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zh:notes:exponential_family [2017/04/24 16:08]
pzczxs [Covariance of $\vec{u}(\vec{x})$]
zh:notes:exponential_family [2022/06/30 11:33] (当前版本)
pzczxs 讨论状态变化了
行 43: 行 43:
 Rearranging,​ and making use again of Eq. \ref{normalization} then gives Rearranging,​ and making use again of Eq. \ref{normalization} then gives
  
-\begin{eqnarray} \label{covariance-middle}+\begin{eqnarray}
   \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) + \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \mathbb{E}[\vec{u}(\vec{x})] + \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \mathbb{E}[\vec{u}(\vec{x})] + \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] & = & 0    \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) + \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \mathbb{E}[\vec{u}(\vec{x})] + \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \mathbb{E}[\vec{u}(\vec{x})] + \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] & = & 0 
 \end{eqnarray} \end{eqnarray}
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 Making use of Eq. \ref{expectation} then gives Making use of Eq. \ref{expectation} then gives
  
-\begin{eqnarray}+\begin{eqnarray} \label{covariance-middle}
   \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - 2 \mathbb{E}[\vec{u}(\vec{x})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}] + \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] & = & 0   \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - 2 \mathbb{E}[\vec{u}(\vec{x})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}] + \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] & = & 0
 \end{eqnarray} \end{eqnarray}
行 58: 行 58:
   \nabla \nabla \ln g(\vec{\eta}) & = & \nabla \frac{\nabla g(\vec{\eta})}{g(\vec{\eta})} \nonumber \\   \nabla \nabla \ln g(\vec{\eta}) & = & \nabla \frac{\nabla g(\vec{\eta})}{g(\vec{\eta})} \nonumber \\
   & = & \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - \left[ \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \right] \left[ \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \right]^{\mathrm{T}} \nonumber \\   & = & \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - \left[ \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \right] \left[ \frac{1}{g(\vec{\eta})} \nabla g(\vec{\eta}) \right]^{\mathrm{T}} \nonumber \\
-  & = & \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] \label{log-second-derivative}+  & = & \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) - \mathbb{E}[\vec{u}(\vec{x})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}] \label{log-second-derivative}
 \end{eqnarray} \end{eqnarray}
  
行 64: 行 64:
  
 \begin{eqnarray} \label{log-second} \begin{eqnarray} \label{log-second}
-  \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) & = & \nabla \nabla \ln g(\vec{\eta}) + \mathbb{E}[\vec{u}(\vec{x}) \vec{u}+  \frac{1}{g(\vec{\eta})} \nabla \nabla g(\vec{\eta}) & = & \nabla \nabla \ln g(\vec{\eta}) + \mathbb{E}[\vec{u}(\vec{x})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}]
 \end{eqnarray} \end{eqnarray}
  
 Inserting Eq. \ref{log-second} into \ref{covariance-middle},​ and then we obtain Inserting Eq. \ref{log-second} into \ref{covariance-middle},​ and then we obtain
  
 +\begin{eqnarray} \label{covariance}
 +  - \nabla \nabla \ln g(\vec{\eta}) & = & \mathbb{E}[\vec{u}(\vec{x}) \vec{u}(\vec{x})^{\mathrm{T}}] - \mathbb{E}[\vec{u}(\vec{x})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}] = \mathrm{cov}[\vec{u}(\vec{x})]
 +\end{eqnarray}
 +
 +~~DISCUSSION:​closed~~
zh/notes/exponential_family.1493021314.txt.gz · 最后更改: 2017/04/24 16:08 由 pzczxs