I am referring to the book "Pattern recognition and machine learning" book by Bishop. Equation 2.104 states that the precision matrix for the joint distribution is

$ \begin{equation} R = \left(\begin{array}{cc} \Lambda+A^TLA & -A^TL\\ -LA & L \end{array}\right) \end{equation} $

and the mean of the joint distribution is

$ \begin{equation} E[z] = \left(\begin{array}{cc} \mu\\ A\mu+b\end{array}\right) \end{equation} $

For conditional Gaussian distributions equation 2.75 states that

$ \mu_{a|b} = \mu_{a} - \Lambda_{aa}^{-1}\Lambda_{ab}(x_{b} - \mu_{b}) $

From the equation for R

$ \Lambda_{aa}^{-1} = (\Lambda+A^TLA)^{-1} $

$ \Lambda_{ab} = -A^TL $

$ \mu_{a} = mu $

and $ \mu_{b} = -A\mu + b $

Substituting in expression for $\mu_{a|b}$ I get

$ E[x|y] = \mu + (\Lambda+A^TLA)^{-1}(A^TL)(y - (A\mu + b)) $ But this expression is different from what is concluded in equation 2.111 which is

$ E[x|y] = (\Lambda+A^TLA)^{-1} [A^TL(y - b) + \Lambda\mu] $

How do I reconcile what I got above and what equation 2.111 reproduced above gives?

  • $\begingroup$ What is the context here (just briefly)? $\endgroup$ Commented Jun 7, 2019 at 10:35
  • $\begingroup$ The goal of the relevant section in the book is to find the mean and covariance of a conditional Gaussian distribution p(x|y) where the joint is partitioned into x and y $\endgroup$ Commented Jun 7, 2019 at 10:38
  • $\begingroup$ You've some typos in $\mu_b$,$\mu_a$. $\endgroup$
    – gunes
    Commented Jun 7, 2019 at 15:39

1 Answer 1


Expanding your solution gives: $$\begin{align}E[x|y]&=\mu+(\Lambda+A^TLA)^{-1}(A^TL)(y-b)\\&-(\Lambda+A^TLA)^{-1}A^TLA\mu\end{align}$$ The second summand here is equal to the first expression in the answer, when expanded. We just need to show that sum of first and third summands is equal to the second expression in the answer: $$\begin{align}e&=\mu-(\Lambda+A^TLA)^{-1}A^TLA\mu=\mu-(\Lambda+A^TLA)^{-1}(A^TLA+\Lambda-\Lambda)\mu\\ &=\mu-(\Lambda+A^TLA)^{-1}(A^TLA+\Lambda)\mu+(\Lambda+A^TLA)^{-1}\Lambda\mu\\&=\mu-\mu+(\Lambda+A^TLA)^{-1}\Lambda\mu\\&=(\Lambda+A^TLA)^{-1}\Lambda\mu\end{align}$$ which is really the second expression in the answer. So, your result is equivalent to the given one.


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