Bruno Luong

Joschua, thanks for the report. I have correct the code and hopefully it resolves these two issues.

Hi Xavier,

You could do this ([lower(i), upper(i)] will compose C):

Aleft=[0 6];
Aright=[2 9];
Bleft=[1 8];
Bright=[7 12];

iitersect = @(i,j) deal(max([Aleft(i) Bleft(j)]),min([Aright(i) Bright(j)]));
[I J]=ndgrid(1:numel(Aleft),1:numel(Bleft));
[left right]=arrayfun(iitersect, I, J);
[lower upper] = MergeBrackets(left, right);

Bruno

Dear Derek,
Great educational writing, I enjoy reading it a lot.
Yes, the complexity of the implemented code is not optimal. It's still perfectly alright for product of less than 10 matrices, which is sufficient in most of the practical use. When I have time I'll take a look at methods based on non-overlapping polygonal partition.

resolved. thanks!

Dear Bruno,

Thank you for your answer. Following is my understanding:

 [X meaneffective sigmaeffective] = TruncatedGaussian(...)
If meaneffective is 5, but I want the mean to be 15.
Then shift=10
X = shift + TruncatedGaussian(sigma, range-shift, ...)

This is how we generate random variables from truncated normal distribution with mean 15, variance sigmaeffective and range [a,b].

But what is the CDF of a random variable from truncated normal distribution?

Let y be the cdf of a truncated normal variable x, then is the following equation right? I am not sure of the mean and variance in normcdf

y = (normcdf(x) - normcdf(a)) ./ (normcdf(b) - normcdf(a))