Finding an interval that covers 95% area under an asymmetrical curve

Question

I want to move from the mode (the red vertical line) of the curve called posterior toward the tails and stop when 95% of the area of the posterior is covered. My desire is to find the shortest interval (in x-axis units) that can do so. The two limit values of such interval are desired?

Note: I have tried the first solution HERE. But that solution DOES NOT work with this current problem!

P.S. Note that my posterior curve is NOT symmetric. Thus the shortest possible 95% is the best choice.

Here are my functions:

     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)

I think OP is right, that solution that Hong Ooi refers to doesn't work with this problem! — , Aug 21 '17 at 15:59

Ben Bolker · Accepted Answer · 2017-08-25T21:13:33.373

I believe the way to solve this problem is analogous to that found in coda::HPDinterval (which works on a density); starting from the peak of the curve, move a horizontal line down; for each level, invert the two halves of the curve to find the intersection points; measure the area between the intersection points.

Setup:

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)

Function inverse of the posterior distribution, one side at a time:

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")

Compute the area corresponding to a given horizontal cutoff:

areafun <- function(h) {
  i1 <- inverse.posterior(h,"left")
  i2 <- inverse.posterior(h,"right")
  return(integrate(posterior,i1,i2)$value)
}

areafun(0.07)

Find the height that gives a specific fraction of the density:

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)
}

Try it out:

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

For your second problem (Cauchy) I decided to encapsulate my solution into a function. tl;dr it works if you make the limits wide enough.

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

(1) I believe it does. Have you tested it? (2) Normalization won't make too much difference; it will set `post.area` to 1 — Ben Bolker, Aug 22 '17 at 12:34
the new version assumes you have normalized the posterior. You're going to have to decide whether to *either* (1) assume you've normalized the posterior *or* (2) assume you've chosen a broad enough range to capture 95% of the area. — Ben Bolker, Aug 26 '17 at 00:51

Finding an interval that covers 95% area under an asymmetrical curve

1 Answers1