# Estimating integral by Monte Carlo method

I'm trying to solve the following problem

Use Monte-Carlo methods to find a $$95 \%$$ confidence interval for the following integral: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\left( \frac{1}{2}\left[x^2 - \frac{x(y-1)}{10} +(y-1)^2\right] \right ) dx dy$$

Now, my first thoughts were that this function looks similar to a bivariate normal density. For example, considering a bivariate normal distribution with parameters $$\mu_1 = 0, \mu_2 = \sigma_1 = \sigma_2 = 1$$ and $$\rho = \frac{1}{20}$$ we would have the density

$$f(x,y) = \frac{1}{2 \pi \sqrt{399/400}} \ \exp\left(-\frac{400}{399} \ g(x,y) \right)$$ where $$g(x)$$ is the function being integrated, that is,

$$g(x)= \exp\left( \frac{1}{2}\left[x^2 - \frac{x(y-1)}{10} +(y-1)^2\right] \right ).$$

I have tried to express the integral as some kind of expected value under the bivariate normal distribution, so as to simulate values under that distribution, but somehow I always end up getting stuck. If anyone has any idea on how to proceed, I would appreciate it very much.

Thanks!

• The normal distribution has the shape $\exp(\mathbf{-} x^2)$, not $\exp(x^2)$. Commented Oct 10, 2020 at 5:04
• Agreed, that's why I said similar. I was still hoping there might be a way to write the expression as an expectation of the bivariate normal density. Commented Oct 10, 2020 at 5:44

Unless there is a typo in the text of the exercise (and this is quite likely!), Monte Carlo methods are superfluous for computing this integral since the integrand diverges at $$\pm\infty$$. Hence $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\left( \frac{1}{2}\left[x^2 - \frac{x(y-1)}{10} +(y-1)^2\right] \right )\, \text{d}x\,\text{d}y=+\infty$$