보통 위시 아트 후손의 유도

나는 Normal-Wishart 후부의 파생을 연구하고 있지만 매개 변수 중 하나 (스케일 매트릭스의 후부, 아래쪽 참조)에 붙어 있습니다.

맥락과 완전성을 위해 다음은 모델과 나머지 파생물입니다.

$$\begin{align} x_i &\sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Lambda})\\ \boldsymbol{\mu} &\sim \mathcal{N}(\boldsymbol{\mu_0}, (\kappa_0 \boldsymbol{\Lambda})^{-1})\\ \boldsymbol{\Lambda} &\sim \mathcal{W}(\upsilon_0, \mathbf{W}_0) \end{align}$$

세 가지 요소 각각의 확장 된 형태는 다음과 같습니다 (최대 비례 상수).

• 가능성 :

  $$\begin{align} \mathcal{N}(\mathbf{x}_i &| \boldsymbol{\mu}, \boldsymbol{\Lambda}) \propto\notag\\ &|\boldsymbol{\Lambda}|^{N/2} \exp{\left(-\frac{1}{2}\sum_{i=1}^N \left( \mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i - 2 \boldsymbol{\mu}^T \boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu}^T\boldsymbol{\Lambda}\boldsymbol{\mu}\right) \right)} \end{align}$$

• 일반적인 이전 :

  $$\begin{align} \mathcal{N}(\boldsymbol{\mu} &| (\boldsymbol{\mu}_0, \kappa_0 \boldsymbol{\Lambda})^{-1}) \propto\notag\\ &|\boldsymbol{\Lambda}|^{1/2} \exp{\left(-\frac{1}{2}\left( \boldsymbol{\mu}^T\kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu} - 2 \boldsymbol{\mu}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0} + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\right) \right)} \end{align}$$

• Wishart 이전 :

  $$\begin{align} \mathcal{W}(\boldsymbol{\Lambda} | \upsilon_0, \mathbf{W}_0) \propto |\boldsymbol{\Lambda}|^{\frac{\upsilon_0-D-1}{2}} \exp{\left(-\frac{1}{2} tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda})\right)} \end{align}$$

우리는 다음 과 같이 분해 될 수있는 사후 Normal-Wishart ( ) 뿐만 아니라 :N ( μ | μ , κ " Λ ) W ( Λ | υ ' , W '$\mathbf{\mu}, \boldsymbol{\Lambda} | \boldsymbol{\mu}', \kappa', \upsilon', \mathbf{W}'$$\mathcal{N}(\boldsymbol{\mu} | \boldsymbol{\mu}, \kappa' \boldsymbol{\Lambda}) \mathcal{W}(\boldsymbol{\Lambda} | \upsilon', \mathbf{W}')$

자유의$\upsilon'$

가능성의 첫 번째 요소와 Wishart를 병합함으로써 우리는 뒤에서 Wishart 요소의 첫 번째 요소를 얻습니다. 따라서 후부의 첫 번째 매개 변수가 있습니다 : υ'=υ0+

$$\begin{align} |\boldsymbol{\Lambda}|^{\frac{\upsilon_0+N-D-1}{2}} \end{align}$$ $$\begin{align} \upsilon' = \upsilon_0 +N \end{align}$$

스케일 팩터$\kappa'$

및 로 둘러 싸인 요소를 식별하여 이전 이 다음 가능성에 의해 업데이트되는 사람을 찾습니다 . 따라서 μ κ 0 Λ μ $\boldsymbol{\mu}^T$$\boldsymbol{\mu}$$\kappa_0 \boldsymbol{\Lambda}$ κ ' = κ 0 +

$$\begin{align} \boldsymbol{\mu}^T\left((\kappa_0 + N) \boldsymbol{\Lambda}\right)\boldsymbol{\mu} \end{align}$$ $$\begin{align} \kappa' = \kappa_0 + N \end{align}$$

