Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm currently trying to evaluate an integral with respect to an MCMC kernel, and I wanted someone else to read what I `proved' to see if it's really correct. It just doesn't feel right.

Suppose I have an MCMC Kernel $K(a'|a)$ with stationary distribution $p(a)$. I want to take the expectation of a function $f(a)$ given a point $a'$ through the kernel like so,

$$ \int_{a} f(a) K(a|a') $$

I use detailed balance,

$$ = \int_{a} f(a) \frac{K(a'|a) p(a)}{p(a')} $$ $$ = \mathbb{E}[f(a)K(a'|a)] \frac{1}{p(a')} $$

Independence (at least I think they should be independent?) $$ = \mathbb{E}[f(a)] \mathbb{E}[K(a'|a)] \frac{1}{p(a')} $$

And detailed balance once again

$$ = \mathbb{E}[f(a)] p(a') \frac{1}{p(a')} $$

Now this just seems incorrect. If I interpret this correctly, it's telling me that if I take a large number of samples from $K(a|a')$ from any starting point $a'$ and integrate a function wrt to it, I might as well have had the original distribution. I must have done something wrong.

share|improve this question
    
Your use of independence is flawed, you can not replace $\mathbb{E}[f(a)K(a'|a)]$ with $\mathbb{E}[f(a)] \mathbb{E}[K(a'|a)]$. Indeed, you are claiming that $g(a)$ and $h(a)$ are independent, for some functions $f$ and $g$... –  Sasha Nov 22 '11 at 20:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.