first claim of Theorem \[t1\]. The proof of the second claim is more elaborate.

The results in the preceding paragraph also show that the error $e$ in the working model is such that ${{\mathbb E}}[e\|x]$ is approximately zero and $\operatorname{Var}[e\|x]$ is approximately constant, provided that $R \in \mathbb U$: We first re-write the error $e$ in a convenient form. Set $\tilde{\theta} = R'\Sigma^{1/2} \theta$ and $\tilde{M} = R' \Sigma^{1/2} M$. Then it is easy to see that $e =\tilde{\theta}'(I_d-P_{\tilde{M}}) \tilde{z} + \epsilon$ and hence $$\begin{aligned}
 \label{tmp1}
\begin{split}
{{\mathbb E}}[e\|x] &\quad=  
    \quad\tilde{\theta}'(I_d-P_{\tilde{M}}) 
    \Big\{{{\mathbb E}}[\tilde{z}\|P_{\tilde{M}}\tilde{z}]  -
        P_{\tilde{M}}\tilde{z}\Big\} \quad\text{and}
    \\
{{\mathbb E}}[e^2\|x] - s^2 &\quad=\quad\\
&\hspace{-1cm}
    \tilde{\theta}'(I_d-P_{\tilde{M}}) \Big\{
    {{\mathbb E}}[\tilde{z}\tilde{z}'\|P_{\tilde{M}}\tilde{z}]
    - ((I_d - P_{\tilde{M}})+P_{\tilde{M}}\tilde{z}\tilde{z}'P_{\tilde{M}})
    \Big\}
    (I_d-P_{\tilde{M}}) \tilde{\theta};
\end{split}\end{aligned}$$ see also –. Our goal is to show that the expressions in the preceding two displays are approximately zero. To this end, we focus on the expressions in curly brackets and use Cauchy-Schwarz: For each $t>0$, we have $$\begin{aligned}
{{\mathbb P}}( |{{\mathbb E}}[e\|x]| > t) &\quad\leq\quad 
    {{\mathbb P}}\left(\Big\|{{\mathbb E}}[\tilde{z}\|P_{\tilde{M}}\tilde{z}] - 
    P_{\tilde{M}}\tilde{z}\Big\|
    > t / \|(I_d-P_{\tilde{M}})\tilde{\theta}\|\right) \quad\text{and}\\
{{\mathbb P}}( |{{\mathbb E}}[e^2\|x] - s^2| > t) &\quad\leq\quad 
\\ & \hspace{-1cm}
    P\left( \Big\| {{\mathbb E}}[\tilde{z}\tilde{z}'\|P_{\tilde{M}}\tilde{z}] -
        ((I_d - P_{\tilde{M}}) + 
            P_{\tilde{M}}\tilde{z}\tilde{z}'P_{\tilde{M}})
    \Big\| > t/\|(I_d-P_{\tilde{M}})\tilde{\theta}\|^2\right).\end{aligned}$$ Now if $R \in 