Monte Carlo Integration: Error Estimate Reliability for Non-Square Integrables

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Monte Carlo integration struggles with functions that are not square integrable, such as 1/sqrt(x) on the interval from 0 to 1, because their integral of the absolute value squared diverges, leading to infinite variance in the estimator. This results in unreliable error estimates, as demonstrated by the significant discrepancy between estimated and actual errors when evaluating the integral of 1/sqrt(x). The discussion also touches on the function exp(-x^2), which raises questions about its probability distribution and the occurrence of NaN (not a number) errors in computations. Understanding the properties of these functions is crucial for effective Monte Carlo integration. The challenges highlight the limitations of this method for certain types of functions.
trelek2
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Hi!

I need help with the monte carlo integration: reliability of the error estimate for functions that are not square integrable.

I'm supposed to investigate this topic.*Hence my first question is what is a function that is not square integrable? I found that such a function is 1/sqrt(x) on the interval 0 to 1. Apparently a function is not square integrable if the integral of its absolute value squared is not finite on that integral... I thought the for f(x)= 1/sqrt(x) that will be -1?
Anyway I evaluated the integrals for 1/sqrt(x) from 0 to 1 (which is 2 analytically) for dofferent number of sample points. Indeed the estimated errors are nowhere close the actual errors...

Can anyone explain why does this happen? And why is 1/sqrt(x) not square integrable?
 
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trelek2 said:
Hi!

I need help with the monte carlo integration: reliability of the error estimate for functions that are not square integrable.

I'm supposed to investigate this topic.*Hence my first question is what is a function that is not square integrable? I found that such a function is 1/sqrt(x) on the interval 0 to 1. Apparently a function is not square integrable if the integral of its absolute value squared is not finite on that integral... I thought the for f(x)= 1/sqrt(x) that will be -1?
Anyway I evaluated the integrals for 1/sqrt(x) from 0 to 1 (which is 2 analytically) for dofferent number of sample points. Indeed the estimated errors are nowhere close the actual errors...

Can anyone explain why does this happen? And why is 1/sqrt(x) not square integrable?

The integral of 1/x is infinite. Therefore the variance of your estimator is infinite. That is why Monte Carlo doesn't work.
 


Will that be also true for exp(-x^2)? I get a NaN as the error estimate oddly only in this case.
 


trelek2 said:
Will that be also true for exp(-x^2)? I get a NaN as the error estimate oddly only in this case.
Several questions:

1. What is the domain of x?
2. Exactly what is the prob. distribution function (or prob. density function)?
3. What is NaN (Sodium Nitride?)?
 


In computing, NaN stands for "not a number", which usually means an infinity or undefined value has popped up.
 


If exp(-x^2) is supposed to be your density function (although there will be a constant attached to it to make it integrate to 1), then the variance is known.
exp(-.5x^2)/ √(2π) is the density for the standard normal with mean 0 and variance 1.
 
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