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From Wikipedia

Let $X:\Omega\to \mathbb{R}$ be a random variable in some probability space $(\Omega,\mathcal{F},P)$. The basic idea of importance sampling is to change the probability $P$ so that the estimation of $E[X;P]$ is easier. Choose a random variable $L\geq 0$ such that $E[L;P]=1$ and that $P$-almost everywhere $L(\omega)\neq 0$. The variate $L$ defines another probability $P^{(L)}=L\, P$ that satisfies $$ \mathbf{E}[X;P] = \mathbf{E}\left[\frac{X}{L};P^{(L)}\right]. $$

I was wondering if $P^{(L)}$ is a probability measure on $\mathbb{R}$ induced by $L$ from $\Omega$?

How shall $\mathbf{E}\left[\frac{X}{L};P^{(L)}\right]$ be understood as an integral?

Why is it true that $ \mathbf{E}[X;P] = \mathbf{E}\left[\frac{X}{L};P^{(L)}\right]$?

Thanks!

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1 Answer 1

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The probability measure $P^{(L)}$ is just the probability measure that has density $L$ with respect to $P$, i.e. $P^{(L)}(A)=\int_A L\;\mathrm{d}P$ for $A\in\mathcal{F}$. The result here is actually trivial when writing it as integrals, because $E\left[\tfrac{X}{L};P^{(L)}\right]$ is nothing but the integral $\int_\Omega \tfrac{X}{L}\mathrm{d}P^{(L)}$. By applying how to integrate with respect to a measure that has density we obtain the equality:

$$ \int_\Omega \tfrac{X}{L}\mathrm{d}P^{(L)}=\int_\Omega\tfrac{X}{L}L\;\mathrm{d}P=\int_\Omega X\;\mathrm{d}P. $$

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Thanks, Stefan! A related question: I saw there are two versions of importance sampling: the one in Wikipedia and the one in Rubinstein's book. I personally think they are not equivalent. What do you think about their relations? Shall I post it a new question, so that it may allow longer replies than comments? –  Tim Dec 9 '12 at 19:58
    
I think you should post it as another question, because I can't quite figure out if they are equivalent. By a first look, it seems that the Rubinstein version is this just with random variables which admits densities wrt. the Lebesgue measure. But on the other hand, the Rubinstein version involves two random variables $X$ and $Y$ whereas this only considers one. –  Stefan Hansen Dec 10 '12 at 7:18

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