Derivation of squared coefficient of variation (Importance Sampling)

Question

Text: Computational Statistics 2E by Givens and Hoetings

Section: 6.3.2.3 Weight Degeneracy, Rejuvenation, and Effective Sample Size

I am having trouble following another result in the text. Below is a screenshot.

I am not sure on how $X$ is to be generated. I believe there are two options: 1) $X$ is generated using Sampling Importance Resampling, and 2) $X$ is generated using Sequential Importance Sampling.

I have ruled out option 2 because there is no notion of time in the equations (i.e. no subscript for $t$).

So I am assuming that there is a sample $X_1, \ldots, X_n$ ~ (approximately) iid from $f$ (the target). Let $g$ be an envelope for $f$.

Then the standardized importance weights are defined as $w(X) = \frac{f(x)/g(x)}{\sum_{i=1}^n f(x_i)/g(x_i)}$.

The authors are claiming that $E(w(X)) = \frac1n$, but this is not clear to me how to arrive there.

I have managed to put together something that is possibly sensible.

We know, from the definition of $w(X)$, that $\sum_{i=1}^n w(X_i) = 1$.

Then, $E \left( \sum_{i=1}^n w(X_i) \right) = 1$.

Since $E \left( \sum_{i=1}^n w(X_i) \right) = \sum_{i=1}^n E \left( w(X_i) \right)$ and there is and (approximate) iid sample we get that $$1 = \sum_{i=1}^n E \left( w(X_i) \right) = n \cdot E \left( w(X_i) \right)$$

My issue is that the authors don't explicitly state how $X$ is generated. Any insights would be helpful.

Answer 1

It turns out that my idea was in the right direction, however my implementation was incorrect. The solution is as follows:

Suppose $f$ is the target distribution and $g$ is the envelope for $f$. Using the SIR method sample $$X_1, \ldots, X_n \overset{\text{iid}}{\sim} g,$$ with importance weights defined by $w^*(X) := w_i^* = \frac{f(x_i)}{g(x_i)}$. Define the standardized importance weights by $w(X) := w_i = \frac{w_i^*}{ \sum_{i=1}^n w_i^* }.$

Then it follows that $\sum_{i = 1}^n w_i = 1,$ and hence we have that $E \left( \sum_{i = 1}^n w_i \right) = \sum_{i = 1}^n E(w_i)= 1$.

Now, although each $w_i$ is NOT independent, they are identically distributed. Hence, $\sum_{i = 1}^n E(w_i) = n E(w_i) = 1$. Therefore we obtain the desired result $$E(w_i) = \frac1n$$