1
$\begingroup$

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 $p(d|f)$ ?

Thanks

  • 0
    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?2011-02-15

0 Answers 0