# Monte Carlo integrate Improper integral

I am trying to use Monte Carlo method to integrate the following improper integral

$${1 \over \sqrt{2\pi}}\int\limits_{-\infty}^{\infty}x^4e^{(-x^2/2)} \ dx$$

Using change of variables, $$y = x^2/2$$, I can transform the integral into gamma function and get the solution as $$3\sqrt{2}\pi$$ (as shown here)

However, the integral still remains improper at $$0$$ to $$\infty$$ and I am not sure how to get that into a proper one so that I can use Monte-Carlo method to arrive at an approximate solution.

Any help is much appreciated.

Thank you

• The point of MC integration is to find a good random variable $Z$ s.t. $E[Z]$ is the desired integral. You can generate standard normal $X$. Generate $X_1,\dots, X_{10000}$ iid following standard normal. Take the mean of fourth power of the corresponding variable. By LLN, it will converge to the relevant mean $E[Z]$ which is the desired integral. You could do substitution by $y=x^2/2$ and consider generating iid exponential random variable as well. You can find the relevant power here as the normal case. Commented Apr 14, 2023 at 21:34
• Thanks @user45765 for your comments. I have just started learning this approach and I am not sure what you meant by "Take the mean of fourth power of the corresponding variable". Can you please explain a bit more? Thank you Commented Apr 14, 2023 at 21:59
• @Yoshiro The mean of the fourth power is $\langle x^4 \rangle$. If your distribution is the normal distribution that is just your integral. Commented Apr 14, 2023 at 22:36

What you have written down is incorrect. $$E[X^4]=3$$. Thus your answer should be close to 3 instead.

I will denote $$p(x)$$ as standard normal density $$1/\sqrt{2\pi}\exp(-x^2/2)$$. Then you want $$\int_{-\infty}^{\infty} x^4p(x)dx$$ which is really $$E[X^4]$$ by definition.

Recall weak law of large number for $$Z_1,\dots, Z_n$$ iid following distribution $$f(Z)$$, then $$\bar{Z}=\frac{Z_1+\dots+Z_n}{n}$$ has for any $$\epsilon>0$$, $$P(|\bar{Z}-E[Z]|>\epsilon)\to 0$$ as $$n\to\infty$$. Here you want to choose $$Z=X^4$$ and by normal MGF, you will see it has $$E[Z]<\infty$$. The only matter is to compute it.

To simulate $$Z_1,\dots, Z_n$$, it suffices to simulate $$X_1,\dots, X_n$$ following $$p(x)$$ density, where $$Z_i=X_i^4$$.

Draw $$X_1,\dots, X_n$$ iid from $$p(x)$$, then compute $$X_i^4=Z_i$$. At last take the mean and apply weak law of large number.

Here is the code for simulation in R.

me_calc=c()
for(i in 1:1000){
x=rnorm(100000,mean = 0,sd=1)
z=x^4
me_calc=c(me_calc,mean(z)-3)
}
hist(me_calc)


Here is the simulation histogram and you can see it is fairly close to the true answer 3.

• It's clear now. Thank you so much for your help. Commented Apr 14, 2023 at 22:45

Explicit Monte Carlo using importance sampling to integrate $$\sqrt{\frac{2}{\pi}}\int\limits_0^\infty x^4e^{\frac{-x^2}{2}}dx$$ Let $$f(x)=e^{-x}dx$$ and $$h(x)=\sqrt{\frac{2}{\pi}} x^4e^{(x\frac{-x^2}{2})}$$.
$$f(x)$$ is the sampling density with the nice property that $$F(x)=\int\limits _0^xf(x)dx$$ is easily invertable . In addition, $$h(x)$$ is bounded.

To get a sample for Monte Carlo. Random number $$R$$ uniform (0.1) Use $$R=F(x)= 1-e^{-x}$$ to get $$x=-ln(1-R)$$. (Note $$1-R$$ is uniform). Sample value is $$h(x)$$

Quota sampling can be used to greatly reduce variance.