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If I know how to simulate the distribution $\pi(\mathbf{x})$, then is there a way to directly generate samples of $(\pi(\mathbf{x}))^\beta$ for some $\beta > 0$ ?(assume that it can be normalised to a density, i.e. $\int (\pi(\mathbf{x}))^\beta d\mathbf{x} < \infty$ ) What I'm interested in is a example like this if $\pi = 0.5 N(\mu_1, \Sigma_1) + 0.5 N(\mu_2, \Sigma_2)$, and I know how to sample $\pi$ from Generating random variables from a mixture of Normal distributions, but how can I sample from $\pi^\beta$ (without using MCMC algorithms).

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ywx
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    I'm not following what you mean by $(\pi(x))^\beta$. If you mean that probability density is taken to the power $\beta$, this is clearly not a well defined probability distribution as it will not integrate to 1. If you mean you want draws that are $x^\beta$, then you can just draw from $\pi(x)$ and then take $x^\beta$ (it's the pdf that needs to be transformed, not the inverse cdf). But I'm not sure what you mean by this question. – Cliff AB May 18 '15 at 16:31
  • Another legitimate interpretation is that "$\pi$" is the CDF (the distribution function itself). The use of "$(\mathbf{x})$" as an argument to $\pi$ is also mysterious, because $\mathbf{x}$ is undefined and unnecessary. – whuber May 18 '15 at 16:57
  • @whube $(\pi(x))^\beta$ is the pdf of x up to normalising constant, assume that it is integrable. My use of bold type $x$ is simply an indication that $x$ may be a random vector. – ywx May 18 '15 at 18:58
  • It's nice that $\mathbf{x}$ is a random vector--but it doesn't appear to have anything at all to do with the question! For instance, in your example of a Normal mixture $\mathbf{x}$ does not appear anywhere on the right hand side and therefore is totally absent from the definition. (Your notation would make some sense for a spatial stochastic process indexed by $\mathbf{x}$, but I doubt that is what you intend.) Because many people will not read through all the comments, your clarification of what the $\beta$ power of a distribution means needs to appear in the question itself. – whuber May 18 '15 at 19:14
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    @whuber I guess not all questions are formated perfectly. I think as far as asking questions here is concerned, it's not a big deal as long as people get their ideas across. – ywx May 19 '15 at 08:53
  • I agree with the implication, but I am concerned about the assumed truth of its antecedent. When a notation is unconventional and contains superfluous undefined characters, our default assumption should be that it is *not* going to get the idea across correctly to many readers. I'm sure *you* understand what you want to ask and that *some* readers will understand as you intended. However, that is insufficient because it creates a risk that other readers--which ultimately may number in the millions--will either not understand or will misunderstand, which is even worse. – whuber May 19 '15 at 12:11

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