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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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