Timeline for On the optimal distribution for importance sampling
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2022 at 0:15 | vote | accept | Symbol 1 | ||
| Jan 3, 2022 at 0:15 | comment | added | Symbol 1 | OK. It seems like I did not aware of that $\hat\mu_h$ is different from $\tilde\mu_h$ (not equivalent by an affine transformation), so it now makes sense that different estimators have different $q$. | |
| Jan 2, 2022 at 22:01 | comment | added | fool | in one sentence: optimality is defined with respect to your estimator. Let me know if I'm still not misunderstanding/not answering the question. | |
| Jan 2, 2022 at 21:59 | comment | added | fool | Ah, thank you for the clarification, looks like I totally misunderstood your question. Hopefully I understand now. So for the specific estimator $\hat \mu_f := (1/n)\sum f(X_i)p(X_i)/q(X_i)$, we have the optimal $q(x) \propto f(x)p(x)$. If you change your test function to say $h = 1-f$, then WITH RESPECT TO $\hat\mu_h$, sampling from the distribution proportional to $h$ is optimal. Alternatively, you can sample proportionally to $f(x)p(x)$, and change your estimator $\hat\mu_h$ to $\tilde \mu_h := 1- \hat\mu_f$, then this will also have 0 variance. | |
| Jan 2, 2022 at 21:34 | comment | added | Symbol 1 | Because, although what I really want is $\mathbb E[f(X)]$, I can always claim that I am interested in $\mathbb E[1-f(X)]$. You then proceed to sample an empirical value of $\mathbb E[1-f(X)]$ for me. So depending on what I claim, you will use different $q$ for simulations, some of them must be better than the others. The question is, how do I know if one $q$ is better than another $q$, if they are all optimal up to some affine transformation of $f$? | |
| Jan 2, 2022 at 21:30 | comment | added | fool | @Symbol1 could you explain why you think it shouldn't depend on a transformation? (So I can understand where you're coming from) | |
| Jan 1, 2022 at 12:35 | comment | added | Symbol 1 | I did get that the optimal estimator is different if I rewrite $f$ using an affine transformation. What I didn't get is why the choice of a sampling distribution, $q$, should depend on the arbitrarily affine transformation I applied to $f$? | |
| Dec 31, 2021 at 21:26 | history | edited | fool | CC BY-SA 4.0 |
Typo
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| Dec 31, 2021 at 19:53 | history | answered | fool | CC BY-SA 4.0 |