Question about accuracy in Monte Carlo integration

Question

Suppose that we want to estimate the integral:
$$\psi=\int_{a}^{b}h(x)dx.$$

Let $\hat{\psi}$ be the Monte Carlo estimator. As far as I know, if we desire an accuracy up to the fourth decimal, we need to have:
$$2\sqrt{\hat{var}(\hat{\psi})} < 10^{-4}\;\;\;\;\;\;\;\;[1]$$
where $\hat{var}(\hat{\psi})$ is the estimated variance of the Monte Carlo estimator. This a consequence of the asymptotic normality of the Monte Carlo estimator $\hat{\psi}$ according to the CLT, although I don't understand the link between this fact and the accuracy rule.

If [1] is satisfied, does it mean that the first, the second, the third and the fourth decimals are exact? For instance, if we get $\hat{\psi}= 0.149415$, can we say that the first digits of the integral are exactly $0.14941$?

Answer 1

You want to estimate $\psi = \int_a^b h(x) \, dx$, using Monte Carlo integration. This means that samples $X_1, X_2, \dots X_n$ were obtained and an appropriate estimator $\hat{\psi}$ was constructed. The CLT then states that

$$\sqrt{n}(\hat{\psi} - \psi) \overset{d}{\to}N(0, nVar(\hat{\psi})) $$

If you desire an accuracy of say $\delta$, that means that you want $\hat{\psi} - \psi < \delta$. But of course, $\hat{\psi}$ is a random quantity and will take different values when different Monte Carlo samples are obtained. Thus guarantying $\hat{\psi} - \psi < \delta$ is impossible for any simulation. However, a way to be fairly confident is to made a confidence interval for $\psi$ around $\hat{\psi}$ that is of size $\delta$.

For example a $95\%$ confidence interval for $\psi$ is $\left(\hat{\psi} - 2\sqrt{Var(\hat{\psi})}, \hat{\psi} + 2\sqrt{Var(\hat{\psi})} \right)$. This interval has half width $2\sqrt{Var(\hat{\psi})}$ and thus accuracy is established when

$$2\sqrt{Var(\hat{\psi})} < \delta. $$

A way to interpret this confidence interval is that if you repeated this experiment many times, on an average 95% of the intervals will contain the true $\psi$. Thus if your $\delta$ is $10^{-4}$ and $\hat{\psi} = 0.149415$, then you are 95% confident that the true $\psi$ has the first four digits $.1494$. You being 95% confident means that if you repeated the Monte Carlo experiment many times, then 95% of them will have $\hat{\psi}$ of the form .1494.....