Equally distributed multidimensional random values with boundaries - how to generate?

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I have to generate a matrix that will have 100 columns. Every row represents a value that can change in defined range. For example, I can describe it by the array:
A=[1 5; 3 7; 1 10]
Where 1 to 5 is first row range, 3 to 7 is the second, and 1 to 10 is the last. If I want to generate the random distribution to cover the range, just for one line, I can do this as follow:
data = lb + rand(1,100) .* ( ub - lb );
Where ub and lb are upper and lower boundary. Now, I can reproduce this in simple for loop:
for i=1:size(A,1)
data2(i,:) = lb + rand(1,100) .* ( ub - lb );
But in this case, every single row is evaluated separately, So I don't have any guarantee that the distribution will be equal in the meaning of comibinations between rows, as every rows changes independent. For example I can encounter situation where I will not have any combination with Row 1 close to 1 and Row 2 close to 7, just because of RNG. Is there any way I can sovle my problem and ensure multidimensional equal random distribution?
John D'Errico
John D'Errico on 13 Feb 2023
Edited: John D'Errico on 13 Feb 2023
I'm sorry, but I think you still misunderstand random numbers, what a uniform distribution means, and, apparently the entire point of my comment.
That you have columns with different ranges is completely irrelevant. Each column will be filled with sets of numbers that are uniformly distributed. And they are independent of other columns, or of previous samples.
For example:
n = 25000;
X = [rand(n,1),rand(n,1)*2 + 1];
So the first column of X (thus X(:,1)) is uniformly distributed, on the open interval (0,1).
X(:,2) is niformly distributed on the open interval (1,3).
These points, if taken as points in the two dimensional box (0,1)x(1,3), will fill that space uniformly. Of course if the sampling is coarse enough, the box will be filled in very well.
If I choose n a bit larger, then the figure turns completely blue, with white showing through at all. And if you count the number of points in any local region of the box, so essentially a 2-dimensional histogram, then you would find that locally the number of points in that region will be proportional to the area of the region you looked at.
For example, histcounts produces that 2-d histogram.
[N,XEDGES,YEDGES] = histcounts2(X(:,1),X(:,2))
N = 10×10
225 253 249 264 235 204 250 255 267 265 250 287 247 221 251 244 277 244 228 256 279 251 236 249 224 258 262 267 263 296 257 243 227 230 266 228 273 232 223 250 267 274 248 252 236 241 268 236 261 255 257 263 264 242 258 255 248 270 220 242 239 272 243 249 253 227 222 248 244 265 240 262 277 253 270 252 259 239 239 262 255 264 257 235 289 258 200 235 231 264 238 245 232 252 251 234 233 254 254 261
XEDGES = 1×11
0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000
YEDGES = 1×11
1.0000 1.2000 1.4000 1.6000 1.8000 2.0000 2.2000 2.4000 2.6000 2.8000 3.0000
And we would expect to see on average, with a 10x10 grid of bins on that domain, we would expect to see 1% of the samples falling in each bin. Indeed, that is what happens. If the sample size were larger, then the counts in each bin will more accurately approach that value of 1% in each bin. We expect to see some degree of variability of course in those bin counts, but as I have said, that will decrease with sample size.
That the different sets of variables live in different intervals is completely irrelevant. (Sorry, I forgot to scale the x and y axes in the 2-d hstogram plot.)
Karol P.
Karol P. on 13 Feb 2023
So do I understand correctly, that, as long as the sample size is high enough, the independed calculation of every row will not lead to any unequalities in distribution? I mean the case where, for example, I will have statistically important surplus of columns where the value of first row will be close to lb while in the second it will be close to up? It is pure RNG so I expected that without further limitations this case is at least possible.

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Accepted Answer

William Rose
William Rose on 13 Feb 2023
@Karol, my new understanding is that you want to find a uniformly distributed random point in a 3D rectangle. The bounds of the rectangle are chosen at random from a discrete set of possibilities. A is 3x2. Column 1 of A has the 3 allowed lower bounds for the edges. Column 2 has the 3 allowed upper values. Am I understanding you correctly? If so you will need two discrete random choices (one each for lower and upper bounds) followed by a 3d uniform random choice.
Karol P.
Karol P. on 14 Feb 2023
Edited: Karol P. on 14 Feb 2023
OK, thank you once again. I think we can cosider the question answered.

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