a Deeper Look into Importance Sampling: Difference between revisions
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Obtaining the second moment, | Obtaining the second moment, | ||
<math>\displaystyle E[(w(x))^2] = \int (\frac{h(x)f(x)}{g(x)})^2</math> | <math>\displaystyle E[(w(x))^2] = \int (\frac{h(x)f(x)}{g(x)})^2 g(x) dx</math> | ||
<math>\displaystyle = \int \frac{h^2(x)f^2(x}}{g^2(x)} g(x) dx | |||
<math>\displaystyle = \int \frac{h^2(x)f^2(x}}{g(x)} dx | |||
We can see that if <math>\displaystyle g(x) \rightarrow 0 </math>, then <math>\displaystyle E[(w(x))^2] \rightarrow \infinite </math> | |||
====Jenson's Inequality==== | ====Jenson's Inequality==== | ||
===[[Continuing on]] - June 4, 2009=== | ===[[Continuing on]] - June 4, 2009=== |
Revision as of 22:28, 3 June 2009
A Deeper Look into Importance Sampling - June 2, 2009
From last class, we have determined that an integral can be written in the form [math]\displaystyle{ I = \displaystyle\int h(x)f(x)\,dx }[/math] [math]\displaystyle{ = \displaystyle\int \frac{h(x)f(x)}{g(x)}g(x)\,dx }[/math] We continue our discussion of Importance Sampling here.
Importance Sampling
We can see that the integral [math]\displaystyle{ \displaystyle\int \frac{h(x)f(x)}{g(x)}g(x)\,dx = \int \frac{f(x)}{g(x)}h(x) g(x)\,dx }[/math] is just [math]\displaystyle{ \displaystyle E_g(h(x)) \rightarrow }[/math]the expectation of h(x) with respect to g(x), where [math]\displaystyle{ \displaystyle \frac{f(x)}{g(x)} }[/math] is a weight [math]\displaystyle{ \displaystyle\beta(x) }[/math]. In the case where [math]\displaystyle{ \displaystyle f \gt g }[/math], a greater weight for [math]\displaystyle{ \displaystyle\beta(x) }[/math] will be assigned. Thus, the points with more weight are deemed more important, hence "importance sampling". This can be seen as a variance reduction technique.
Problem
The method of Importance Sampling is simple but can lead to some problems. The [math]\displaystyle{ \displaystyle \hat I }[/math] estimated by Importance Sampling could have infinite standard error.
Given [math]\displaystyle{ \displaystyle I= \int w(x) g(x) dx }[/math] [math]\displaystyle{ = \displaystyle E_g(w(x)) }[/math] [math]\displaystyle{ = \displaystyle \frac{1}{N}\sum_{i=1}^{N} w(x_i) }[/math] where [math]\displaystyle{ \displaystyle w(x)=\frac{f(x)h(x)}{g(x)} }[/math].
Obtaining the second moment, [math]\displaystyle{ \displaystyle E[(w(x))^2] = \int (\frac{h(x)f(x)}{g(x)})^2 g(x) dx }[/math] [math]\displaystyle{ \displaystyle = \int \frac{h^2(x)f^2(x}}{g^2(x)} g(x) dx \lt math\gt \displaystyle = \int \frac{h^2(x)f^2(x}}{g(x)} dx We can see that if \lt math\gt \displaystyle g(x) \rightarrow 0 }[/math], then [math]\displaystyle{ \displaystyle E[(w(x))^2] \rightarrow \infinite }[/math]