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John D'Errico
on 23 Jun 2016

John D'Errico
on 26 Jun 2016

Edited: John D'Errico
on 26 Jun 2016

Sigh. I think this is a misunderstanding of what uniformly random means. A fairly common one too. Since you seem not interested in showing your code, one can only guess the problem though. There can be no certain answer if I do not see your code. (addendum: The OP has since added a comment that indicates my conjecture is exactly on target.)

You are generating numbers uniformly random with a target interval [1e-6,1].

A uniform distribution implies that for ANY sub-interval of fixed width that is contained in the global window [1.e-6,1] (so assume a sub-interval [a,b]) where we have

1e-6 <= a <= b <=1

then the expected number of events we will observe should be:

(b - a)/(1 - 1e-6)

If you will generate N samples, then the expected number of events in the sub-interval is:

N*(b - a)/(1 - 1e-6)

So expect to see a number of events that are proportional to the sub-interval width. You won't seee exactly that many, since this is a random sampling.

So a uniform random sampling on the interval [0,1] would have roughly 10% of the samples in each bin [0,0.1], [0.1,0.2], [0.2,0.3], etc.

Now in your case, you are sampling on the interval [1e-6,1]. You find that very few samples occur right down at the bottom end, say between [1e-6,1e-5].

Lets use the rule above to see what fraction of the samples SHOULD occur in that interval. Lets say that we generate a sample size of 1000 values in the overall interval. Seems pretty big to me.

1000*(1e-5 - 1e-6)/(1 - 1e-6)

ans =

0.009

Hmm. I only expect to see 0.009 samples in that sub-interval, whereas I would have expected to see

1000*(1 - 0.9)/(1 - 1e-6)

ans =

100

So 100 events in the subinterval [0.9,1].

Is this truly uniform sampling? YES!!!!!!!!! Of course it is! You need to understand that the first interval I showed is a terribly tiny interval.

If you asked to generate a sampling that is uniformly probable over that region, but what you REALLY wanted was some sort of sampling that is uniform in a log space, then you needed to use a proper random sampling scheme!

For example, try this:

R = 10.^(rand(1,1000)*6 - 6);

Look at some percentiles of this sampling scheme:

Min 1.005e-06

1.0% 1.144e-06

5.0% 2.036e-06

10.0% 4.529e-06

25.0% 4.117e-05

50.0% 0.001265

75.0% 0.03185

90.0% 0.2554

95.0% 0.5309

99.0% 0.8607

Max 0.9928

It is NOT uniform, at least not in the domain [1e-6,1]. But the log10 of those numbers WILL be uniformly distributed. So, we will expect roughly 50% of the log10 values to be less than -3.

Min -5.998

1.0% -5.941

5.0% -5.691

10.0% -5.344

25.0% -4.385

50.0% -2.898

75.0% -1.497

90.0% -0.5928

95.0% -0.275

99.0% -0.06513

Max -0.003133

Again, it won't be perfect. But a sample size of 1000 is not really that huge. These predictions only become valid in the limit as N grows to a really large number.

Again, it is just a wild guess.

John D'Errico
on 26 Jun 2016

Did you read my response? If not, then why not? I spent, what, an hour writing that response to you. Then you asked exactly the same question that I just answered.

READ MY ANSWER. In that answer, I explained why you are confused, why you are not getting the kind of sampling that you want to see. I then show how to achieve the sampling that you want. But if you will ask a question and won't bother to read the answers you get and think about what they say, how can I do more? (Sorry if I seem frustrated.)

Roger Stafford
on 26 Jun 2016

Edited: Roger Stafford
on 26 Jun 2016

It seems clear from his most recent comment that where Pankaj says “uniform” he actually means a "logarithmic" distribution where there would be as many samples in the interval [10^(-6),10^(-5)] as in the interval [10^(-1),10^(0)], and indeed in any interval [10^(-k),10^(-k+1)], -6<=k<=-1. If that is the case, the proper code would be:

r = 10.^(-6*rand(1,n));

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