How to estimate an expectation under expontial random measure by Importance sampling ? Thanks

Question

Say I have to estimate an expectation of a function $X^{0.9}$ under exponential random measure(with density $f$), i.e., $$E_f(X^{0.9})= \int_0^ \infty x^{0.9} e^{-x} dx $$by importance sampling. The exponential tilted measure could be easily derived as $ f_{\theta }(x)={(1-\theta)e^{({1-\theta})x}} $. The optimal tilting parameter would be found by solute the equation

$$ \nabla E_{f_{ \theta }} [(X^{0.9}e^{-\theta X+ \psi (\theta)})^2] = \nabla E_f [X^{1.8}e^{-\theta X+ \psi (\theta)}]=0$$ where $\psi (\theta) = ln \Psi (\theta)$ is the cumulant function, and the moment generating function of X $\Psi (\theta) = \int f(x) e^{\theta x} dx = \int e^{-x} e^{\theta x} dx = \frac{1}{\theta-1}$. Therefore, $\theta^*$ is the root of the following nonlinear equation, $$\nabla \psi(\theta)= \frac{E_f [X^{1.8} X e^{-\theta X}]} {E_f [X^{1.8} e^{-\theta X}]} $$. However, we still need to evaluate two expectations which must be calculated via crude Monte Carlo method.In this case, I do not think importance sampling is superior to crude MC like $E_f(X^{0.9})= \int_0^ \infty x^{0.9} e^{-x} dx \sim \frac{1}{N} \sum_{n=1}^N X_n^{0.9} $ under directly sampling Exponential random variable X.

My questions are:

1. Is my approach correct ?

It would be very kind if someone can comment on determination of optimal tilting parameter.

2. Is this importance sampling for this integral better than crude MC method?

Thanks.

Answer 1

Is the goal to use importance sampling or to evaluate the integral? If the latter, I think numerical integration in R might work better:

integrand = function(x){x^.9*exp(-x)}
integrate(integrand, 0, Inf)
## 0.9617659 with absolute error < 4.2e-05

Not surprising because $\int_0^\infty xe^{-x}\,dx = 1.$

Crude MC sampling (with error estimate):

m = 10^7;  y = rexp(m); mean(y^.9)
## 0.9618752
2*sd(y^.9)/sqrt(m)
## 0.0005483819

Not sure what accuracy you need or what measure of efficiency you're using.