In the Laplace cum Gauss case, t would take roughly the same time, presumably. I was rather thinking of an unsuspected series resolution, when one is unaware of the series being a computable and invertible function…

]]>Yes, it is hard to fathom a counterexample where inverting the cdf is less costly than drawing first the component index… Unless the inverse cdf F⁻¹ of an infinite mixture is available in closed form while the mixture amounts to a series decomposition of the original F, possibly with negative coefficients to make it harder?!

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