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From Rubinstein's Simulation Monte Carlo Method, assume r.v. $X$ has density function $f$, $H$ is a measurable function, and $g$ is another density function. If $g$ dominates $Hf$ in the sense that $g(x) =0$ implies $H(x) f(x) =0$, then $$ E_{X \sim f} H(X) = \int H(x) f(x) dx = \int \frac{H(x) f(x)}{g(x)} g(x) dx = E_{Y \sim g} \frac{H(Y) f(Y)}{g(Y)} $$

  1. I was wondering why $g$ is required to dominate $Hf$? For example, under that condition, $\frac{H(x) f(x)}{g(x)}$ will never be $\frac{a}{0}$ with $a \neq 0$, which is $+\infty$ or $-\infty$, but may be $\frac{0}{0}$, which I think is undefined even for integrand in Lebesgue integral? How does that affect the above equation for importance sampling?

  2. Does a different condition "$g$ being non-zero a.s." suffice for importance sampling? Why not $g$ being non-zero a.s. but $g$ dominating $Hf$? Also see the comments after Stefan Hansen's reply.

Thanks and regards!

1 Answers 1

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Because then $H(x)f(x)\neq 0$ implies $g(x)\neq 0$, and so we are allowed to multiply and divide by $g(x)$ $$ \begin{align*} \int H(x)f(x)\,\mathrm dx&=\int_{\{H(x)f(x)\neq 0\}} H(x)f(x)\,\mathrm dx=\int_{\{H(x)f(x)\neq 0\}}\frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx\\ &=\int_{\{g(x)\neq 0\}}\frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx. \end{align*} $$

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    +1 Thanks! (1) But how is the RHS $\int_{\{g(x)\neq 0\}}\frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx = \int \frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$? (2) Does this single condition "$g$ being non-zero a.s." make the equality in my post hold? Why not $g$ being non-zero a.s. in place of $g$ dominating $Hf$?2012-12-09
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    Note that by $g$ being non-zero a.s., I mean it with respect to the Lebesgue measure.2012-12-09
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    I'm pretty sure that $\int \frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$ means exactly $\int_{\{g(x)\neq 0\}} \frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$ because outside $\{g(x)\neq 0\}$ the integrand is not defined (as you've pointed out yourself).2012-12-10
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    And yes, if $g$ is non-zero a.s., then there is no problem when the integrals are understood to be taken over $\{g(x)\neq 0\}$. I think one wants $g$ to dominate $Hf$ because this is more broad than requiring $g$ to be non-zero a.s.2012-12-10
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    Thanks! (1) In the definition of Lebesgue integral, the integrand can take infinity value, as long as the region where it is infinity has measure zero, and in that case, the definition of Lebesgue integral defines the integral over that region to be zero. But if the region has positive measure, the Lebesgue integral may encounter difficulty and take infinite. That is why I think $g$ is nonzero a.s. will be the right condition, instead of $g$ dominates $Hf$. Let me know what do you think? Thanks!2012-12-10
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    $g$ being non-zero a.s. definitely will do the job, but I still think it is unnecessary. You're right that if $g$ is zero on a set with positive Lebesgue measure, then $\int \frac{H(x)f(x)}{g(x)}\,\mathrm dx$ will be $+\infty$ or maybe even undefined (it depends on $H$). But we are indeed looking at the integral $\int \frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$ and I believe that this means exactly $\int_{\{g(x)\neq 0\}}\frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$. I've encountered similar notation in measure theory such as $\frac{1}{h(x)}1_{\{h(x)\neq 0\}}$ which in principle is undefined when $h(x)=0$.2012-12-10
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    But this is understood to be $\frac{1}{h(x)}$ for $x$ such that $h(x)\neq 0$ and $0$ otherwise.2012-12-10
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    Similarly, when $g$ dominates $Hf$, does $\frac{H(x)f(x)}{g(x)}$ also take $0$ where $g(x)=0$, even though $\frac{H(x)f(x)}{g(x)} = \frac{0}{0}$, so that $\int \frac{H(x)f(x)}{g(x)} dx$ can be well-defined?2012-12-10
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    But that integral $\int \frac{H(x)f(x)}{g(x)\,\mathrm dx$ doesn't appear anywhere in my comment.2012-12-10
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    It appeared in Rubinstein's book though, which is the source of my question. From the discussion here, I guess his requirement of $g$ dominating $Hf$ is not a proper condition for importance sampling.2012-12-10
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    Did $\int \frac{H(x)f(x)}{g(x)}\,\mathrm dx$ appear in the book?2012-12-10
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    If I understand the book correctly, yes, it appears as $E_g \frac{H(X)f(X)}{g(X)} $. See this snapshot http://i.stack.imgur.com/hYF9R.png2012-12-10
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    That is equal to $\int \frac{H(x)f(x)}{g(x)}g(x)\,\mathrm dx$, which must be over the set where $g(x)\neq 0$.2012-12-10
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    Sorry, I missed multiplying $g(x)$ in the integrand. So do you mean that because $g(x)$ appears as the denominator of the integrand, the integral region must be $\{g \neq 0\}$ instead of $\mathbb{R}$ in order to make the integrand well-defined? $E_g\frac{H(X)f(X)}{g(X)}$ also implies the integral region is $\{g \neq 0\}$ not $\mathbb{R}$?2012-12-10
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    Yes, that's what I mean. $E_g[\frac{H(X)f(X)}{g(X)}]$ must mean $\int_{\{\omega: g(X(\omega))\neq 0\}} \frac{h(X)f(X)}{g(X)}\,\mathrm dP$ which in turn equals $\int_{\{g(x)\neq 0\}}\frac{h(x)f(x)}{g(x)}g(x)\,\mathrm dx$.2012-12-10