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