Suppose that we want to estimate the integral:

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 $0.14941$?

  • $\begingroup$ The confidence on the first digit is higher than the confidence on the second, which is higher than the confidence on the third, which &tc... $\endgroup$
    – Xi'an
    Apr 15 '16 at 6:38

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.....


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.