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? $$ \mathbb{E}(J(X,Y)^2)-\mathbb{E}(\hat{J}(X)^2) = \mathbb{E}\left(\left(J(X,Y)-\hat{J}(X)\right)^2\right) $$



2 Answers 2


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.


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.


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