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Pierre-antoine Comby 2019-05-03 14:08:02 +02:00
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9
451-Signal_Image/Cours/chap4.tex
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@ -160,7 +160,10 @@ On remarque que $DA = I_n$.
\subparagraph{Méthode 2} Pour $A^TMA>0$.
\[
J_{MC} = \underbracket{(D(Y-m_B)-\Theta)^TA^TMA(D(Y-m_B)-\theta)}_ {J_1(Y,\theta)} + \underbracket{(Y-m_B)^T(M-D^TA^TMAD)(Y-m_B)}_{J_2(Y)}
\begin{aligned}
J_{MC} &= \underbracket{(D(Y-m_B)-\Theta)^TA^TMA(D(Y-m_B)-\theta)}_ {J_1(Y,\theta)}\\
&+ \underbracket{(Y-m_B)^T(M-D^TA^TMAD)(Y-m_B)}_{J_2(Y)}
\end{aligned}
\]
Alors $\nabla J_{MC} = 0 \implies J_1 = 0 \implies D(Y-m_B) = \hat{\theta}_{MC}$
@ -376,8 +379,8 @@ On considère un cout uniforme.
\begin{defin}
En prenant:
\begin{align*}
E_{\theta|Y}[C(\hat{\theta},\theta)] &= \int_{\R^m}(1-\Pi_{\Delta}(\tilde{\theta}))f_{\theta|Y=y}(\theta)d\theta
&= 1 - \int_{\hat{\theta}-\Delta/2}^{{\hat{\theta}+\Delta/2}}f_{\theta|Y=y}(\theta)d\theta
E_{\theta|Y}[C(\hat{\theta},\theta)] &= \int_{\R^m}(1-\Pi_{\Delta}(\tilde{\theta}))f_{\theta|Y=y}(\theta)d\theta \\
&= 1 - \int_{\hat{\theta}-\Delta/2}^{{\hat{\theta}+\Delta/2}}f_{\theta|Y=y}(\theta)d\theta \\
&\simeq 1- \Delta^nf_{\theta|Y=y}(\hat{\theta})
\end{align*}
Soit \[