Question on Importance Sampling (Monte Carlo method)

[kasraa]

Hi,

Suppose I have N iid samples from a distribution q, and I want to estimate another distributin, p, using those samples (Importance Sampling).

By "standard importance sampling", I mean the case where samples (prior samples. i.e. samples from q) have equal weights ($$w_i = 1/N$$).

In the case of "standard importance sampling", I should perform these steps:

1) compute (unnormalized) weights for those sample according to $$p(s_i)/q(s_i)$$ ($$s_{i}$$ is the i'th sample from q)
2) normalize those weights
3) then an estimate of p would be this:
$$\hat{p} = \sum_{i=1}^N w_{i} \delta(i)$$

(w_i are normalized weights computed at step 2. delta(i) is the Dirac delta function at s_i)


Now consider the case where samples (prior samples, i.e. samples from q) are weighted (differnt weights, and normalized. for example $$u_i$$).

Is it enough (justified) to change the (unnormalized) weights (computed at step 1) to $$p(s_i)u_{i}/q(s_i)$$?
(multiplying prior weights and "standard importance sampling" weights together?)


Thanks in advance.

 

[bpet]

... Is it enough (justified) to change the (unnormalized) weights (computed at step 1) to $$p(s_i)u_{i}/q(s_i)$$?
(multiplying prior weights and "standard importance sampling" weights together?)

Possibly not, because any Monte Carlo simulation (including importance sampling) is essentially based on approximating the (cumulative) distribution by the empirical distribution, i.e.

$$P(x) = Prob[X\le x] = E_P[I[X\le x]] \approx \frac{1}{N}\sum_{i=1}^N I[X_i\le x]$$

where I is the Boolean indicator function and the $$X_i$$ are taken from distribution P. To change the weights from (1/N) to other numbers you'd need to change the sampling method to ensure that the "weighted" empirical distribution remains a good approximation to the CDF.

However if you do find a way to overcome that, the new importance sampling formula would easily follow from the change of measure formula, with

$$P(x) = E_P[I[X\le x]] = E_Q\left[I[X\le x]\frac{dP}{dQ}\right] = E_Q\left[I[X\le x]\frac{p(X)}{q(X)}\right] \approx \sum_{i=1}^N w_i I[X_i\le x]\frac{p(X_i)}{q(X_i)}$$

where the $$X_i$$ are samples such that

$$\sum_{i=1}^N w_i I[X_i\le x]$$

closely approximates Q(x).

 

[kasraa]

Thanks.