The equivalence of the Stosszahlansatz and the usual Boltzmann entropy arguments? Question
Below I show one can use the Boltzmann Stosszahlansatz to independently arrive at the Maxwell Boltzmann distribution without using the usual phase space arguments (Assuming $*$ equation has a unique solution). This makes me that suspect there is a deeper relation between the two? Can one show that both statements are strangely equivalent or mutually dependent?
The Boltzmann Stosszahlansatz
We state a version of the Stosszahlansatz as follows:
"The relative velocities of a pair of gas particles are uncorrelated to the relative velocities of the other gas particles."
We shall show this is sufficient to derive the a Maxwell Boltzmann distribution having only one species of gas.
Derivation
Now,
\begin{equation}
 P(\tilde \beta \vec \lambda) = \text{Probability of measuring } \vec \lambda
\end{equation}
we introduce $\tilde \beta$ a constant to ensure the argument remains dimensionless. Consider the below diagram consisting of $4$ particles all separated by distance of order $\delta \to 0$:

For pictorial purposes we have depicted non-zero separation. Now, we ask ourselves the following question. If,
\begin{equation}
\vec v_{AB} = \vec \lambda - \vec k
\end{equation}
\begin{equation}
\vec v_{BC} = \vec 0
\end{equation}
\begin{equation}
\vec v_{CD} = \vec k
\end{equation}
Then we know,
\begin{equation}
\vec v_{AD} = \vec \lambda
\end{equation}
Now, we try to ask ourselves what is the probability of the above set-up? We begin by writing the expression for $\vec v_{AB}$ and $\vec v_{CD}$
\begin{equation}
P(\tilde \beta \vec v_{AB}) = P(\tilde \beta (\vec \lambda - \vec k ))
\end{equation}
Similarly for $\vec v_{CD}$:
\begin{equation}
P(\tilde \beta \vec v_{CD}) = P(\tilde \beta  \vec k )
\end{equation}
We multiply the above and get:
\begin{equation}
P(\tilde \beta \vec v_{AB}) P(\tilde \beta \vec v_{CD})  =  P(\tilde \beta (\vec \lambda - \vec k )) P(\tilde \beta  \vec k )
\end{equation}
This represents the probability of $2$ independent pairs with with velocities $\vec \lambda - \vec k$ and $\vec k$.

Now, we recall if the labelling is completely random. What is the probability one will label a $B$ such that $\vec v_{AB} = \vec \lambda - \vec k$ and another one $\vec v_{BD} = \vec k$. Thus:
\begin{equation}
P(\tilde \beta \vec v_{AB}) P(\tilde \beta \vec v_{BD})  =  P(\tilde \beta (\vec \lambda - \vec k )) P(\tilde \beta  \vec k )
\end{equation}
Let us put everything together:
\begin{equation}
P(\tilde \beta \vec v_{AB}) P(\tilde \beta \vec v_{BD}) P(\tilde \beta \vec v_{AB}) P(\tilde \beta \vec v_{CD}) =  \Big ( P(\tilde \beta (\vec \lambda - \vec k )) P(\tilde \beta  \vec k ) \Big )^2
\end{equation}
What about our final piece $P(\tilde \beta \vec v_{AD}) $?
\begin{equation}
P(\tilde \beta \vec v_{AD}) = P(\tilde \beta \vec \lambda) 
\end{equation}
Obviously that is independent of the other relative velocities.  Thus,
\begin{equation}
\label{MB 3D}
P(\tilde \beta \vec \lambda) = A \tilde \beta^3 \int_{-\infty}^\infty \int_{-\infty}^\infty  \int_{-\infty}^\infty \Big ( P(\tilde \beta (\vec \lambda - \vec k )) P(\tilde \beta  \vec k ) \Big )^2 d k_x d k_y d k_z
\end{equation}
where $k_x$, $k_y$ and $k_z$ are projections of the $\vec k$ vector along the $x$, $y$ and $z$ axis and $A$ is a normalization constant.
In one dimension we expect this to reduce to:
\begin{equation}
\label{MB 1D}
P(\tilde \beta  \lambda) = A \tilde \beta \int_{-\infty}^\infty \Big ( P(\tilde \beta ( \lambda -  k )) P(\tilde \beta   k ) \Big )^2 d k \tag{*}
\end{equation}
Verification using the Maxwell Boltzmann Distribution
We already know the one dimensional Maxwell distribution in one dimension is:
\begin{equation}
f(v)  = D \exp(- \frac{m v^2}{2 k_B T}) 
\end{equation}
This represents the probability of a particle going with velocity $v$. Now, the probability of observing another particle of velocity $ v +d$ is:
\begin{equation}
f(v + d)  = D \exp(- \frac{m (v+d)^2}{2 k_B T}) 
\end{equation}
Thus, the probability of witnessing a relative velocity $d$ upto a normalization constant is:
\begin{equation}
\int_{-\infty}^\infty f(v) f(v+d) dv \propto \int_{-\infty}^\infty  \exp(- \frac{m (v+d)^2}{2 k_B T}) \exp(- \frac{m v^2}{2 k_B T}) dv 
\end{equation}
This is of the form:
\begin{equation}
P(\tilde \beta d) = A' \exp(-\tilde \beta d^2/2)
\end{equation}
where $A'$ is a normalization constant. Note this also satisfies equation $*$.
 A: The question is, if the Stosszahlansatz and the usual phase space treatment independently lead to to the MB distribution, is there a deeper connection between the two approaches?
Let's begin by pointing out that the MB distribution can be derived in a number of ways that are quite different in their details. Two of them are due to Maxwell himself. An excellent account is given by Uffink, which is my source for what follows.
Maxwell Proof #1 (1860) Assume that:

*

*the components of the velocity are independent so that the distribution of velocities is of the form $f(\mathbf{v}) = f(v_x,v_y,v_z)$, where $g$ is some function, the same for all three components of velocity; and


*$f(\mathbf{v}$ is spherically symmetric so that $f(\bar v) = f(v_x^2+v_y^2+v_z^2)$.
Under these conditions Maxwell showed that $g(v_i)$ must be of the form
$$g(v_i) = a e^{-b v_i^2}$$
which leads to the MB distribution.
Maxwell's proof #2 (1867) This proof is more involved and takes into consideration the Stosszahlansatz proposition. The details are given in Uffink
Boltzmann's $H$ theorem (1872) This is Boltzmann's derivation and is based on the definition of functional $H$ defined as
$$H = \int f(\mathbf{v}) \log f(\mathbf{v}) \, d\mathbf{v}$$
Boltzmann showed that $dH/dt\leq 0$ and the stationary condition $dH/dt = 0 $ is satisfied by the same disribution as that obtained by Maxwell. With his proof Boltzmann attached his name to the distribution that is now known as Maxwell-Boltzmann.
The conclusion is that arguments that vary in sophistication and degree of physical realism can still lead to the same answer. One can argue that Maxwell's first derivation is almost completely devoid of physics, there are no particles, no collisions, just intuitive expectations about the form of the result – yet, it gives the right answer. And while Boltzmann's $H$ theorem is based on a physical model of matter in the form of colliding particles, it assumes no inter-particle interactions and in this sense it is a weaker proof.
The most general proof is the one based on phase space and is independent of the strength of interactions. One could say that the Stosszahlansatz is built-into the phase treatment because its basic premise is that every feasible microstate is a probable microstate.
