composition versus inversion

While trying to convey to an OP on X validated why the inversion method was not always the panacea in pseudo-random generation, I took the example of a mixture of K exponential distributions when K is very large, in order to impress (?) upon said OP that solving F(x)=u for such a closed-form cdf F was very costly even when using a state-of-the-art (?) inversion algorithm, like uniroot, since each step involves adding the K terms in the cdf. Selecting the component from the cumulative distribution function on the component proves to be quite fast since using the rather crude

x=rexp(1,lambda[1+sum(runif(1)>wes)])

brings a 100-fold improvement over

Q = function(u) uniroot((function(x) F(x) - u), lower = 0, 
    upper = qexp(.999,rate=min(la)))[1] #numerical tail quantile
x=Q(runif(1))

when K=10⁵, as shown by a benchmark call

         test elapsed
1       compo   0.057
2      Newton  45.736
3     uniroot   5.814

where Newton denotes a simple-minded Newton inversion. I wonder if there is a faster way to select the component in the mixture. Using a while loop starting from the most likely components proves to be much slower. And accept-reject solutions are invariably slow or fail to work with such a large number of components. Devroye’s Bible has a section (XIV.7.5) on simulating sums of variates from an infinite mixture distribution, but, for once,  nothing really helpful. And another section (IV.5) on series methods, where again I could not find a direct connection.

beta HPD

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)
[1] 0.4504879
> f(ff(.95))
[1] 0.5580267

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