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Can't integrate the subtraction of two Gaussian CDFs

Posted 11 years ago
3 Replies

Thank you!

I was getting crazy with this error! Now it works perfectly!

Now it works, specify d in the integration like d_?NumericQ forcing it into numeralization

meaningful integration

see the notebook for a more conventional treatment of R and L and a more explicit treatment of variables, too. Gosh!

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POSTED BY: Udo Krause

This is extraordinarily dull, the result is not the integral, but the length of the integration interval, if one integrates {d, 0.01, 4.2} - guess what - the result is 4.2 - 0.0.1 = 4.19 and so on

In[116]:= NIntegrate[
 Product[If[R[priors, sigma, d, c] < L[priors, sigma, d, c], 
   0, (CDF[NormalDistribution[(c - 1) d, sigma], 
      R[priors, sigma, d, c]] - 
     CDF[NormalDistribution[(c - 1) d, sigma], 
      L[priors, sigma, d, c]])], {c, 1, k}], {d, 0.01, 4.2}, 
 PrecisionGoal -> 30, WorkingPrecision -> 60]

Out[116]= 4.19000000000000017742751712290782961645163595676422119140625

In[117]:= NIntegrate[
 Product[If[R[priors, sigma, d, c] < L[priors, sigma, d, c], 
   0, (CDF[NormalDistribution[(c - 1) d, sigma], 
      R[priors, sigma, d, c]] - 
     CDF[NormalDistribution[(c - 1) d, sigma], 
      L[priors, sigma, d, c]])], {c, 1, k}], {d, 0.01, 2.7}, 
 PrecisionGoal -> 30, WorkingPrecision -> 60]

Out[117]= 2.69000000000000017742751712290782961645163595676422119140625

because the max value of the CDF is 1 that means that somehow under integration the functions do not evaluate or have always their maximum value and so the result of integration is just 1 * (dmax - dmin).

Your usage of the Do is at least strange. It could be rewarding to make variables explicit - the dependency from e is working only through the priors table ...

POSTED BY: Udo Krause
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