Integration of norminv function

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Mariia Vasileva 2017-4-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/337247-integration-of-norminv-function

评论： Mariia Vasileva 2017-4-28

采纳的回答： Andrew Newell

在 MATLAB Online 中打开

Is there any way to integrate norminv function for big number of dimensions?

For example in case of 3-dimensional integration integration:

fun3=@(x,y,z) exp(-norminv(x,0,1)).*(abs(sin(norminv(y,0,1)))+abs(sin(norminv(z,0,1)))+abs(sin(norminv(x,0,1))));
q3 = integral3(fun3,0,1,0,1,0,1);
vpa(q3)

Matlab returns the following warnings:

Warning: Reached the maximum number of function evaluations (10000). The result passes the global error test. 
> In integral2Calc>integral2t (line 136)
  In integral2Calc (line 9)
  In integral3/innerintegral (line 146)
  In integralCalc/iterateScalarValued (line 314)
  In integralCalc/vadapt (line 132)
  In integralCalc (line 75)
  In integral3 (line 121)
  In ICDFd (line 20)

Can somebody please give me an advise?

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Answer 1

Andrew Newell 2017-4-25

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https://ww2.mathworks.cn/matlabcentral/answers/337247-integration-of-norminv-function#answer_264484

编辑：Andrew Newell 2017-4-25

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The problem is that norminv(x,0,1) goes to -Inf as x goes to zero and Inf as x goes to 1 (and ditto for y and z), so it's hard to integrate accurately. If you can manage with a larger tolerance,

q3 = integral3(fun3,0,1,0,1,0,1,'AbsTol',1e-3);

does not return any warnings (the default tolerance is 1e-10). If you need higher precision, you'll just need to be patient, as the warnings are not fatal. However, I don't know if the precision is actually met under such circumstances.

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Andrew Newell 2017-4-27

编辑：Andrew Newell 2017-4-27

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The source of the error is your attempt to use norminv with symbolic variables. This gives the same errors:

syms a
norminv(a,0,1)

Trying to come up with your own substitute for integral3 is tricky, and probably not worth the effort as integral3 will almost certainly do a better job.

Mariia Vasileva 2017-4-28