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The context of this question is ordinary least squares. $X$ denotes the design matrix.

I would like a proof of the claim – or a corrected version thereof – made in this other question that the exogeneity assumption $E[\epsilon|X] = 0$ implies that $E\!\left[h(X)^T\epsilon\right] = 0$ for any function $h$ on $X$. (By "any function" I guess what is meant is an arbitrary function $\mathbb{R}^{n\times p}\to \mathbb{R}^{n\times p}$, where $n\times p$ is the shape of the matrix $X$.)

The claim for the special case $h(X)=X$ says that the errors are uncorrelated with the regressors, which we can be prove using the expected value of the product of two random variables:

\begin{align} E\!\left[X^T\epsilon\right] &= E_X\!\left[\vphantom{\sum}X^T E[\epsilon|X]\right] \\ &= E_X\!\left[X^T 0\right] \\ &= 0 \end{align}

However, if we try the same proof for the general case, then we seem to run into trouble:

\begin{align} E\!\left[h(X)^T\epsilon\right] &= E_{h(X)}\!\left[\vphantom{\sum}h(X)^T E[\epsilon|h(X)]\right] \\ &\overset{?}{=} E_{h(X)}\!\left[\vphantom{\sum}h(X)^T E[\epsilon|X]\right] \\ &= E_{h(X)}[h(X)^T 0] \\ &= 0 \end{align}

Does the equality $E[\epsilon|h(X)] = E[\epsilon|X]$ hold if and only if $h$ is injective?

dwolfeu
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2 Answers2

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This result is a consequence of the law of iterated expectation. Applying that law for any function $h$ gives:

$$\begin{align} \mathbb{E}(h(X)^\text{T} \epsilon) &= \mathbb{E} \Big( \mathbb{E}(h(X)^\text{T} \epsilon | X ) \Big) \\[6pt] &= \mathbb{E} \Big( h(X)^\text{T} \mathbb{E}(\epsilon | X) \Big) \\[6pt] &= \mathbb{E} \Big( h(X)^\text{T} \mathbf{0} \Big) \\[6pt] &= \mathbb{E} (\mathbf{0}) = \mathbf{0}. \\[6pt] \\[6pt] \end{align}$$

Ben
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$E[h(X)^T\epsilon] = 0$ holds for any $h(X)$ because $h(X)$ can be treated as a constant when conditional on $X$. Using law of iterated expectations:

$E[h(X)^T\epsilon] = E[E[h(X)^T\epsilon|X]] = E[h(X)^TE[\epsilon|X]] = 0$

RevealedPreference
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