I am trying to understand the identity in 2.6 of the following article, in which we have random variables $X \in \mathbb{R}, Y\in \mathbb{R}$ and $Z \in \mathbb{R}^{p-1}$. They define the conditional mean function $$ \mu^* := \mu^*(X,Z) = \mathbb{E}[Y|X,Z] $$ and some estimate of $\mu^*$, denoted $\mu := \mu(X,Z)$. They then claim that \begin{align*} \mathbb{E} [\text{Cov}(\mu^*, \mu|Z)] = \mathbb{E}[Y(\mu - \mathbb{E}[\mu|Z])] \end{align*} which I would like to prove. Using the law of iterated covariances, we have for random variables $A,B,C$ \begin{align*} \text{Cov}(A,B) = \mathbb{E}(\text{Cov}(A,B|C)) - \text{Cov}(\mathbb{E}(A|C), \mathbb{E}(B|C)), \end{align*} and so we can rewrite \begin{align*} \mathbb{E} [\text{Cov}(\mu^*, \mu|Z)] &= \text{Cov}(\mu, \mu^*) + \text{Cov}((\mathbb{E}(\mu^*|Z), \mathbb{E}(\mu|Z))) \end{align*}
Then, considering the terms separately. We have \begin{align*} \text{Cov}(\mu, \mu^*) &= \mathbb{E} ( \mathbb{E}[Y|X,Z] \mu) - \mathbb{E}[\mathbb{E}[Y|X,Z]] \mathbb{E}[\mu]\\ &=\mathbb{E}[Y \mu] - \mathbb{E}[Y]\mathbb{E}[\mu] \end{align*} For the second term, we have \begin{align*} \text{Cov}((\mathbb{E}(\mu^*|Z), \mathbb{E}(\mu|Z))) &=\mathbb{E}( \mathbb{E}[\mu^*|Z] \mathbb{E}[\mu|Z]) - \mathbb{E}[\mathbb{E}[\mu^*|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]]\\ &=\mathbb{E}( \mathbb{E}[Y|Z] \mathbb{E}[\mu|Z]) - \mathbb{E}[\mathbb{E}[Y|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]]\\ &=\mathbb{E}( Y \mathbb{E}[\mu|Z]) - Y\mathbb{E}[\mathbb{E}[\mu|Z]]\\ &=\mathbb{E}( Y \mathbb{E}[\mu|Z]) - Y\mathbb{E}[\mu]\\ \end{align*}
So putting these together yields \begin{align*} \mathbb{E}[Y \mu] - \mathbb{E}[Y]\mathbb{E}[\mu] +\ \mathbb{E}( Y \mathbb{E}[\mu|Z]) - Y\mathbb{E}[\mu] \end{align*} which doesn't give the right answer. Can someone point out where I've gone wrong, or suggest an easier proof?
Your error occurs in simplification of the expression $$\mathbb{E}( \mathbb{E}[Y|Z] \mathbb{E}[\mu|Z]) - \mathbb{E}[\mathbb{E}[Y|Z]]\mathbb{E}[\mathbb{E}[\mu|Z]].$$ The first factor in the rightmost term simplifies to $\mathbb{E}[Y]$, not $Y$. Fixing this should yield the claimed result.
A quicker proof applies the definition of covariance, but conditioned on $Z$: $$ \operatorname{Cov}(\mu,\mu^*\mid Z)=\mathbb{E}(\mu^*\mu\mid Z) - \mathbb{E}(\mu^*\mid Z)\mathbb{E}(\mu\mid Z).$$ Taking expectation of both sides yields $$\mathbb{E}[\operatorname{Cov}(\mu,\mu^*\mid Z)]=\mathbb{E}[Y\mu]-\mathbb{E}[Y\mathbb{E}[\mu\mid Z)],$$ using the same arguments as you've made, but avoiding having to add and subtract $\mathbb{E}[Y]\mathbb{E}[\mu]$.