MATLAB: Is that possible to take numerical integration over a semi-definite integral

integrationnumerical integration

Suppose one is going to calculate an integral over, say, [s, 5]. Although s is regarded a constant for a moment, but later one has to take another integration over s. The function inside the integral is very complicated and so one will have to use numerical techniques. In sum, I need a function of variable s from the second integral, so I can take another integral over it. But, as I know, numerical integration makes sense when the limits are numbers, but here I have a parameter (s).

Best Answer

If a symbolic solution cannot be found for the inner integral, you can use the matlab function ‘integral2’ for your task. If you read the instructions for it at:

 https://www.mathworks.com/help/matlab/ref/integral2.html,

you will see that it allows the inner integral limits to be functions of the outer variable rather than fixed numerical values - in your case the lower inner limit is s itself. Hence, even though your integrand is a function only of s’, it should still arrive at a solution for the double integral.

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MATLAB: How to do integration

Please read help about functions MATLAB:

quad, quad2d, dblquad, quadgk, quadl, quadv, trapz, triplequad,int

MATLAB: How to make covariance matrix positive semi-definite (PSD)

No, there is not a way. At least there is no constructive, unambiguous, intelligent way.

For example, consider the covariance matrix that arises from

X = repmat(rand(10,1),1,2);
C = cov(X);

Which column causes it to be not positive definite? Column 1 or column 2? What about column 2 makes it more a factor in that zero eigenvalue? I could as easily argue for column 1.

Or, how about this one:

X = rand(10,2);
X = [X,-mean(X,2)];
C = cov(X);

Here, I can delete any of the three columns and end up with a positive definite result, and each column is as "important" in contributing to the zero eigenvalue.

If you wish, I can keep going. How about this one?

X = randn(10,11);
C = cov(X);

Again, each column is as equally random as any other. And since they were randomly generated, we can write any column as a linear combination of the remaining columns. With probability essentially 1, there will be no zero coefficients employed in that linear combination. So which column is the offender? And if you say the last column, then I'll just randomly permute the columns and get a different answer. So effectively, your answer would be to just choose a random column.

Just use a good tool that will yield a positive definite matrix, and do so efficiently. nearestSPD is such a tool.

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