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This might be a trivial question, but I've come across a paper where some expectation is said to be taken with respect to some pdf. See example:

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How am I to interpret this, and is there some notation to indicate that it should be taken with respect to that specific pdf? I assume that "with respect to" means that it's the function to use within the product inside the summation, but I'm not sure.

For example I interpret ${E}[\log q(z)]$ as ${E}[\log X]$ where $X$ is a random variable distributed according to q, and then make use of the law of the unconcious statitician, where $g(X)$ is $log(X)$, and similarly with ${E}[\log p(z|x)]$, given that x are observed variables. So given my assumption above, the first case should be with respect to $q(z)$, but in the second case it should be with respect to $p(z | X = x)$, so my assumption seems to fail.

Bonus question: Can they break $\log p(x)$ out of the expectation because it is treated as a constant?

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This isn’t quite right, $\mathbb{E}[\log p(X)]$ is the expected value of the log of the pdf of the random variable. So if you have a random variable, work out what the density of that value for $p$ is and log it, that’s the expectation you’re considering. Here it explicitly says the expectation is with respect to $q$ but often you will see it in a subscript after the $\mathbb{E}$ when it’s not clear ($\mathbb{E}_q$). And by “respect to” we’re saying that’s how the random variable is distributed and so when we integrate we use $q$.

If you’re wondering about why this is a useful quantity you should have a look at Information Theory, the book by MacKay for instance, and in particular the entropy of a probability distribution.

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  • $\begingroup$ So, another super trivial question, what is the difference of saying $q(z)$ and that Z is distributed according to q(z) i.e (Z ~ some pdf that is q(z)), because I'm assuming that the lowercase z is not set? Furthermore, what you're saying is that $E[\log q(z)]=\int \log q(z)*q(z) \mathrm{dz}$ and $E[\log p(z|x)]=\int \log p(z|x)*q(z) \mathrm{dz}$ $\endgroup$ – NotoriousFunk Aug 10 at 8:55
  • $\begingroup$ q(z) is just a pdf, it’s a function, whereas if you say Z is distributed according to q(z) you’re telling me properties about the random variable Z. And yes those two equations are right. $\endgroup$ – XanderJC Aug 10 at 13:40

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