From 8cae2bbf6d8d92caa7fc0d2fe31ffd76baed28e5 Mon Sep 17 00:00:00 2001 From: Pierre-antoine Comby Date: Fri, 3 May 2019 14:08:02 +0200 Subject: [PATCH] remove overfull hbox --- 451-Signal_Image/Cours/chap4.tex | 9 ++++++--- 1 file changed, 6 insertions(+), 3 deletions(-) diff --git a/451-Signal_Image/Cours/chap4.tex b/451-Signal_Image/Cours/chap4.tex index afc9362..f5ff58a 100644 --- a/451-Signal_Image/Cours/chap4.tex +++ b/451-Signal_Image/Cours/chap4.tex @@ -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 \[