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

• trelek2
In summary: The variance of the sample mean is then 1/n, and the variance of the sample sum is n/n or 1. The central limit theorem says the distribution of the sample sum approaches normality, but your sample sum isn't approaching anything.In summary, the Monte Carlo integration method may not work for functions that are not square integrable, such as 1/sqrt(x) and exp(-x^2). This is because the integral of such functions may be infinite, resulting in an infinite variance of the estimator and unreliable error estimates.
trelek2
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?

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.

Last edited:

## What is Monte Carlo Integration?

Monte Carlo Integration is a numerical method for approximating the value of a definite integral. It involves randomly sampling points within a specific range and using the average of their function values to estimate the integral.

## How does Monte Carlo Integration work?

Monte Carlo Integration uses random sampling to generate points within the range of the definite integral. These points are then used to calculate the average function value, which is then multiplied by the range of the integral to estimate the integral value.

## What is the error estimate reliability for non-square integrables?

The error estimate reliability for non-square integrables refers to the accuracy of the estimated integral value using Monte Carlo Integration when the function being integrated is not a perfect square. This accuracy is affected by the number of sample points and the range of the integral.

## How do you calculate the error estimate for Monte Carlo Integration?

The error estimate for Monte Carlo Integration can be calculated by taking the standard deviation of the function values at the sampled points and dividing it by the square root of the number of samples. This value can then be multiplied by a constant, usually referred to as the confidence level, to determine the estimated error.

## What are the limitations of Monte Carlo Integration?

Monte Carlo Integration can be computationally expensive, especially for higher dimensional integrals. It also relies on random sampling, so the accuracy of the estimated integral value is dependent on the number of sample points and the range of the integral. Additionally, it may not work well for functions with a high degree of variability or sharp changes in the interval being integrated.

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