On characterization of MRE estimators

Question

I have some trouble understanding the second equality in the proof of theorem 6;

Answer 1

By S. Catterall;

To understand this proof, think very carefully about the difference between min and argmin. If the proof involved minimising using min then there would be no $\delta_0$ outside (or a minus sign).

Argmin works somewhat differently. The first equality in the proof states that $\hat\delta=\delta'$ (say) where $\delta'\in\Delta$ and $E_0 (L(0,\delta'(X)))=\min_{\delta\in\Delta} E_0 (L(0,\delta (X)))$. (This follows from the definition of argmin.) Because $\delta'\in\Delta$, we can write it as $\delta'=\delta_0 -v'$ where $v'$ is the function associated with the minimiser $\delta'$. We can then write $v'=\arg\min_{v} E_0 (L(0,\delta (X)))$ where we write $\delta$ as $\delta_0 -v$ and remember that minimising over $\delta$ is equivalent to minimising over $v$ and then writing $\delta=\delta_0 -v$. This is why the second equality in the proof is true.