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I am trying to do:

  1. find 1000 points that represent samples from distribution X with parameters $(a,b,c,\ldots, d)$
  2. be guaranteed that the MLEs for those 1000 points are $(a,b,c,\ldots, d)$ with $99\%$ probability or whatever

Is there a methodology to generate 1000 points for a distribution, where when working backwards, the MLEs fit the original distribution with a particular certainty?

Motivation: Instead of using the inverse CDF method to generate random samples from a distribution, I am trying to pre-generate them to save computation time. As my software runs, I would be picking from the 1000 pre-generated points using a $U\sim[0,1)$ as input, which is computationally faster than generating new samples.

Note: method of moments might work instead of MLE?

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    $\begingroup$ The point to pre-generating a sample is unclear. What do you hope it would accomplish? $\endgroup$
    – whuber
    Commented Dec 10, 2022 at 17:19
  • $\begingroup$ I would want 1000 points so that when I calculated the MLEs on those points, I would get the original parameters. My question (OP) is about how to generate those 1000 points. I think it's a good question, do you agree? $\endgroup$ Commented Dec 10, 2022 at 18:26
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    $\begingroup$ MLEs can be biased, so I’m not sure that’s really what you want. That is, if you’re as accurate as you say you want to be in calculating the MLE, that might suggest incorrect generation of the synthetic data. $\endgroup$
    – Dave
    Commented Dec 10, 2022 at 18:35
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    $\begingroup$ Using MLEs of parameters for some distribution does not serve to prove observations came from the distribution they're MLE's for. $\endgroup$
    – Glen_b
    Commented Dec 11, 2022 at 5:10
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    $\begingroup$ 1. That you used some model and computed ML estimates under that model doesn't mean the model is correct; surely you can posit that data might have come from distribution F (albeit unknown to the statistician), but estimation was conducted under a model that's distribution G. 2. In general you cannot prove that observations come from some distribution. I feel like we're not getting at the root of your difficulty here at all. It may be best to post a new question. $\endgroup$
    – Glen_b
    Commented Dec 11, 2022 at 23:00

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