Logical understanding of the canonical probability distribution (canonical ensemble) I am having problems in understanding the logic of this distribution:
$P(\Psi_{j})=\displaystyle\frac{e^{-E_{j}/kT}}{\displaystyle\sum_{j'}e^{-E_{j'}/kT}}$
The book I am studying use the case of a sample in contact with a reservoir at thermal equilibrium to derive this distribution. I understand the derivation, but I don't understand the logic of the distribution itself. The aspect I'm having problem with is the fact that the lowest the energy of the sample, the highest the probability. What I don't understand is why this happens even thought there is an average energy given by the temperature which I thought should be more probable then any energy lower than this for a given particle. This doubt implies that I am looking at $P(\Psi)$ as the probability for a given particle, wich does not seem to be the case, but if I think of it as the probability for the hole sample, it makes even less sense for me since the energy should be totaly given by the temperature, so it wouldn't make sense to make a distribution of it if the temperature is considered constant. 
Thanks in advance
 A: Maybe your intuition about energy and temperature need to be revisited. Your system can exchange energy with the reservoir at a given temperature. The system+reservoir will iterate through all microstates with equal probability (total energy being fixed), but you can show by using entropy arguments, that the probability of the system being in a state with energy $E$ is given by your first equation. The average energy of your system is then
$\langle E \rangle = \sum_i E_i P(E_i) $.
That is not the same as the most probable energy, which is $E_0$.
When deriving the Boltzmann factor from the reservoir argument, there are corrections to the factor when the reservoir is finite. You can write, for the system in uniquely labelled states $i$ and $j$,
$\frac{P(i)}{P(j)} = \frac{\Omega_R(i)}{\Omega_R(j)}$
where $\Omega_R(i)$ is the number of microstates of the reservoir when the system is in state $i$. Writing this in terms of entropy gives a more fundamental form
$\frac{P(i)}{P(j)} = e^{-\frac{S_R(i) - S_R(j)}{k_B} }$
If you label the states of the system by energy, and allow it to be continuous, then you can Taylor expand $S_R(E_i)$ around $E=0$, since by assumption the energy of the reservoir is vastly bigger than that of the system. Changing variables to the energy of the reservoir $U_{res}$, the linear term is
$-\frac{\partial S_R}{\partial U_{res}} E_i = -\frac{E_i}{T}$,
from which the Bolzmann factor follows.
