# Finding Volume of Monte Carlo Integration

Suppose $\mathbf{X}\in\mathrm{R}^n$ is an $n-$ dimensional random vector having joint Gaussian distribution i.e. $\mathbf{X}\sim\mathcal{N}\left(\boldsymbol\mu,\boldsymbol\Sigma\right)$, where, $\boldsymbol\mu$ and $\boldsymbol\Sigma$ are mean and covariance of $\mathbf{X}$. Now, let us define another random vector $\mathbf{Y} = \mathbf{f}\left(\mathbf{X}\right)$, where $\mathbf{Y}\in\mathrm{R}^n$ and $\mathbf{f}\left(\mathbf{X}\right)$ is a function of $\mathbf{X}$. I want to calculate $I = \int_{-\infty}^{\infty}g\left(\mathbf{Y}\right)dy_1dy_2\cdots dy_{n}$. Here, $g\left(\mathbf{Y}\right)$ is a function of $\mathbf{Y}$ and the integration is performed with respect to all elements of vector $\mathbf{Y}$. In order to perform Monte Carlo (MC) integration, I am using the following steps:

1. Generate MC samples for $\mathbf{X}$ using its mean and covariance. Suppose N samples are generated.

2. Propagate these samples through $\mathbf{Y} = \mathbf{f}\left(\mathbf{X}\right)$ to get N samples of $\mathbf{Y}$.

3. Now, $I' = \frac{1}{N}\sum_{i=1}^Ng\left(\mathbf{Y}\left(i\right)\right)$. where, $\mathbf{Y}\left(i\right)$ is the $i^{th}$ sample of $\mathbf{Y}$.

4. Finally, $I = V*I'$, where $V$ is the volume of integration.

My query is "How to find $V$"?

Is it simply $\left(\max\left(y_{1i}\right) - \min\left(y_{1i}\right)\right)\times \left(\max\left(y_{2i}\right) - \min\left(y_{2i}\right)\right)\times \cdots \times\left(\max\left(y_{ni}\right) - \min\left(y_{ni}\right)\right)$

where, $y_{1i}$ is the $ith$ sample of $y_1$ and maximum and minimum is taken over entire samples.