Using Randn with min and max matrices

Hi all,

Is it possible to use randn with the min and max values being matrices?

For Example:

 A = [1x28]   %The numbers vary within this matrix

 B = [1x28]   %The numbers vary within this matrix

 r = A + (B-A).*randn(500,28);

I'm trying to understand if this is mathematically possible, considering there are multiple min and max values. Will Matlab allocate each column a separate min and max value (this is what I am hoping it will do)? Or should this problem simply be tackled using a different approach?

What I am looking for is to create a random [500x28] matrix with varying min and max values that are specified in matrices A and B. I have used randn before for integer values and it has worked beautifully. I just don't know if it is mathematically correct to use it in this context. I would love some thoughts and information on the best approach to take.

Thanks in advance.

Answer 1

[Walter Roberson]

You wrote

   r = A + (B-A).*randn(500,28);

but the form you are using does not apply to randn(). randn() has infinite tails.

If you switch to

   r = A + (B-A).*rand(500,28);

then that works fine in R2016b or later. In R2016a or earlier you would need

r = bsxfun(@plus, A, bsxfun(@times, B-A, rand(500,28)) );

[Zena Assaad]

Thank you very much. I am using R2015a so I went with the second option and it worked very well. Out of curiosity, is Matlab allocating separate min max for each column?

[Walter Roberson]

With the bsxfun, it would copy the per-column (A) and ((B-A)) to all of the rows

[Zena Assaad]

OK that makes sense. Thank you.

Answer 2

[Image Analyst]

You could do this:

A = randi(99, 1,28)   % The numbers vary within this matrix
B = randi(99, 1,28)   % The numbers vary within this matrix
r = zeros(500, 28);
for col = 1 : size(A, 2)
  r(:, col) = A(col) + (B(col) - A(col)) .* rand(500, 1);
end

You might want to make sure B is more than A if that's what you expect.

[Zena Assaad]

Thank you for this. I tried this option and Walters option and both worked well. I went with Walters option because it was only one line.

[Image Analyst]

OK, though it can be a bit cryptic for beginners. That's why I went with something I thought would be easier for you to understand. Though if you can remember how bsxfun works, it can be handy.

[Zena Assaad]

It was actually helpful to see it laid out line by line. I did a bit of reading on bsxfun before using it and between the two answers it made sense :)