I am trying to generate a set of N random numbers where the set has a normal distribution.

I'm currently using a brute force approach:

  1. Randomly select N numbers from a normal distribution.
  2. Check the set's standard deviation (more important than mean).
  3. If it is the best set so far, keep it.
  4. Repeat 10000 times, and use the best set.

Is there any better approach? Anyone know where I could look?

Thanks in advance!

  • 1
    $\begingroup$ If you stop after step 1, then you already have a sample from a normal distribution. I don't think it makes any sense to say that any particular set has a normal distribution. I guess you are trying to maximize the likelihood that this particular sample came from a normal distribution, given some meta-distribution of possible distributions? $\endgroup$ Commented Oct 11, 2011 at 4:36
  • 3
    $\begingroup$ Look at this article for ideas. The usual pseudo-random number generators give you, more or less, random variables uniformly distributed on $[0,1]$. Given such a generator, you can fiddle with it to get a standard normal. The Box-Muller method is good, not hard to implement. To simulate a normal with mean $\mu$, standard deviation $\sigma$, multiply by $\sigma$, add $\mu$. Repeat $5000$ times to get your simulated sample. $\endgroup$ Commented Oct 11, 2011 at 4:48
  • $\begingroup$ André: That's currently what I'm doing. I was just curious if there was a way to avoid the "repeat 5000 times" step. $\endgroup$
    – sharoz
    Commented Oct 11, 2011 at 4:58
  • 2
    $\begingroup$ @sharoz, I don't think it makes sense. Using Box-Muller in step 1, you have a sample from a normal distribution. Rejecting and iterating gives you a sample from some other distribution no matter what criteria are used. What good is it for it to appear normal if it is actually not? (Although now I am curious exactly what distribution does this procedure produce?) $\endgroup$ Commented Oct 11, 2011 at 5:36
  • 1
    $\begingroup$ If you sample 10 values... and get all 0s, the normality is dubious. // Since we are considering sample variances, I might refer you to this, if only to stress that the empirical variance of a sample is itself random, and liable, in principle, to take any positive value whatsoever. $\endgroup$
    – Did
    Commented Oct 11, 2011 at 6:50

1 Answer 1


Assume we are talking about a standard normal distribution with zero mean and unit variance. For the observer to be able to answer the question "is this sample from a standard normal distribution?" with a high probability of correctness, he needs to know the distribution of distributions from which the sample may have been generated. The probability that the observer will guess "yes" is maximized when the sample is generated from a standard normal distribution, assuming that is possible. So according to my interpretation, you should use the values generated by Box-Muller in step 1 without inspecting them.

  • $\begingroup$ Repeating comment from above: I should say that this is for a perception experiment. If you sample 10 values from a normal distribution and get all 0s, it is technically a random set (albeit unlikely). However, someone looking at it would say that it appears to have low variance. See figure 1 here $\endgroup$
    – sharoz
    Commented Oct 11, 2011 at 6:41
  • $\begingroup$ I skimmed the paper and I think my answer still applies. If you are trying to determine the perceptual threshold for normally-distributed orientation variance, then you should stop at step 1. Otherwise you are using a distribution that is not normal. Also, you should generate a new random image for each perception experiment to maximize the SNR. $\endgroup$ Commented Oct 11, 2011 at 6:48
  • $\begingroup$ Very well. One step it is. Thanks for the help and info! $\endgroup$
    – sharoz
    Commented Oct 11, 2011 at 6:52
  • $\begingroup$ You are welcome sir. $\endgroup$ Commented Oct 11, 2011 at 6:53

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