Expectation of ratios of probability density functions

Question

I'm trying to solve/simply the expression below:-

$\large \mathbb{E_{x \sim b(x)}} B\ [log\left(1 - \frac{A\ a(x)}{2\ c(x)}\right)]$,

or

$B \large \int_{x}b(x)log\left(1 - \frac{A\ a(x)}{2\ c(x)}\right)dx$

where

• $A + B = 1, 0 \le A\le 1 \text{ and } 0 \le B\le 1$
• $a, b\text{ and } c\text { are probability density functions. So } \int_x a(x) = \int_x b(x) = \int_x c(x) = 1.0$
• And most importantly: $\textbf{c(x) = A a(x) + B b(x)}$, or essentially $c$ is the finite mixture of $a$ and $b$, where A and B represents proportions or weights for pdf $a$ and $b$.

My actual problem is really really big (which I've almost solved already), so I don't think explaining all of that would be easy here. I just want to get rid of the integral in the above expression, so that I can continue solving the rest of my problem.

It would be much better if the integral can be transformed in the form of KL/JS divergence, or any other distance metric.

Thanks for your time

Answer 1

I hope the following might be what you want.

$$B\int_{x}b(x)log(1-\frac{Aa(x)}{c(x)})dx = B\int_{x}b(x)log(\frac{c(x)-Aa(x)}{c(x)})dx$$

$$=B\int_{x}b(x)log(\frac{Bb(x)}{c(x)})dx = B\int_{x}b(x)[log(B)+log(\frac{b(x)}{c(x)})]dx$$

$$=Blog(B)+B\int_{x}b(x)log(\frac{b(x)}{c(x)})dx= B \ log(B)+B\times KL(b||c)$$

In the second line I used your third bullet property $c(x)=Aa(x)+Bb(x).$