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# Marginalization of a chain of Gaussian distributions

2024/01/14に公開

\begin{align*} p(y|\theta) &= \int p(y,f|\theta) df = \int p(y|f) p(f|\theta) df \\ & p(y|f) = \mathcal{N}(y|f, \sigma^2I) \\ & p(f|\theta) = \mathcal{N}(f|0, K_\theta) \\ \end{align*}

When

\begin{align*} p(y, f|\theta) &= \mathcal{N}\left( \begin{pmatrix} y \\ f \end{pmatrix} \bigg\vert \begin{pmatrix} \mu_y \\ \mu_f \end{pmatrix} , \begin{pmatrix} \Sigma_{yy} & \Sigma_{yf}\\ \Sigma_{fy} & \Sigma_{ff} \end{pmatrix} \right) \\ \end{align*}

p(y|\theta) = \mathcal{N}(y|\mu_y, \Sigma_{yy}) , so we have to obtain \mu_y, \Sigma_{yy}.

Let \Lambda = \Sigma^{-1},

\begin{align*} \ln p(y,f|\theta) &\propto - \frac{1}{2} \left\{ \begin{pmatrix} y - \mu_y \\ f - \mu_f \end{pmatrix}^\top \begin{pmatrix} \Lambda_{yy} & \Lambda_{yf} \\ \Lambda_{fy} & \Lambda_{ff} \\ \end{pmatrix} \begin{pmatrix} y - \mu_y \\ f - \mu_f \end{pmatrix} \\ \right\} \\ &\propto - \frac{1}{2} \left\{ \underbrace{ \begin{pmatrix} y \\ f \end{pmatrix}^\top \begin{pmatrix} \Lambda_{yy} & \Lambda_{yf} \\ \Lambda_{fy} & \Lambda_{ff} \\ \end{pmatrix} \begin{pmatrix} y \\ f \end{pmatrix} }_{\text{second order term}} \\ -2 \underbrace{ \begin{pmatrix} \mu_y \\ \mu_f \end{pmatrix}^\top \begin{pmatrix} \Lambda_{yy} & \Lambda_{yf} \\ \Lambda_{fy} & \Lambda_{ff} \\ \end{pmatrix} \begin{pmatrix} y \\ f \end{pmatrix} }_{\text{linear term}} \tag{1}\\ \right\} \end{align*}

On the other hand,

\begin{align*} \ln p(y,f|\theta) &= \ln p(y|f) p(f|\theta) \\ &= \ln \mathcal{N}(y|f, \sigma^2I) \mathcal{N}(f|0, K_\theta) \\ &\propto - \frac{1}{2} \left\{ (y-f)^\top \frac{1}{\sigma^2} I (y-f) + f^\top K_\theta^{-1} f \right\} \\ &= - \frac{1}{2} \left\{ \frac{1}{\sigma^2} y^\top y - 2 y^\top \frac{1}{\sigma^2} f + f^\top (K_\theta^{-1} + \frac{1}{\sigma^2} I) f \right\} \\ &= -\frac{1}{2} \begin{pmatrix} y \\ f \end{pmatrix}^\top \underbrace{ \begin{pmatrix} \frac{1}{\sigma^2} & -\frac{1}{\sigma^2} \\ -\frac{1}{\sigma^2} & (K_\theta^{-1} + \frac{1}{\sigma^2}) \end{pmatrix} }_{\Lambda = \Sigma^{-1}} \begin{pmatrix} y \\ f \end{pmatrix} \tag{2} \end{align*}

According to comparing the second order term in (1) and (2)

\begin{align*} \Sigma = \Lambda^{-1} &= \begin{pmatrix} \frac{1}{\sigma^2}I & -\frac{1}{\sigma^2}I \\ -\frac{1}{\sigma^2}I & (K_\theta^{-1} + \frac{1}{\sigma^2}) \end{pmatrix}^{-1} \\ &= \begin{pmatrix} \Sigma_{yy} & \Sigma_{yf} \\ \Sigma_{fy} & \Sigma_{ff} \\ \end{pmatrix} \\ \end{align*}

Because the linear term in (2) does not exist, means are

\begin{align*} \begin{pmatrix} \mu_y \\ \mu_f \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \end{align*}

Here we have to remember the following matrix identity.

\begin{align*} \begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1} &= \begin{pmatrix} M & -M B D^{^-1} \\ -D^{-1} C M & D^{-1} + D^{-1} C M B D^{-1} \end{pmatrix} \\ M &= (A - B D^{-1} C)^{-1} \end{align*}

or

\begin{align*} \begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1} &= \begin{pmatrix} A^{-1} + A^{-1} B M C A^{-1} & -A^{-1} B M \\ -M C A^{-1} & M \end{pmatrix} \\ M &= (D - C A^{-1} B)^{-1} \end{align*}

In this case the second identity is better because we can easily obtain A^{-1}.

\begin{align*} M &= \left( (K_\theta^{-1} + \frac{1}{\sigma^2}) - (-\frac{1}{\sigma^2}I) (\frac{1}{\sigma^2}I)^{-1} (-\frac{1}{\sigma^2}I) \right)^{-1} = K_\theta \end{align*}
\begin{align*} \Sigma_{yy} &= (\frac{1}{\sigma^2}I)^{-1} + (\frac{1}{\sigma^2}I)^{-1} (-\frac{1}{\sigma^2}I) K_\theta (-\frac{1}{\sigma^2}I) (\frac{1}{\sigma^2}I)^{-1} \\ &= K_\theta + \sigma^2 I \end{align*}

\begin{align*} \therefore p(y|\theta) = \mathcal{N}(y | 0, K_\theta + \sigma^2 I) \end{align*}