## beta HPD

Posted in Books, R, Statistics, Uncategorized, University life with tags , , , , , , , on October 17, 2013 by xi'an

While writing an introductory chapter on Bayesian analysis (in French), I came by the issue of computing an HPD region when the posterior distribution is a Beta B(α,β) distribution… There is no analytic solution and hence I resorted to numerical resolution (provided here for α=117.5, β=115.5):

f=function(p){

# find the symmetric
g=function(x){return(x-p*((1-p)/(1-x))^(115.5/117.5))}
return(uniroot(g,c(.504,.99))$root)} ff=function(alpha){ # find the coverage g=function(x){return(x-p*((1-p)/(1-x))^(115.5/117.5))} return(uniroot(g,c(.011,.49))$root)}


and got the following return:

> ff(.95)
 0.4504879
> f(ff(.95))
 0.5580267


which was enough for my simple book illustration… Since (.450,558) is then the HPD region at credible level 0.95.

## A slice of infinity

Posted in R, Statistics, University life with tags , , , , , , , on July 28, 2011 by xi'an

Peng Yu sent me an email about the conditions for convergence of a Gibbs sampler:

The following statement mentions convergence. But I’m not familiar what the regularity condition is.

“But it is necessary to have a finite probability of moving away from the current state at all times in order to satisfy the regularity conditions on which the whole MCMC theory depends.”

Slice sampler is discussed in your book Monte Carlo Statistical Methods. I think that the “regularity condition” may have been discussed in your book. If so, would you please let me know where it is? Thanks and look forward to hearing from you!

The quote is from Martyn Plummer and deals with a stopping rule in JAGS implementation of the slice sampler. (The correct wording should be “strictly positive probability” rather than “finite probability”, I think.) However, this has nothing to do with a “regularity condition” on the irreducibility of a Markov chain: if a slice sampler is implemented for an unbounded density target, say a Beta(1/2,1/2), there is no irreducibility condition connected with the infiniteness of the density. In theory, (a) the chain never visits the “state” where the density is infinite (if only because we are dealing with a continuous state space) and (b) after visiting a value x with a large density f(x), the slice sampler allows for a move away from it since the slice involves a uniform simulation over (0,f(x)). Deeper properties of the slice sampler (like geometric ergodicity) are explored in, e.g., this JRSS B paper by Gareth Roberts and Jeff Rosenthal and this one in the Annals of Statistics by Radford Neal. In practice, the problem is caused by values of f(x) that cannot be computed and hence produce an error message like

Singularity in likelihood found by Slicer.

If those singularities can be localised, a neighbourhood excluding them should be introduced. (More easily said than done, obviously!)

Here is an example of a slice sampler with the Beta(1/2,1/2) distribution:

#graphics
dote=function(x,y) points(x,y,col="gold",pch=19,cex=.4)
mote=function(x,y,z,w) lines(c(x,z),c(y,w),col="gold",lwd=.5)
cst=dbeta(.5,.5,.5)*.5 #normalising constant
#inverting f(x)=d, 2nd degree equation
hitden=function(d) .5+.5*sqrt(1-4*( cst/ max(d,dbeta(.5,.5,.5)))^2)*c(-1,1)
#output
curve(dbeta(x,.5,.5),0,1,ylab="density",lwd=2,col="steelblue",n=1001)
x=runif(1);u=runif(1)*dbeta(x,.5,.5);dote(x,u)
for (t in 1:100){ #100 slice steps
bo=hitden(u)
nx=sample(c(runif(1,0,bo),runif(1,bo,1)),1)
nu=runif(1)*dbeta(nx,.5,.5)
mote(x,u,nx,nu)
x=nx;u=nu;dote(x,u)
}


which clearly explores the whole area under the Beta(1/2,1/2) density. Even when started at a large density value like f(.999999), it eventually leaves the vicinity of this highly improbable value.

## Asher’s enigma

Posted in R, Statistics with tags , , , on July 26, 2010 by xi'an

On his Probability and statistics blog, Matt Asher put a funny question (with my rephrasing):

Take a unit square. Now pick two spots at random along the perimeter, uniformly. For each of these two locations, pick another random point from one of the three other sides of the square and draw the segment. What is the probability the two segments intersect? And what is the distribution for the intersection points?

The (my) intuition for the first question was 1/2, but a quick computation led to another answer. The key to the computation is to distinguish whether or not both segments share one side of the square. They do with probability $\dfrac{2}{4}\times 1 + \dfrac{2}{4}\times\dfrac{2}{3} = \dfrac{5}{6},$

in which case they intersect with probability 1/2. They occupy the four sides with probability 1/6, in which case they intersect with probability 1/3. So the final answer is 17/36 (as posted by several readers and empirically found by Matt). The second question is much more tricky: the histogram of the distribution of the coordinates is peaked towards the boundaries, thus reminding me of an arc-sine distribution, but there is a bump in the middle as well. Computing the coordinates of the intersection depending on the respective positions of the endpoints of both segments and simulating those distributions led me to histograms that looked either like beta B(a,a) distributions, or like beta B(1,a) distributions, or like beta B(a,1) distributions… Not exactly, though. So not even a mixture of beta distributions is enough to explain the distribution of the intersection points… For instance, the intersection points corresponding to segments were both segments start from the same side and end up in the opposite side are distributed as $\dfrac{u_1(u_4-u_3)-u_3(u_2-u_1)}{u_4-u_3-u_2+u_1}$

where all u‘s are uniform on (0,1) and under the constraint $(u_2-u_1)(u_4-u_3)<0$. The following graph shows how well a beta distribution fits in that case. (Not perfectly, though!) The R code is

u=matrix(runif(4*10^5),ncol=4)
u[,c(1,3)]=t(apply(u[,c(1,3)],1,sort))
u[,c(2,4)]=-t(apply(-u[,c(2,4)],1,sort))
y=(u[,1]*(u[,4]-u[,3])-u[,3]*(u[,2]-u[,1]))/(u[,1]+u[,4]-u[,2]-u[,3])

Similarly, if the two segments start from the same side but end up on different sides, the distribution of one coordinate is given by $\dfrac{u_1(1-u_3)-u_3u_4(u_2-u_1)}{1-u_3-u_4(u_2-u_1)}$

under the constraint $u_3. The outcome is once again almost distributed as a beta: The corresponding R code is

u=matrix(runif(4*10^5),ncol=4)
u[,c(1,3)]=-t(apply(-u[,c(1,3)],1,sort))
y=(u[,1]*(1-u[,3])-u[,3]*u[,4]*(u[,2]-u[,1]))/(1-u[,3]-u[,4]*(u[,2]-u[,1]))