Make sense because a integrand is a whole continuous thing and estimator can be sampled and broken down
pw123
How would you find V(Omega)?
CMUScottie
what is V(Omega) here? there is no V(Omega) in the integral one.
keenan
$V(\Omega)$ is the volume of the domain $\Omega$. So, you can think of the Monte Carlo estimate on the right as "compute the average (rather than total) value of $f$ over the whole domain, then multiply the average by the volume to get the integral."
The question of how to compute the volume is itself a good question. In some cases it's obvious: for instance, if you're taking an integral over, say, an interval $[a,b]$ or a hemisphere, then you know this value exactly. Otherwise... you can use Monte Carlo! For instance, you could perform rejection sampling: uniformly sample random points in a bounding box around your domain, count the fraction that land inside the domain, and multiply by the box volume.
Make sense because a integrand is a whole continuous thing and estimator can be sampled and broken down
How would you find V(Omega)?
what is V(Omega) here? there is no V(Omega) in the integral one.
$V(\Omega)$ is the volume of the domain $\Omega$. So, you can think of the Monte Carlo estimate on the right as "compute the average (rather than total) value of $f$ over the whole domain, then multiply the average by the volume to get the integral."
The question of how to compute the volume is itself a good question. In some cases it's obvious: for instance, if you're taking an integral over, say, an interval $[a,b]$ or a hemisphere, then you know this value exactly. Otherwise... you can use Monte Carlo! For instance, you could perform rejection sampling: uniformly sample random points in a bounding box around your domain, count the fraction that land inside the domain, and multiply by the box volume.