## open problem

Posted in R, Statistics with tags , , , , , on October 24, 2013 by xi'an

On the plane back from Warwick, I was reading an ABC arXived paper by Umberto Picchini and Julie Forman, “Accelerating inference for diffusions observed with measurement error and large sample sizes using Approximate Bayesian Computation: A case study” and came upon this open problem:

“A closed-form expression for generating percentiles from a fi nite-components Gaussian mixture is unavailable.” (p.5)

which means solving $\alpha\Phi(x)+(1-\alpha)\Phi(\{x-\mu)/\sigma) = \beta$

is not possible in closed form. (Of course it could also be argued that the equation Φ(x)=β is unavailable in closed-form ie that the analytic solution x=Φ-1(β) is formal…) While I can think of several numerical approaches, a few minutes with a sheet of paper let me convinced that indeed this is not solvable (hence not an open problem, contrary to the title of the post!).

Just for R practice (and my R course students!), here is a basic R code:

mixant=function(alpha=0.5,beta=0.95,mu,sig=1,prec=1/10^4){
onmal=1-alpha
qbeta=qnorm(beta)

# initial bounds
omb=min(qbeta,mu+sig*qbeta)
omB=max(qbeta,mu+sig*qbeta)
if (beta<alpha){
omB=min(omB,qnorm(beta/alpha))
}else{
omb=max(omb,mu+sig*qnorm((beta-alpha)/onmal))}
if (beta<onmal){
omB=min(omB,mu+sig*qnorm(beta/onmal))
}else{
omb=max(omb,qnorm((beta-onmal)/alpha))}

# iterations
for (t in 1:5){
ranj=seq(omb,omB,len=17)
cfs=alpha*pnorm(ranj)+onmal*pnorm((ranj-mu)/sig)
omb=max(ranj[cfs<=beta])
omB=min(ranj[cfs>=beta])

if ((omB-omb)<prec)
break()}
return(.5*(omb+omB))}