# Does $\hat{x}$ always mean normalized version of a vector $x$?

"...a maximum-a-posteriori $(MAP_{x,k}^{\,\,\,\,\,1})$ estimation, seeking a pair $(\hat{x}, \hat{k})$ maximizing: $$p(x, k\mid y) \propto p(y|x, k)p(x)p(k).$$
Are $\hat{x}$ and $\hat{k}$ the normalized version of vectors $x$ and $k$, having length $1$? If so, could someone explain me why this does make sense?
This is related to statistics, and $(\hat{x}, \hat{k})$ are not normalized-things but estimators, found by optimizing a given function. See here: en.wikipedia.org/wiki/Maximum_a_posteriori_estimation –  Jean-Claude Arbaut Mar 13 '13 at 11:28
If $x$ is the observation variable then $\hat{x}$ denotes the predicted (or fitted) value. –  Stéphane Laurent Mar 13 '13 at 11:49
Why do we write bayes's rule in this form and not like this: $P(A_n|B) \propto P(B|A_n) P(A_n)$ –  user1095340 Mar 20 '13 at 7:18