Why rand function is not uniform in large intervals?

I am using rand function to generate uniformly distributed random numbers in the interval [10e-6 and 1.] But the function generates the nos. which are close to 1 (RATHER THAN BEING UNIFORM IN THE ENTIRE INTERVAL]. I have tried with 10 nos. and 100 nos. But I found that most the nos. generated are close to 1. Then, how it will be a uniform distribution??

[Walter Roberson]

Please show your code.

[Adam]

rand always generates numbers between 0 and 1. What you then do with those to get them into a range you are interested in is entirely up to you.

[John D'Errico]

Rand IS uniform, and it generates numbers in the range from 0 to 1. If you are mis-using the results of rand in some way, then expect strange results. So show what you wrote.

Answer 1

[John D'Errico]

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.

[pankaj singh]

I am really sorry for late reply. Actually I thought that the code was not needed (as it really simple). My code is

 % 
N = 10;
dmin = 10e-6;
dmax = 1 ;
d = dmin + (dmax-dmin).*rand(1,N); 
%

The random nos. generated are close to the upper value (or in the upper intervals) and no nos. are generated in the lower intervals (e.g. near 1e-4, 1e-5, or 1e-6). How to resolve it. I want uniform samples from every interval randomly (apart from calling rand in every interval) ??

one result of above code is d =

    0.0246    0.0287    0.3213    0.0176    0.0474    0.0255    0.6637    0.1558    0.1694    0.6470

[Stephen23]

@pankaj singh: Instead of ten samples (which is too few for showing any kind of probability distribution trend), here is your code with one million samples:

N = 1e6;
dmin = 10e-6;
dmax = 1 ;
d = dmin + (dmax-dmin).*rand(1,N); 
hist(d)

and a histogram of those random values:

Does this look like a uniform distribution to you? If not, what would you expect it to look like ?

[John D'Errico]

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.)

Answer 2

[Roger Stafford]

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));

[Roger Stafford]

Oops! I didn't notice the same answer given by John earlier on.

Answer 3

[pankaj singh]

Dear all,

Thanks for all your responses and time... I think using 'logarithmic' distribution would probably be more suitable to use in my application rather than 'uniform' distribution.