Case 3:

$$\int \exp\left( - (Y - B F)^{\mathrm{\textsf{T}}}V^{-1} (Y - B F) \right) \; dF = \int \exp\left( - Y^{\mathrm{\textsf{T}}}V^{-1} Y + 2 Y^{\mathrm{... ...1} B F - F^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} B F \right) \; dF$$

$$= \vert\det(K)\vert \int \exp\left( - Y^{\mathrm{\textsf{T}}}V^{-1}... ...sf{T}}}K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} B K G \right) \; dG$$

$$= \vert\det(K)\vert \int \exp\left( - Y^{\mathrm{\textsf{T}}}V^{-1}... ... 2 Y^{\mathrm{\textsf{T}}}V^{-1} B K G - G^{\mathrm{\textsf{T}}}G \right) \; dG$$

$$= \vert\det(K)\vert \int \exp\left( - (G - K^{\mathrm{\textsf{T}}}B... ...{T}}}(G - K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} Y) \right) \; dG$$

$$\qquad \qquad \qquad \exp\left( - Y^{\mathrm{\textsf{T}}}V^{-1} Y... ...f{T}}}V^{-1} B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} Y \right)$$

$$= \vert\det(K)\vert \, (\pi)^{M/2} \exp\left( - Y^{\mathrm{\textsf{... ... B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1/2}) V^{-1/2} Y \right)$$

where $\dim(Y) = N$, $\dim(F) = \dim(G) = M \le N$, $F = KG$ and $K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} B K = I$. This assumes that $\ensuremath{\mathrm{rank}}(B^{\mathrm{\textsf{T}}}V^{-1} B) = M$ so that $K = (B^{\mathrm{\textsf{T}}}V^{-1} B)^{-1/2}$.

For $M = N = \ensuremath{\mathrm{rank}}(B)$

$\ensuremath{\mathrm{rank}}(B^{\mathrm{\textsf{T}}}V^{-1} B) = N$ implies that $\ensuremath{\mathrm{rank}}(B) = N$ and $\ensuremath{\mathrm{rank}}(V) = N$. Consequently, using the SVD decompositions $B=U_B D_B W_B^{\mathrm{\textsf{T}}}$ and $V=U_V D_V U_V^{\mathrm{\textsf{T}}}$ gives $K = W_B D_B^{-1} U_B^{\mathrm{\textsf{T}}}U_V D_V^{1/2}$. Therefore $B K = U_V D_V^{1/2}$ and so $V^{-1/2} B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1/2} = I$, so that the integral is given by:

$$\int \exp\left( - (Y - B F)^{\mathrm{\textsf{T}}}V^{-1} (Y - B F)... ...ght) \; dF = \vert\det(B^{\mathrm{\textsf{T}}}V^{-1} B)\vert^{-1/2} (\pi)^{M/2}$$ (17)

For $\ensuremath{\mathrm{rank}}(B) = M < N$,

$B^{\mathrm{\textsf{T}}}V^{-1} B$ is taken to have full rank, but $B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}$ is not, as $D_B$ is not a square matrix, so that $D_B^{-1}$ does not exist. Instead, $K = W (D^{\mathrm{\textsf{T}}}D)^{-1/2}$ where $V^{-1/2} B = U D W^{\mathrm{\textsf{T}}}$ by SVD decomposition. As $D$ has dimensions $N$ by $M$ (same as $B$) then $D^{\mathrm{\textsf{T}}}D$ is $M$ by $M$ and hence invertible (by virtue of $B^{\mathrm{\textsf{T}}}V^{-1} B$ being full rank). Therefore, $K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} B K = (D^{\mathrm{\textsf{T}}}D)^{-1/2} (D^{\mathrm{\textsf{T}}}D) (D^{\mathrm{\textsf{T}}}D)^{-1/2} = I$ as desired, but $V^{-1/2} B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1/2} = U D (D^{\mathrm{\textsf{T}}}D)^{-1} D^{\mathrm{\textsf{T}}}U^{\mathrm{\textsf{T}}}= P_w$ which is a projection matrix. Consequently, $R_w = I - P_w = I - V^{-1/2} B K K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}... ...{-1/2} B (B^{\mathrm{\textsf{T}}}V^{-1} B)^{-1} B^{\mathrm{\textsf{T}}}V^{-1/2}$ is not zero, but is the residual projection matix (onto the null space of $V^{-1/2} B$ - the prewhitened version of $B$). The integral can then be written as:

$$\int \exp\left( - (Y - B F)^{\mathrm{\textsf{T}}}V^{-1} (Y - B F) \right) \; dF = \vert\det(K)\vert (\pi)^{M/2} \exp\left( - Y^{\mathrm{\textsf{T}}}V^{-1/2} R_w V^{-1/2} Y \right)$$

$$= \vert\det(B^{\mathrm{\textsf{T}}}V^{-1} B)\vert^{-1/2} (\pi)^{M/2} \exp\left( - Y^{\mathrm{\textsf{T}}}R_c Y \right)$$ (18)

where $R_c = V^{-1/2} R_w V^{-1/2}$ is the residual projection matrix in the coloured space, and $R_w = I - V^{-1/2} B (B^{\mathrm{\textsf{T}}}V^{-1} B)^{-1} B^{\mathrm{\textsf{T}}}V^{-1/2}$ is the residual projection matrix in the whitened space.

For $\ensuremath{\mathrm{rank}}(B) < M$,

In this case the matrix $B$ will have many linearly dependent columns and $K^{\mathrm{\textsf{T}}}B^{\mathrm{\textsf{T}}}V^{-1} B K = I$ cannot be achieved. Instead, let the number of independent columns be $N_1 \le N$. Furthermore, let $V^{-1/2} B = U D W^{\mathrm{\textsf{T}}}$ by SVD, where

$$D = \left[ \begin{array}{cccc} D_1 & 0 \\ 0 & 0 \end{array} \right]$$

where $D_1$ is an $N_1$ by $N_1$ diagonal matrix with all diagonal entries being non-zero. Integration of the parameters associated with the zero singular values can be carried out if they have finite extents. Letting $F = KG$, and $K = W$ gives

$$\int \exp\left( - (Y - B F)^{\mathrm{\textsf{T}}}V^{-1} (Y - B F) \right) \; dF = \int \exp\left( - (V^{-1/2} Y - U D G)^{\mathrm{\textsf{T}}}(V^{-1/2} Y - U D G) \right) \; dG$$

$$= L^{M-N_1} \vert\det(D_1^{\mathrm{\textsf{T}}}D_1)\vert^{-1/2} (\pi)^{M/2} \exp\left( - Y^{\mathrm{\textsf{T}}}R_c Y \right)$$ (19)

where the null parameters are integrated over $[0,L]$, and $R_c = V^{-1} - V^{-1/2} U D_0 (D_1^{\mathrm{\textsf{T}}}D_1)^{-1} D_0^{\mathrm{\textsf{T}}}U^{\mathrm{\textsf{T}}}V^{-1/2}$, with

$$D_0 =\left[ \begin{array}{c} D_1 \\ 0 \end{array} \right]$$

which is an $N$ by $N_1$ matrix.