3
$\begingroup$

from https://arxiv.org/abs/1401.0118

If we have a function $J(X,Y)$ of two random variables $X$ and $Y$ and we want to compute the expectation $\mathbb E_{p(X,Y)}[J(X,Y)]$.

We define $\hat J(X)= \mathbb E_{p(Y)}[J(X,Y)\mid X]$.

Note that: $$\mathbb E_{p(X,Y)}[J(X,Y)] = \mathbb E_{p(X)} [\hat J(X)]$$

So we can use $\hat J(X)$ instead of $J(X, Y)$ in a Monte-Carlo Estimate.

For the variance holds (variance reduction) $$ var(\hat J(X)) = var(J(X,Y)) - \mathcal E[(J(X,Y) - \hat J(X))^2] $$

Here I have one questions:

  • How can we prove that (variance reduction)? In the paper, no proof is given.

Here is one proof of this question (from this link):

The variance reduction follows from the the law of total variance. Suppose that $W,Z$ are two random variables, then it follows that $$ \mathbb{V}(W)=\mathbb{V}(\mathbb{E}(W\vert Z))+\mathbb{E}(\mathbb{V}(W\vert Z)) $$ then, replace $W$ by $J(X,Y)$ and $\mathbb{E}(W\vert Z)$ by $\hat{J}(X)$ and we obtain: $$ \mathbb{V}(J(X,Y))=\mathbb{V}(\hat{J}(X))+\mathbb{E}(\mathbb{V}(J(X,Y)\vert X)) $$ Notice that the second summand on the right hand side is given by $$\mathbb{V}(J(X,Y)\vert X)=\mathbb{E}(J(X,Y)^2\vert X)-(\mathbb{E}J(X,Y)\vert X)^2=\mathbb{E}(J(X,Y)^2\vert X)-\hat{J}(X)^2 $$ plug into the ANOVA identity, solve with respect to $\mathbb{V}(\hat{J}(X)$ to obtain $$ \mathbb{V}(\hat{J}(X))=\mathbb{V}(J(X,Y))-\left(\mathbb{E}(J(X,Y)^2)-\mathbb{E}(\hat{J}(X)^2)\right)=\mathbb{V}(J(X,Y))-\mathbb{E}\left(\left(J(X,Y)-\hat{J}(X)\right)^2\right) $$ as desired.

======================== end proof ==========================

But the can't follow the idea of the last line: $$ \mathbb{V}(J(X,Y))-\left(\mathbb{E}(J(X,Y)^2)-\mathbb{E}(\hat{J}(X)^2)\right)=\mathbb{V}(J(X,Y))-\mathbb{E}\left(\left(J(X,Y)-\hat{J}(X)\right)^2\right) $$ why the following equality holds? $$ \mathbb{E}(J(X,Y)^2)-\mathbb{E}(\hat{J}(X)^2) = \mathbb{E}\left(\left(J(X,Y)-\hat{J}(X)\right)^2\right) $$

Thanks.

$\endgroup$
1
$\begingroup$

Another form of answer can be the following, based on the proof outlined in the post you've linked:

If you open up the square form: $$\mathbb E[(J(X,Y)-\hat J(X))^2]=\mathbb E[J(X,Y)^2]-2\mathbb E[J(X,Y)\hat J (X)]+\mathbb E[\hat J(X)^2]$$

The middle term can be written as (using Law of Total Expectation):

$$\mathbb E[J(X,Y)\hat J (X)]=\mathbb E[\mathbb E[J(X,Y)\hat J (X)|X]]=\mathbb E[\hat J(X)\mathbb E[J(X,Y)|X]]=\mathbb E[\hat J(X)^2]$$

When substituted, you've the equality.

$\endgroup$
0
$\begingroup$

The Pythagorean decomposition of the variance $$\mathbb{V}(W)=\mathbb{V}(\mathbb{E}(W\vert Z))+\mathbb{E}(\mathbb{V}(W\vert Z))$$ shows that the original variance is a sum of two positive terms, hence $$\mathbb{V}(W)\ge\mathbb{V}(\mathbb{E}(W\vert Z))$$ showing that both versions have the same expectation and that the conditional expectation has a smaller variance.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.