평균$\boldsymbol{\mu}'$

세 번째 매개 변수는 내부의 내용을 식별하는 데 있습니다 : 따라서 세 번째 매개 변수가 있습니다 : 2 μ T ( Λ N ¯ x + κ 0 Λ μ 0$2\boldsymbol{\mu}^T...$μ ' =

$$\begin{align} 2\boldsymbol{\mu}^T\left(\boldsymbol{\Lambda} N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu}_0}\right) &= 2\boldsymbol{\mu}^T \kappa'\boldsymbol{\Lambda} \boldsymbol{\mu}'\\ \left(\boldsymbol{\Lambda} N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu}_0}\right) &= \kappa'\boldsymbol{\Lambda} \boldsymbol{\mu}'\\ \left(N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\mu}_0}\right) &= \kappa' \boldsymbol{\mu}' \end{align}$$ $$\begin{align} \boldsymbol{\mu}' = \frac{1}{k'}(N\mathbf{\overline{x}} + \kappa_0 \boldsymbol{\mu}_0) \end{align}$$

스케일 매트릭스$\boldsymbol{W}'$

그리고 네 번째 매개 변수는 나머지 매개 변수에 대한 작업에서 비롯됩니다.

$$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) &= tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda}) + \sum_{i=1}^{N}\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\\ &= tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda}) + \sum_{i=1}^{N}tr(\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i) + tr(\boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0})\\ &= tr\bigg(\mathbf{W}_0^{-1} \boldsymbol{\Lambda} + \sum_{i=1}^{N}\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\bigg) \end{align}$$

여기에서 계속 진행하고 (지금까지 실수하지 않은 경우) 대한 표준 솔루션을 얻는 방법은 무엇입니까?$\mathbf{W}'$

편집 1 :

이제 우리는 표준 솔루션에서와 같이 두 개의 제곱을 얻기 위해 항을 재정렬하고 몇 가지 요소를 더하고 뺍니다.

$$\begin{align} tr(\mathbf{W}'^{-1}\boldsymbol{\Lambda}) =\;& tr\bigg(\mathbf{W}^{-1} \boldsymbol{\Lambda}\\ &+ \sum_{i=1}^N (\mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{x}_i + \overline{\mathbf{x}}^T \boldsymbol{\Lambda} \overline{\mathbf{x}} -2 \mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{\overline{x}})\\ &+\kappa_0 (\boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \boldsymbol{\overline{x}}^T \boldsymbol{\Lambda}\boldsymbol{\overline{x}} - 2\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0) \notag\\% Substract added factors &-\sum_{i=1}^{N}\overline{\mathbf{x}}^T \boldsymbol{\Lambda} \overline{\mathbf{x}} +2\sum_{i=1}^{N}\mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{\overline{x}} - \kappa_0\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda} \boldsymbol{\overline{x}} + 2\kappa_0\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda}\boldsymbol{\mu}_0\bigg)\\ =\;& tr\bigg( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})\boldsymbol{\Lambda}(\mathbf{x}_i - \overline{\mathbf{x}})^T + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)\boldsymbol{\Lambda}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \notag\\% Substract added factors &-N\overline{\mathbf{x}} \boldsymbol{\Lambda} \overline{\mathbf{x}}^T +2N\mathbf{\overline{x}} \boldsymbol{\Lambda}\mathbf{\overline{x}}^T - \kappa_0\boldsymbol{\overline{x}}\boldsymbol{\Lambda}\boldsymbol{\overline{x}}^T + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\Lambda} \boldsymbol{\mu}_0^T \bigg) \end{align}$$

우리는 사각형에 남아있는 요소들을 단순화합니다 :

$$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) =\;& tr( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})^T\boldsymbol{\Lambda}(\mathbf{x}_i - \overline{\mathbf{x}}) + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T\boldsymbol{\Lambda}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0) \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}}^T\boldsymbol{\Lambda} \mathbf{\overline{x}} + 2\kappa_0\boldsymbol{\overline{x}}^T\boldsymbol{\Lambda}\boldsymbol{\mu}_0 \bigg) \end{align}$$

편집 2 ( @bdeonovic의 답변 덕분에 후속 조치 )

