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


*

*Is my approach correct ? 


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


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


Thanks.
 A: 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.
A: In calculating $\Psi(\theta)$ it seems like you're already assuming you know the $\lambda$-parameter of the distribution you're measuring. If that's the case, then I don't understand why you even bother with estimation?
Instead, let's assume $X \sim Exp(\lambda)$ such that $\Psi(\theta)=\frac{\lambda}{\lambda-\theta}$ and $\psi^{'}(\theta) = \frac{1}{\lambda-\theta}$.
Let $a=1.8$ then we then have to solve
$$\frac{1}{\lambda-\theta}
= \frac{E[X^{a+1}e^{-\theta X}]}{E[X^{a}e^{-\theta X}]}
= \frac{1+a}{\lambda-\theta}
\implies \theta = \frac{a\lambda}{2+a}
.
$$
Of course we don't know $\lambda$, so we have to make an educated guess.
If you guess way off the real value, maybe the tilted sampling approach will be worse, but if you guess somewhere close to the truth, it should help you.
