Exponential form of Boltzmann Distribution I am trying to understand why the Boltzmann distribution is of the specific form:
$$ f(E) = A e^{-E/B} $$
for some constants $A$ and $B$. I am following the discussion here:
https://courses.physics.ucsd.edu/2017/Spring/physics4e/boltzmann.pdf
The author gives an example of a six particle system with eight discrete energy levels. I am happy with the argument that:
$$ n(E_i) = \sum_j n_{i,j} \, p_j $$
where $p_j$ is the probability of the system being in the $j$-th microstate, $n_{i,j}$ is the number of particles with energy $E_i$ in the $j$-th microstate and $n(E_i)$ is the expected number of particles to be found with energy $E_i$.
I have checked numerically that the distribution does have an approximately negative exponential shape in this small example.
I get confused however at the next step of the argument. Let $f(E_i)$ denote the probability that a given particle has energy $E_i$. Then, since there are six particles in our system we have:
$$ f(E_i) = \frac{n(E_i)}{6} $$
Clearly $f(E)$ has the same shaped distribution as $n(E)$, since they are proportional. Consider now the probability $h(E_0 + E_2)$ that a pair of particles has combined energy $E_0 + E_2$.
By my calculation:
$$ h(E_0 + E_2) = 2f(E_2)f(E_0) + f(E_1)^2 $$
Why? Because either one particle has energy $E_0$ while the other has energy $E_2$ or both particles have energy $E_1$.
This is now, however, what the author has written. He claims that:
$$ h(E_0 + E_2) = f(E_0)f(E_2) $$
I agree that if we could show that:
$$ f(E_i + E_j) = f(E_i)f(E_j)$$
Then the exponential form of the distribution would follow immediately. However I fail to see the relation between the probability $f(E_i + E_j)$ that a single particle has energy $E_i + E_j$ and the probability $h(E_i + E_j)$ that a pair of particles has combined energy $E_i + E_j$.
Obviously I am doing something wrong. Any clarification would be greatly appreciated. 
 A: You seem to be slightly misunderstanding the text.  $h(E_1+E_2)$ is not the probability that two particles have a total energy $E_1+E_2$.
The argument goes like this: Imagine that you have a bag of particles with a discrete, equally-spaced set of allowed energies.  The probability of reaching into your bag and pulling out a particle with energy $E_i$ is $f(E_i)$, which means that the probability of reaching into your bag and pulling out two particles with respective energies $E_0$ and $E_2$ is $f(E_0)\times f(E_2)$.
Notice that this is not the probability of pulling out two particles with total energy $E_0+E_2$.  To get the latter, we would need to consider all of the possible ways (read: microstates) in which we might get that total.  Neglecting subtle arguments about particle distinguishability, we could have


*

*Particle A has energy $E_0$ while particle B has energy $E_2$

*Particle A has energy $E_2$ while particle B has energy $E_0$

*Particle A and particle B both have energies $E_1$


However, on the bottom of page 2:

If  we  now  make  the  reasonable  assumption  that all  microstates occur  with  the same probability [...]

This is the crucial assumption, and is actually a very deep one.  If we make it, however, it says that each of the three possibilities enumerated above occurs with the same probability.  It is this probability which we call $h(E_0+E_2)$.
If this is true, then in particular, we have that $h(E_0+E_2)=f(E_0)\times f(E_2)^\dagger$, which is the first of the three possible microstates.  The only $f$ which could satisfy such a relationship is an exponential function, as stated in the text.
Of course, this logic holds if the allowed energies are not evenly spaced; I only included that part to mirror your question.  The fundamental assumption is that each energetically allowed microstate occurs with equal probability.

$^\dagger$ Your linked text has a typo - on the top of page 23, that should be $f(E_1)\times f(E_2)$, not $f(E_1)+f(E_2)$.
