How do I genenate random (or pseudo random) numbers from a list with a specified distribution
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How do I generate random numbers from a list, with the condition that the numbers must have a given distribution?
What Ive done so far:
I was able to generate random numBers with a normal distribution with
y = 2 + .2*randn(10000,1)
Now I want to pick 3 samples from these numbers with size 50 each, each having normal, lognormal and exponential distributions. How will I do it?
Thanks in advance
回答 (2 件)
Walter Roberson
2011 年 5 月 31 日
0 投票
Usually you cannot do that. Suppose for example that the list was just 0's and 1's representing flips of a coin. You are not going to be able to get an exponential distribution from such a list.
7 件のコメント
Oleg Komarov
2011 年 5 月 31 日
@Raymundo: maybe you're trying to sample from a given distribution by taking the inverse cdf of a uniform [0 1]?
Walter Roberson
2011 年 5 月 31 日
I agree, that would make much more sense to do.
Raymundo Addun
2011 年 5 月 31 日
Walter Roberson
2011 年 5 月 31 日
Sub-selections are more likely than not to have a different mean. But getting the mean different _enough_ to be significant might be difficult; my intuition is that it would sometimes be impossible, but I would need to think about that more.
Walter Roberson
2011 年 5 月 31 日
What if you "reverse engineer" kstest2() in order to determine how you would have to subsample in order to get a sufficiently high confidence for your purposes?
Oleg Komarov
2011 年 5 月 31 日
@Raymundo: the uniform distribution is just a trick to sample from any distribution you chose since it enters the cdf from the y axes and returns the values on x axes according to the distribution you've chosen: http://en.wikipedia.org/wiki/Inverse_transform_sampling
Raymundo Addun
2011 年 6 月 1 日
Laura Proctor
2011 年 5 月 31 日
Although you won't be able to generate the numbers from the exact data you have in y, you can still generate numbers using the data to gather the parameters, like so:
[muhat,sigmahat] = normfit(y)
rn = muhat*ones(50,1) + chol(sigmahat)*randn(50,1);
parmhat = lognfit(y);
rl = lognrnd(parmhat(1),parmhat(2),50,1);
muhat = expfit(y);
re = exprnd(muhat,50,1)
2 件のコメント
Walter Roberson
2011 年 5 月 31 日
Though this does break the fundamental premise of the question that the numbers must be generated "from a list" (i.e., a given list of values must be sub-selected from to achieve the given distribution.)
Raymundo Addun
2011 年 5 月 31 日
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