Dérivation de Normal-Wishart postérieur

Je travaille sur la dérivation d'un postérieur de Normal-Wishart mais je suis bloqué sur l'un des paramètres (le postérieur de la matrice d'échelle, voir en bas).

Juste pour le contexte et l'exhaustivité, voici le modèle et le reste des dérivations:

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

Les formes étendues de chacun des trois facteurs sont (jusqu'à une constante de proportionnalité) sont:

• Probabilité:

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

• Priorité normale:

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

• Priorité de Wishart:

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

Nous voulons le Normal-Wishart postérieur ( ) qui peut être décomposé ainsi que 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}')$

Dégressivité $\upsilon'$

En fusionnant les premiers facteurs de la probabilité et le Wishart, nous obtenons le premier facteur du facteur Wishart dans la partie postérieure: et donc on a le premier paramètre de la postérieure: υ′=υ0+

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

Facteur d'échelle $\kappa'$

Nous identifions les éléments entourés de et μ pour trouver qui l'a priori κ 0 Λ est mis à jour par la vraisemblance: μ T ( ( κ 0 + N ) Λ ) μ et nous avons donc obtenu le deuxième paramètre: κ ′ = κ 0 + $\boldsymbol{\mu}^T$$\boldsymbol{\mu}$$\kappa_0 \boldsymbol{\Lambda}$

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

Moyenne $\boldsymbol{\mu}'$

$2\boldsymbol{\mu}^T...$

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

$\boldsymbol{W}'$

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

$\mathbf{W}'$

Modifier 1 :

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

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

Edit 2 ( suivi grâce à la réponse de @bdeonovic )

$tr(ABC) = tr(BCA) = tr(CAB)$

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

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

$tr(ABC) = tr(BCA) = tr(CAB)$$tr(A+B) = tr(A) + tr(B)$$\Lambda$

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

La probabilité avant est $\times$

$$|\boldsymbol{\Lambda}|^{N/2} \exp\left\{-\frac{1}{2}\left(\sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - N\boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} N\boldsymbol{\bar{x}} + N\boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{(\nu_0 - D - 1)/2} \exp\left\{-\frac{1}{2} tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{1/2} \exp \left\{ -\frac{\kappa_0}{2} \left( \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T\boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) \right\}.$$ Cela peut être réécrit sous la forme Nous pouvons réécrire $$|\boldsymbol{\Lambda}|^{1/2} |\boldsymbol{\Lambda}|^{(\nu_0 + N - D - 1)/2} \\ \times \exp \left\{ -\frac{1}{2} \left( \left(\kappa_0 + N\right) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right) \right\}$$ $$(\kappa_0 + N) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right)$$ comme suit en ajoutant et en soustrayant un terme: $$(\kappa_0 + N) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$$ Les deux premières lignes factorisent désormais $$(\kappa_0 + N) \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right).$$

Ajouter et soustraire , comme suit: peut être réécrit comme $N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}$

$$- \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right)$$ $$\sum_{i=1}^N \left( \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{x}_i + \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} \right) + N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$$ Le terme somme $$\sum_{i=1}^N \left( \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{x}_i + \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} \right)$$ est égal à Maintenant peut être développé comme $$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right).$$ $$N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})$$ $$N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0^2 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}_0 + N \kappa_0 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N^2 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}),$$ ce qui équivaut à $$\frac{N\kappa_0}{\kappa_0 + N} \left( \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) = \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$$

Les deux termes suivants sont des scalaires: Et tout scalaire est égal à sa trace, donc

$$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right), \thinspace \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)$$ peut être réécrit en Puisque , la somme ci-dessus est égale à $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \right).$$ $tr(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}) = tr(\boldsymbol{C} \boldsymbol{A} \boldsymbol{B})$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right).$$ En utilisant le fait que , nous pouvons réécrire la somme comme $tr(\boldsymbol{A} + \boldsymbol{B}) = tr(\boldsymbol{A}) + tr(\boldsymbol{B})$ $$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right) = tr \left( \left( \boldsymbol{W}_0^{-1} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \right) \boldsymbol{\Lambda} \right).$$

En mettant tout cela ensemble, si nous laissons nous avons que la probabilité avant est égale à $\boldsymbol{S} = \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T$$\times$

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