3.2.5 Marginal distribution 公式3.71的理解$$ \begin{equation} \be

\begin{equation} \begin{split} & \frac{1}{2} \left[ \Lambda_{bb} \mu_b - \Lambda_{ba} (\mathbf{x}_a - \mu_a) \right]^T \Lambda_{bb}^{-1} \left[ \Lambda_{bb} \mu_b - \Lambda_{ba} (\mathbf{x}_a - \mu_a) \right] \\ & \quad - \frac{1}{2} \mathbf{x}_a^T \Lambda_{aa} \mathbf{x}_a + \mathbf{x}_a^T \left( \Lambda_{aa} \mu_a + \Lambda_{ab} \mu_b \right) + \text{const} \\ & = - \frac{1}{2} \mathbf{x}_a^T \left( \Lambda_{aa} - \Lambda_{ab} \Lambda_{bb}^{-1} \Lambda_{ba} \right) \mathbf{x}_a \\ & \quad + \mathbf{x}_a^T \left( \Lambda_{aa} - \Lambda_{ab} \Lambda_{bb}^{-1} \Lambda_{ba} \right) \mu_a + \text{const} \end{split} \tag{3.71} \end{equation}

第一行里的后面三项 $- \frac{1}{2} \mathbf{x}_a^T \Lambda_{aa} \mathbf{x}_a + \mathbf{x}_a^T \left( \Lambda_{aa} \mu_a + \Lambda_{ab} \mu_b \right) + \text{const}$ 一开始没有理解这是从何而来，仔细看原文

This integration is easily performed by noting that it is the integral over an unnor

malized Gaussian, and so the result will be the reciprocal of the normalization coef-

ficient. We know from the form of the normalized Gaussian given by (3.26) that this

coefficient is independent of the mean and depends only on the determinant of the

covariance matrix. Thus, by completing the square with respect to xb, we can inte

grate out xb so that the only term remaining from the contributions on the left-hand

side of (3.68) that depends on xa is the last term on the right-hand side of (3.68) in

which m is given by (3.69). Combining this term with the remaining terms from

(3.54) that depend on xa, we obtain

尤其是"Combining this term with the remaining terms from (3.54) that depend on xa"

-\frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu}) = \\ -\frac{1}{2} (\mathbf{x}_a - \boldsymbol{\mu}_a)^T \boldsymbol{\Lambda}_{aa} (\mathbf{x}_a - \boldsymbol{\mu}_a) \\ -\frac{1}{2} (\mathbf{x}_a - \boldsymbol{\mu}_a)^T \boldsymbol{\Lambda}_{ab} (\mathbf{x}_b - \boldsymbol{\mu}_b) \\ -\frac{1}{2} (\mathbf{x}_b - \boldsymbol{\mu}_b)^T \boldsymbol{\Lambda}_{ba} (\mathbf{x}_a - \boldsymbol{\mu}_a) \\ -\frac{1}{2} (\mathbf{x}_b - \boldsymbol{\mu}_b)^T \boldsymbol{\Lambda}_{bb} (\mathbf{x}_b - \boldsymbol{\mu}_b). \tag{3.54}

此外我们还需要看一下（3.68)

\begin{equation} -\frac{1}{2} \mathbf{x}_b^T \boldsymbol{\Lambda}_{bb} \mathbf{x}_b + \mathbf{x}_b^T \mathbf{m} = -\frac{1}{2} (\mathbf{x}_b - \boldsymbol{\Lambda}_{bb}^{-1} \mathbf{m})^T \boldsymbol{\Lambda}_{bb} (\mathbf{x}_b - \boldsymbol{\Lambda}_{bb}^{-1} \mathbf{m}) + \frac{1}{2} \mathbf{m}^T \boldsymbol{\Lambda}_{bb}^{-1} \mathbf{m} \tag{3.68} \end{equation}

公式3.68是3.54中关于 $\mathbf{x}_b$ 的二次项以及一次项，二次项可以在积分中直接应用高斯分布得到结果，所以重点是后面一次项，以及3.54中不包含 $\mathbf{x}_b$ 的项目，这些便是 $- \frac{1}{2} \mathbf{x}_a^T \Lambda_{aa} \mathbf{x}_a + \mathbf{x}_a^T \left( \Lambda_{aa} \mu_a + \Lambda_{ab} \mu_b \right) + \text{const}$ 的由来