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Topic: Monte Carlo Integration
Replies: 1   Last Post: Sep 10, 2012 9:48 AM

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Steven Lord

Posts: 17,944
Registered: 12/7/04
Re: Monte Carlo Integration
Posted: Sep 10, 2012 9:48 AM
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"kumar vishwajeet" <kwzeet@gmail.com> wrote in message
news:k2hia8$so5$1@newscl01ah.mathworks.com...
> I am integrating the following function between L and U using dblquad. I
> get "NaN" and a warning of singularity. F =
> real(constantForM.*((x-L).^(A(1)-1).*(y-L).^(A(2)-1).*(U-L-((x-L)+(y-L))).^(A(3)-1)).^2
>
> where, constantForM = 3.378137617443966e-038.
> L = 1e-7
> U = 3.2e5
> A = [3.75 0.25 0.02]
>
> In order to check for singularity, I generated 1e6 points between L and U
> and evaluated F at each of the points. I get a sharp peak(singularity) at
> halfway. But the value at that point is still less than 1. In fact the
> maximum value of F along these points is 0.00116 and minimum is close to
> zero. Then why do I get NaN??


When y = L (assuming you're integrating over the square [L, U] x [L, U] the
middle term in your integrand is 0^(A(2)-1) = 0^(-0.75). That's Inf.
If x is also equal to L, the first term in the product is 0^(2.75) = 0. Then
0*Inf = NaN and once one NaN value creeps into a calculation, it propagates.

> Another thing:
> I used Monte Carlo Integration using the following method:
> 1. Evaluated F at all 1e6 points.
> 2. Found the average of all those values.
> 3. Multiplied by the volume i.e. (U-L)*(U-L).
> 4. I get 119 as the answer.
>
> Which of these two methods is correct??


Your integrand isn't defined in the lower-left corner of your region of
integration, if my assumption about your region of integration is correct.
Therefore I'd say the NaN result you received is correct.

--
Steve Lord
slord@mathworks.com
To contact Technical Support use the Contact Us link on
http://www.mathworks.com




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