추적은 주기적이므로 입니다. 그런 다음 다음 : $tr(ABC) = tr(BCA) = tr(CAB)$

$$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) =\;& tr\bigg( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T\boldsymbol{\Lambda} + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T\boldsymbol{\Lambda} \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}} \mathbf{\overline{x}}^T\boldsymbol{\Lambda} + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \bigg) \end{align}$$ $$\begin{align} tr(\mathbf{W}'^{-1}) =\;& tr\bigg( \mathbf{W}^{-1}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}} \mathbf{\overline{x}}^T + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\mu}_0^T \bigg) \end{align}$$

거의! 그러나 여전히 거기에 없습니다. 목표는 다음과 같습니다.

$$\begin{align} \mathbf{W}^{-1} + \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T + \frac{\kappa_0 N}{\kappa_0 + N}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \end{align}$$

추적은 주기적이므로 입니다. 또한 트레이스는 되도록 덧셈을 분산 합니다. 이러한 사실 로 인해 추적 용어에서 용어를 뒤로 순환 하고 추적 용어를 결합 할 수 있어야 합니다. 결과는$tr(ABC) = tr(BCA) = tr(CAB)$$tr(A+B) = tr(A) + tr(B)$$\Lambda$

$$W'^{-1} = W^{-1} + \sum_{i=1}^N x_i x_i^\intercal + \mu_0\mu_0^\intercal$$

가능성 전에 IS $\times$

$$|\boldsymbol{\Lambda}|^{N/2} \exp\left\{-\frac{1}{2}\left(\sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - N\boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} N\boldsymbol{\bar{x}} + N\boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{(\nu_0 - D - 1)/2} \exp\left\{-\frac{1}{2} tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{1/2} \exp \left\{ -\frac{\kappa_0}{2} \left( \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T\boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) \right\}.$$ 로 다시 쓸 수 있습니다. 다시 쓸 수있다 $$|\boldsymbol{\Lambda}|^{1/2} |\boldsymbol{\Lambda}|^{(\nu_0 + N - D - 1)/2} \\ \times \exp \left\{ -\frac{1}{2} \left( \left(\kappa_0 + N\right) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right) \right\}$$ $$(\kappa_0 + N) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right)$$ 다음과 같이 용어를 더하고 빼서 : $$(\kappa_0 + N) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$$ 상단 두 줄은 이제 $$(\kappa_0 + N) \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right).$$

가산 및 감산 다음 : 는 다음과 같이 다시 쓸 수 있습니다. $N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}$

$$- \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right)$$ $$\sum_{i=1}^N \left( \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{x}_i + \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} \right) + N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$$ 합계 기간 $$\sum_{i=1}^N \left( \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{x}_i + \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} \right)$$ 에 해당 이제 는 다음과 같이 확장 할 수 있습니다. $$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right).$$ $$N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})$$ $$N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0^2 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}_0 + N \kappa_0 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N^2 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}),$$ 이는 $$\frac{N\kappa_0}{\kappa_0 + N} \left( \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) = \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$$

다음 두 용어는 스칼라입니다. 스칼라도 그 트레이스와 같으므로

$$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right), \thinspace \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)$$ 로 다시 쓸 수 있습니다. 이후 , 상기 합산 같음 $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \right).$$ $tr(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}) = tr(\boldsymbol{C} \boldsymbol{A} \boldsymbol{B})$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right).$$ 이라는 사실을 사용하여 다음 과 같이 합을 다시 쓸 수 있습니다. $tr(\boldsymbol{A} + \boldsymbol{B}) = tr(\boldsymbol{A}) + tr(\boldsymbol{B})$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right) = tr \left( \left( \boldsymbol{W}_0^{-1} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \right) \boldsymbol{\Lambda} \right).$$

우리가 할 수있는 경우, 즉 함께 모든 퍼팅 우리는 이전 의 가능성과 같습니다 $\boldsymbol{S} = \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T$$\times$

$$|\boldsymbol{\Lambda}|^{1/2} \exp \left\{-\frac{\kappa_0 + N}{2} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{(\nu_0 + N - D - 1)/2} \exp \left\{ - \frac{1}{2} tr \left( \left( \boldsymbol{W}_0^{-1} + \boldsymbol{S} + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \right) \boldsymbol{\Lambda} \right) \right\},$$ 입니다.