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Intuition of the Independence Lemma

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asked Apr 15 in BUS 4028F - Financial Economics by Pandy (2,770 points)

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!

1 Answer

+1 vote
answered Apr 16 by ChrisS (400 points)
selected Apr 16 by Pandy
 
Best answer

The Independence Lemma is telling us that the only information we need to determine that particular expectation is the random variable, X. Put a little more simply: it is telling us that the conditional expectation given those criteria is a function which depends only on X.

Note also that, in our definition of g(z), we are evaluating the expectation at a specific point, z, over the whole range of Y.

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 \(X\).

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