Set specific standard deviation limit on randn matrix?

I want to create a randn matrix, but I want all the values to be within 2 standard deviations away from the mean.

So I make the randn(a,b) matrix, but I'm confused as to how to set the limit of the standard deviation when creating the matrix itself.

Answer 1

[Wayne King]

With the Statistics Toolbox:

   pd = makedist('Normal','mu',0,'sigma',2);
    t = truncate(pd,0,4);
    % generate the random matrix - here 100x100
    R = random(t,100,100);

The above truncates to [0,4], if you want [-2*sd,2*sd]

    t = truncate(pd,-4,4);
    R = random(t,100,100);

[Paul]

what would I do if I want all the values to be within 2 standard deviations away from the mean? Could I still use the above formulas?

[Wayne King]

Yes, I've shown you how to do that:

    sigma = 3;
    mu = 2;
    pd = makedist('Normal','mu',mu,'sigma',sigma);
    t = truncate(pd,-2*sigma,2*sigma);
    % just 10x10 here
    R = random(t,10,10);

Now look at the values of R.

[Paul]

thank you very much.

is makedist a new function of sort? I have an older version of matlab, and it doesn't seem to be able to recognize it.

[Wayne King]

yes, it's pretty new. I would have to look at the release notes to see when it was introduced

[Paul]

Thanks again.

I'm just a little confused since the other poster said that it's not possible to create a randn matrix with limits on the SD, but that you would have to continue to resample until your conditions are met.

So when using truncate(pd,-2,2), my pd is a normal distribution, and then the -2 and 2 are the limits of distribution. Must a sigma value be provided before? Or is there a way to leave sigma as a variable.

[Walter Roberson]

Wayne used the new-fangled random() call, not randn().

Truncation is quite new, either this new release or the prior one if I recall.

Answer 2

[Walter Roberson]

Are you aware that numbers generated by such a system will never be randomly distributed? The Normal distribution inherently requires infinite distribution.

If you generate by randn(), you can discard values whose abs() > 2, but beware that the result will not be normally distributed and will not have a standard distribution of 1.

[Paul]

yep that makes sense, but I was hoping to just have no values in my "normal distribution" that were above the 2 SD.

I was initially going to use the abs() > 2 , but then would that leave me with missing values in my matrix? Is there a way to create the matrix with the limit set before, so the matrix that is created initially will have no values above 2 SD?

[Walter Roberson]

No, you cannot do that. You will need to keep creating values until you have enough for your purpose. For example instead of generating 10 values, generate 15, throw out the ones that are out of range, and see if you are left with at least 10; if not, then generate again, but if so then return the first 10.

Answer 3

[Roger Stafford]

(Since you say 'randn' the mean will be zero and the standard deviation one.) You will have to call on 'randn' for more elements than you wish to have in your final matrix so as to select those which satisfy your condition.

 x = [];
 n = 0;
 while n<a*b
  t = randn(ceil(1.03*(a*b-n)),1);
  x = [x;t(t<2)];
  n = length(x);
 end
 x = reshape(x(1:a*b),a,b);

To not exceed two standard deviations if you use 1.03 above, you will usually have to go through the while-loop only once. With a looser condition, the multiplicative factor needs to be greater.

EDIT:

To accomplish this task for arbitrary mean, mu, and standard deviation sigma, replace the last line by:

 x = sigma*reshape(x(1:a*b),a,b)+mu;