Weird numerical integration behavior
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    Hussein Ammar
      
 il 24 Set 2019
  
    
    
    
    
    Commentato: John D'Errico
      
      
 il 24 Set 2019
            Hello all,
Please check this simple example:
d1 = 8;
d2 = 1.4142;
apPDF = @(x1) (1./sqrt(2*pi)) .* exp(-power((d2.*x1./sqrt(2)) - d1,2) ./ d2^2);
integral(apPDF, 0, Inf)
myVarBoundary = 10^24;
integral(apPDF, 0, myVarBoundary)
My Matlab output:
>> d1 = 8;
d2 = 1.4142;
apPDF = @(x1) (1./sqrt(2*pi)) .* exp(-power((d2.*x1./sqrt(2)) - d1,2) ./ d2^2);
integral(apPDF, 0, Inf)
myVarBoundary = 10^24;
integral(apPDF, 0, myVarBoundary)
ans =
    1.0000
ans =
     0
The correct answer that I should get from the second integration should be one, right?
What is wrong in my integration? Should I replace myVarBoundary with Inf after some value.
Best Regards,
0 Commenti
Risposta accettata
  John D'Errico
      
      
 il 24 Set 2019
        
      Modificato: John D'Errico
      
      
 il 24 Set 2019
  
      No. You should NOT make it inf. Even 1e24 is incredibly large, wild overkill.
As to why you are using integral to compute the area under what loos like a normal PDF is completely beyond me.
d1 = 8;
d2 = 1.4142;
apPDF = @(x1) (1./sqrt(2*pi)) .* exp(-power((d2.*x1./sqrt(2)) - d1,2) ./ d2^2);
You need to understand what integral does when it sees a function with limits that wide. It evaluates the function at a variety of points in the interval. Lets try a few, just for kicks.
apPDF(0)
ans =
      5.04917109360107e-15
>> apPDF(1e24)
ans =
     0
>> apPDF(1e24 / 2)
ans =
     0
>> apPDF(1e24 / 100000)
ans =
     0
>> apPDF(1e24 / 10000000000000)
ans =
     0
Do you see anything significant? Do you see a function that seems to be everywhere zero on the interval [0,1e24]? And even when not identically zero, it deviats from zero on the order of the convergence tolerance. Should it somehow, magically know that in effectively a tiny corner of that HUGE interval, it is non-zero?
apPDF(5)
ans =
       0.00443082846737379
So now if we do this:
integral(apPDF,0,100)
ans =
                         1
What a surprise! It integrates to 1.
3 Commenti
  John D'Errico
      
      
 il 24 Set 2019
				Steve makes an excellent point. Here, you might need to be looking at the ground using the Hubble space telscope though, to get the necessary resolution.
As I showed, the function was zero above x1=100. So out of an interval of width 1e24, it is zero on only the fraction (well) below 100. 1 part in 1e22? 
1e22 is a number almost as large as Avogadro's number. Big.
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