I'm learning about importance sampling from p139 of this book

which has the following derviation:

enter image description here

What I am confused about is the second step in the derivation, though the rest makes sense to me. I understand that you apply Bayes rule and that derive the bottom fraction by conditioning. What I am stuck on is why the initial integral doesn't go over the entire thing? i.e why dont we have

$$ \int \normalsize{\frac{g(x)p(y_{1:T}|x)p(x)}{\int p(y_{1:T}|x)p(x) dx}} dx$$

My only guess is that you can, because $p(y)$ is independent of $x$, do this: \begin{align*} \int \normalsize{\frac{g(x)p(y_{1:T}|x)p(x)}{p(y_{1:T})}} dx &\stackrel{?}{=} \frac{1}{p(y_{1:T})} \int \normalsize{g(x)p(y_{1:T}|x)p(x)} dx \\ &= \normalsize{\frac{\int g(x)p(y_{1:T}|x)p(x) dx}{\int p(y_{1:T}|x)p(x) dx}} \end{align*}

is this true? If so, it could seem that the follow is true, in general

$$ \int \frac{f(x,y)}{\int g(y|x)p(x) dx} dx = \frac{\int f(x,y) dx}{\int g(y|x)p(x) dx} $$

But its not obvious to me that this is the case. Is there something am I missing or is the final statement true?


1 Answer 1


Indeed $$\int_\mathcal X \frac{f(x,y)}{\underbrace{\int_\mathcal X g(y|x)p(x) \text{d}x}_\text{constant in $x$}} \text{d}x = \frac{\int_\mathcal X f(x,y) \text{d}x}{\int_\mathcal X g(y|x)p(x) \text{d}x}$$ since the integral is a linear operator.


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