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Let $X$ be a r.v. with absolutely continuous distribution and continuous strictly positive density $f: \mathbb{R} \rightarrow [0, \infty)$ and let $g$ a further given continuous density function.

Set $$ Y = \frac{g(X)}{f(X)}. $$ What can we say about the $\mathbb{E}Y$?

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This representation is the principle behind importance sampling.

If $X\sim f(x)$, then [since $Y$ is a transform of $X$, by the law of the unconscious statistician] $$𝔼^Y[Y]=𝔼^X[\frac{g(X)}{f(X)}=∫\frac{g(x)}{f(x)}f(x)\text{d}x$$ which simplifies into $$∫_{\text{supp}(f)}g(x)\,\frac{f(x)}{f(x)}\text{d}x=∫_{\text{supp}(f)}g(x)\,\text{d}x$$ where $\text{supp}(f)$ denotes the support of $f$, that is the closure of the collection of points $x$ when $f(x)>0$. In the event when $\text{supp}(g)\subset\text{supp}(f)$ the integral is equal to one.

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