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30:00 The most intuitive explaination for stochastic optimization I have ever heard so far.
At 44:43 - why does the score function have expectation of zero?
\begin{equation}\begin{aligned}\mathbb{E}_q [abla_u g(z;u)] &= \mathbb{E}_q [abla_u \log p(x,z) - abla_u \log q(z;u)]\\&= - \mathbb{E}_q [abla_u \log q(z;u)] ~ \textrm{($\log p(x,z)$ is not a function of $u$)}\\&= - \int q(z; u) abla_u \log q(z;u) \\&= - \int abla_u q(z;u)~\textrm{(Log Derivative Trick)}\\&= 0 ~\textrm{($q(z;u)$ is a continuous probability distribution)}\end{aligned}\end{equation}
Chain rule and dominated convergence theorem
At 1:27:30 => I didn't really get how you derive the Auxiliary variational bound. Is there a good source where it's explained more thoroughly?
lo
30:00 The most intuitive explaination for stochastic optimization I have ever heard so far.
At 44:43 - why does the score function have expectation of zero?
\begin{equation}
\begin{aligned}
\mathbb{E}_q [
abla_
u g(z;
u)] &= \mathbb{E}_q [
abla_
u \log p(x,z) -
abla_
u \log q(z;
u)]\\
&= - \mathbb{E}_q [
abla_
u \log q(z;
u)] ~ \textrm{($\log p(x,z)$ is not a function of $
u$)}\\
&= - \int q(z;
u)
abla_
u \log q(z;
u) \\
&= - \int
abla_
u q(z;
u)~\textrm{(Log Derivative Trick)}\\
&= 0 ~\textrm{($q(z;
u)$ is a continuous probability distribution)}
\end{aligned}
\end{equation}
Chain rule and dominated convergence theorem
At 1:27:30 => I didn't really get how you derive the Auxiliary variational bound. Is there a good source where it's explained more thoroughly?
lo