Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the book "Markov Random Field Modeling In Image Analysis" by Stan. Z Li, 3d Edition. Page 15.

An observation d of a field f (with m nodes) with i.i.d. gaussian noise:

$d_i = f_i + e_i$, there $e_i$ ~ $N(\mu, \sigma^2)$

$p(d|f) = \frac{1}{\prod_1^m\sqrt{2\pi\sigma^2}} e^{-U(d|f)}$

$U(d|f) = \sum_{i=1}^{m} \frac{(f_i - d_i)^2}{2\sigma^2}$

I would have thought that $f_i - d_i$ should be $d_i - f_i - \mu$?

Why does $\mu$ not impact $p(d|f)$ ?


share|cite|improve this question
$e_i$ looks like error term, usually $\mu=0$ for the error term – mpiktas Feb 15 '11 at 9:19
mpiktas, yes e is the error in the observation, if $\mu$ was zero it would all be OK. In case $\mu$ was nonzero would $d_i - f_i - \mu$ be correct? – j-a Feb 15 '11 at 9:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.