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Intuition of the Independence Lemma

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asked 6 days ago in BUS 4028F - Financial Economics by Pandy (2,150 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 5 days ago by ChrisS (260 points)
selected 5 days ago 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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