Conditional Multivariate Gaussian Distribution - Section 2.3, equation 2.74 Please refer section 2.3 (page 86-87) in Pattern Recognition and Machine Learning - Bishop

$$\mathit{N}(\mathbf{x}|\mathbf{\mu}, \Sigma)$$
where

$
\mathbf{x} = 
\begin{bmatrix}
x_{a} \\
x_{b}
\end{bmatrix}
$,
$
\mathbf{\mu} = 
\begin{bmatrix}
\mu_{a} \\
\mu_{b}
\end{bmatrix}
$,
$
\mathbf{\Sigma} = 
\begin{bmatrix}
\Sigma_{aa} &amp;&amp; \Sigma_{ab}\\
\Sigma_{ba} &amp;&amp; \Sigma_{bb}
\end{bmatrix}
$

The equation below expresses the Quadratic term of exponent in the Bivariate Gaussian probability distribution (ref. book eq. 2.70; ignoring $-\frac{1}{2}$):

$$(x - \mu)^{T}\Sigma^{-1}(x - \mu) = (x_a - \mu_a)^{T}\Lambda_{aa}(x_a - \mu_a)
+(x_a - \mu_a)^{T}\Lambda_{ab}(x_b - \mu_b)
+(x_b - \mu_b)^{T}\Lambda_{ba}(x_a - \mu_a)
+(x_b - \mu_b)^{T}\Lambda_{aa}(x_b - \mu_b)$$


  Author states "..conditional distribution $p(x_{a} | x_{b})$ can be
  evaluated from the the joint distribution $p(x) = p(x_{a}, x_{b})$ by
  fixing $x_b$ to the observed value and normalizing the resulting
  expression to obtain the valid probability distribution over $x_{a}$
  .." and then filters out only the linear terms with $x_{a}$ and $x_{a}^{T}$
  component. The result is presented in equation 2.74 as:


$$x_{a}^{T}\{\Lambda_{aa}\mu_{a} - \Lambda_{ab}(x_b - \mu_b)\}$$

I follow the steps, and see that above is due to $x_{a}^{T}$, $x_{a}$ in the first 2 terms of equation 2.70 - where "term(s)" refer to the components in 2.70 added using $+$ operator.


  However, I seem to be getting two additions elements in $x_a$, in
  excess of the those present in 2.74 - due to the linear $x_a$ factor
  in $3^{rd}$ term (of equation 2.70). Excess term(s):  $$ (x_{b} -
 \mu_{b})^{T}\Lambda_{ba}x_a $$


Where am I going wrong? Please guide.


  Note: I am taking the $x_a$ too to be linear component, whereas 2.74 is (seemingly)
  considering only the $x_{a}^{T}$ as linear term. Seems like I am losing track
  somewhere in the middle (perhaps in equation 2.71 - related extract,
  below):


$$ -\frac{1}{2}(x - \mu)^{T}\Sigma^{-1}(x - \mu) = -\frac{1}{2}{x^{T}\Sigma^{-1}x} + {x^{T}\Sigma^{-1}\mu} + constant \ldots eq. 2.71$$

...where 'const' denotes terms which are independent of ${x}$, and we have made use of the symmetry of $\Sigma$.
\begin{align}&amp;-\frac12(x - \mu)^{T}\Sigma^{-1}(x - \mu) \\&amp;= -\frac12(x_a - \mu_a)^{T}\Lambda_{aa}(x_a - \mu_a) \tag{1}
\\&amp;-\frac12(x_a - \mu_a)^{T}\Lambda_{ab}(x_b - \mu_b) \tag{2}
\\&amp;-\frac12(x_b - \mu_b)^{T}\Lambda_{ba}(x_a - \mu_a) \tag{3}
\\&amp;-\frac12(x_b - \mu_b)^{T}\Lambda_{aa}(x_b - \mu_b) \tag{4}\end{align}

Now let's focus on $(1)$,  the linear part in $x_a$ are 

$$-\frac12 (-x_a^T\Lambda_{aa}\mu_a - \mu_a^T\Lambda_{aa}x_a)=x_a^T\Lambda_{aa}\mu_a \tag{5}$$

Now focus on $(2)$, the linear part in $x_a$ is 

$$-\frac12x_a^T\Lambda_{ab}(x_b-\mu_b) \tag{6}$$

Now focus on $(3)$,  the linear part in $x_a$ is 

$$-\frac12(x_b-\mu_b)^T\Lambda_{ba}x_a=-\frac12x_a^T\Lambda_{ab}(x_b-\mu_b) \tag{7}$$

There is no linear term of $x_a$ in $(4)$, 

Adding $(5)$ to $(7)$, we have 

\begin{align}x_a^T\Lambda_{aa}\mu_a-\frac12x_a^T\Lambda_{ab}(x_b - \mu_b)-\frac12x_a^T\Lambda_{ab}(x_b - \mu_b)&amp;=x_a^T\Lambda_{aa}\mu_a-x_a^T\Lambda_{ab}(x_b - \mu_b)\\
&amp;=x_a^T(\Lambda_{aa}\mu_a -\Lambda_{ab}(x_b-\mu_b))\end{align}