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Contents
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]I = \displaystyle\int h(x)f(x)\,dx [/math] [math]= \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\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 E_g(h(x)) \rightarrow[/math]the expectation of h(x) with respect to g(x), where [math]\displaystyle \frac{f(x)}{g(x)} [/math] is a weight [math]\displaystyle\beta(x)[/math]. In the case where [math]\displaystyle f \gt g[/math], a greater weight for [math]\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 \hat I [/math] estimated by Importance Sampling could have infinite standard error.
Given [math]\displaystyle I= \int w(x) g(x) dx [/math] [math]= \displaystyle E_g(w(x)) [/math] [math]= \displaystyle \frac{1}{N}\sum_{i=1}^{N} w(x_i) [/math] where [math]\displaystyle w(x)=\frac{f(x)h(x)}{g(x)} [/math].
Obtaining the second moment,
 [math]\displaystyle E[(w(x))^2] [/math]
 [math]\displaystyle = \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]
 [math]\displaystyle = \int \frac{h^2(x)f^2(x)}{g(x)} dx [/math]
We can see that if [math]\displaystyle g(x) \rightarrow 0 [/math], then [math]\displaystyle E[(w(x))^2] \rightarrow \infty [/math]. This occurs if [math]\displaystyle g [/math] has a thinner tail than [math]\displaystyle f [/math] then [math]\frac{h^2(x)f^2(x)}{g(x)} [/math] could be infinitely large. The general idea here is that [math]\frac{f(x)}{g(x)} [/math] should not be large.
Remark 1
It is evident that [math]\displaystyle g(x) [/math] should be chosen such that it has a thicker tail than [math]\displaystyle f(x) [/math]. If [math]\displaystyle f[/math] is large over set [math]\displaystyle A[/math] but [math]\displaystyle g[/math] is small, then [math]\displaystyle \frac{f}{g} [/math] would be large and it would result in a large variance.
Remark 2
It is useful if we can choose [math]\displaystyle g [/math] to be similar to [math]\displaystyle f[/math] in terms of shape. Ideally, the optimal [math]\displaystyle g [/math] should be similar to [math]\displaystyle \left h(x) \rightf(x)[/math], and have a thicker tail. Analytically, we can show that the best [math]\displaystyle g[/math] is the one that would result in a variance that is minimized.
Remark 3
Choose [math]\displaystyle g [/math] such that it is similar to [math]\displaystyle \left h(x) \right f(x) [/math] in terms of shape. That is, we want [math]\displaystyle g \propto \displaystyle \left h(x) \right f(x) [/math]
Theorem (Minimum Variance Choice of [math]\displaystyle g[/math])
The choice of [math]\displaystyle g[/math] that minimizes variance of [math]\hat I[/math] is [math]\displaystyle g^*(x)=\frac{\left h(x) \right f(x)}{\int \left h(s) \right f(s) ds}[/math].
Proof:
We know that [math]\displaystyle w(x)=\frac{f(x)h(x)}{g(x)} [/math]
The variance of [math]\displaystyle w(x) [/math] is
 [math]\displaystyle Var[w(x)] [/math]
 [math]\displaystyle = E[(w(x)^2)]  [E[w(x)]]^2 [/math]
 [math]\displaystyle = \int \left(\frac{f(x)h(x)}{g(x)} \right)^2 g(x) dx  \left[\int \frac{f(x)h(x)}{g(x)}g(x)dx \right]^2 [/math]
 [math]\displaystyle = \int \left(\frac{f(x)h(x)}{g(x)} \right)^2 g(x) dx  \left[\int f(x)h(x) \right]^2 [/math]
As we can see, the second term does not depend on [math]\displaystyle g(x) [/math]. Therefore to minimize [math]\displaystyle Var[w(x)] [/math] we only need to minimize the first term. In doing so we will use Jensen's Inequality.
Aside: Jensen's Inequality
If [math]\displaystyle g [/math] is a convex function ( twice differentiable and [math]\displaystyle g''(x) \geq 0 [/math] ) then [math]\displaystyle g(\alpha x_1 + (1\alpha)x_2) \leq \alpha g(x_1) + (1\alpha) g(x_2)[/math]
Essentially the definition of convexity implies that the line segment between two points on a curve lies above the curve, which can then be generalized to higher dimensions:
 [math]\displaystyle g(\alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n) \leq \alpha_1 g(x_1) + \alpha_2 g(x_2) + ... + \alpha_n g(x_n) [/math] where [math] \alpha_1 + \alpha_2 + ... + \alpha_n = 1 [/math]
Proof (cont)
Using Jensen's Inequality,
 [math]\displaystyle g(E[x]) \leq E[g(x)] [/math] as [math]\displaystyle g(E[x]) = g(p_1 x_1 + ... p_n x_n) \lt = p_1 g(x_1) + ... + p_n g(x_n) = E[g(x)] [/math]
Therefore
 [math]\displaystyle E[(w(x))^2] \geq (E[\left w(x) \right])^2 [/math]
 [math]\displaystyle E[(w(x))^2] \geq \left(\int \left \frac{f(x)h(x)}{g(x)} \right g(x) dx \right)^2 [/math]
and
 [math]\displaystyle \left(\int \left \frac{f(x)h(x)}{g(x)} \right g(x) dx \right)^2 [/math]
 [math]\displaystyle = \left(\int \frac{f(x)\left h(x) \right}{g(x)} g(x) dx \right)^2 [/math]
 [math]\displaystyle = \left(\int \left h(x) \right f(x) dx \right)^2 [/math] since [math]\displaystyle f [/math] and [math]\displaystyle g[/math] are density functions, [math]\displaystyle f, g [/math] cannot be negative.
Thus, this is a lower bound on [math]\displaystyle E[(w(x))^2][/math]. If we replace [math]\displaystyle g^*(x) [/math] into [math]\displaystyle E[g^*(x)][/math], we can see that the result is as we require. Details omitted.
However, this is mostly of theoritical interest. In practice, it is impossible or very difficult to compute [math]\displaystyle g^*[/math].
Note: Jensen's inequality is actually unnecessary here. We just use it to get [math]E[(w(x))^2] \geq (E[w(x)])^2[/math], which could be derived using variance properties: [math]0 \leq Var[w(x)] = E[w(x)^2]  (E[w(x)])^2 = E[(w(x))^2]  (E[w(x)])^2[/math].