zh:notes:exponential_family
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zh:notes:exponential_family [2017/04/24 15:56] 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 |
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| \begin{eqnarray} | \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} | ||
| 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} | ||
| Now, let's first consider the following | Now, let's first consider the following | ||
| - | \begin{eqnarray} | + | \begin{eqnarray} |
| \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}}] | + | & = & \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} | ||
| + | |||
| + | Rearranging Eq. \ref{log-second-derivative}, we obtain | ||
| + | |||
| + | \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})] \mathbb{E}[\vec{u}(\vec{x})^{\mathrm{T}}] | ||
| + | \end{eqnarray} | ||
| + | |||
| + | 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.1493020579.txt.gz · 最后更改: 2017/04/24 15:56 由 pzczxs