# Intuition of the Independence Lemma

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asked Apr 15, 2018

The slides say the Independence Lemma (stated below) is an "intuitive result" but I'm struggling to understand the intuition.

The Lemma states:

For $$X, Y, f(X, Y) \in L^{1}(\Omega, \mathcal{F}, \mathbb{P})$$, with $$X \space \mathcal{G}$$-measureable and $$Y$$ independent of $$\mathcal{G}$$, for some sub-sigma-algebra $$\mathcal{G}$$. Define $$g(z)= \mathbb{E}^{\mathbb{P}}\left[f(z, Y) \right]$$ for all $$z \in \mathbb{R}$$. Then: $$\mathbb{E}^{\mathbb{P}}\left[f(X, Y)|\mathcal{G} \right]=g(X)$$

Any help with the intuition would be appreciated!

I am sure that you would have seen the less general case: $$g(X) = \mathbb{E}[XY]$$ ie where $$f(X,Y) = XY$$. Now applying the independence lemma, given that $$X$$ is measurable by $$\mathcal{G}$$ and that $$Y$$ is independent of $$\mathcal{G}$$: $$\mathbb{E}[XY|\mathcal{G}] = X\mathbb{E}[Y|\mathcal{G}]$$ (Measurablity). Now, $$\mathbb{E}[Y|\mathcal{G}] = \mathbb{E}[Y]$$ as $$Y$$ is independent of $$\mathcal{G}$$. Thus: $$\mathbb{E}[XY|\mathcal{G}] = X\mathbb{E}[Y]$$. But $$\mathbb{E}[Y]$$ is a constant, so we can conclude that $$\mathbb{E}[XY|\mathcal{G}]$$ gives a function which only depends on $$X$$.