0.001) {
mode = optimize(posterior, interval = c(lwr, upr), maximum = TRUE, tol = 1e-12)[[1]]
inverse.posterior <- function(x,side="left") {
target <- function(y) posterior(y)-x
ur <- switch(side,
left=try(uniroot(target,interval=c(lwr,mode))),
right=try(uniroot(target,interval=c(mode,upr))))
if (inherits(ur,"try-error")) stop("inverse.posterior failed: extend limits?")
return(ur$root)
}
areafun <- function(h) {
i1 <- inverse.posterior(h,"left")
i2 <- inverse.posterior(h,"right")
return(integrate(posterior,i1,i2)$value)
}
post.area <- integrate(posterior,lwr,upr)$value
if (post.area<level) stop("limits don't encompass desired area: a=",round(post.area,3))
find.lims <- function(a) {
ur <- uniroot(function(h) areafun(h)/post.area-a,
c(eps,posterior(mode)-eps))
return(ur$root)
}
f <- find.lims(level)
return(c(inverse.posterior(f,"left"),
inverse.posterior(f,"right")))
}
get.HPDinterval(posterior)
posterior2 = function(x) dcauchy(x)
get.HPDinterval(posterior2,-10,10) ## limits don't encompass desired area
get.HPDinterval(posterior2,-15,15) ## inverse.posterior failed: extend limits?
get.HPDinterval(posterior2,-20,20) ## -7.83993 7.83993
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.