# Estimating normalization constant with Monte Carlo integration

Be $f(x)$ a function. Suppose that $f(x)$ integrates to a finite value $k$:
$$\int_{-\infty}^{\infty}f(x)dx=k$$
The normalization constant of $f(x)$ is $1/k$. Monte Carlo integration can give an estimate $\hat{k}$ of $k$, so we can estimate the normalization constant simply by taking the reciprocal of this estimate ($1/\hat{k}$). According to the CLT, the Monte Carlo estimator has an asymptotic normal distribution, so a 95% confidence interval for $k$ is simply: $$[\hat{k}-1.96*SE,\hat{k}+1.96*SE]$$ The question is: how can I calculate a confidence interval for the normalization constant? Can I just calculate the reciprocal of the previous confidence interval?

## 1 Answer

Yes: the defining property of a 95% confidence interval $[L, U]$ for an unknown parameter $k$ is that $P(L < k < U) = 95\%$. Since $L < k < U \iff 1/U < 1/k < 1/L$, you are guaranteed that $P(1/U < 1/k < 1/L) = 95\%$, so $[1/U, 1/L]$ is a 95% CI for $1/k$.

As an alternative, you could also use the Delta Method, but I would only do this if I thought it would improve the asymptotic approximation (make the CLT "kick in faster").