My question comes from a computation in the paper Central limit theorem for Maxwellian molecules and truncation of Wild expansion. Specially, consider the following Boltzmann equation $$\frac{\partial f}{\partial t} = Q^+(f,f) - f \label{1}\tag{1}$$ with $f(t=0) = F$. Introduce the notation $f \circ f = Q^+(f,f)$ which is commonly referred to as the Wild convolution, where $Q^+(\cdot,\cdot)$ is some bilinear collision operator. If we define a map $f \mapsto \Phi(f)$ by $$\Phi(f)(t) = \mathrm{e}^{-t}F + \int_0^t \mathrm{e}^{-(t-s)}f\circ f(s)\mathrm{d}s \label{2}\tag{2},$$ then $f(t)$ solves the Boltzmann equation \eqref{1} exactly when $\Phi(f) = f$. Thus, to construct a solution to \eqref{1} with $f(t=0) = F$, the aforementioned paper suggests that we put $f_{(0)} = 0$, and then define $f_{(j+1)} = \Phi\left(f_{(j)}\right)$ for all $j\geq 1$. It is easy to check that $f_{(1)}(t) = \mathrm{e}^{-t}F$ and $$f_{(2)}(t) = \mathrm{e}^{-t}F + \int_0^t \mathrm{e}^{-(t-s)}f_{(1)}\circ f_{(1)}(s)\mathrm{d}s = \mathrm{e}^{-t}F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})F\circ F.$$ However, the authors also computed $f_{(3)}$ and claimed that $$f_{(3)}(t) = \mathrm{e}^{-t}F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})F\circ F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})^2\left(\frac{1}{2}F\circ (F\circ F) + \frac{1}{2}(F\circ F)\circ F\right),$$ which confuses me a lot since I do not understand why in the expression for $f_{(3)}$ there is no term involving $(F\circ F)\circ (F\circ F)$. Any help is greatly appreciated!

Remark 1: Given Professor Terrence Tao's comment, it is **unlikely** that the authors made a typo in the computation of $f_{(3)}$, which in my personal opinion should also include a term $$\frac{1}{3}\mathrm{e}^{-t}(1-\mathrm{e}^{-t})^3 (F\circ F)\circ (F\circ F) \label{3}\tag{3} $$ I reach out to this claim based on a statement in the same paper (near formula (1.15) in the version provided in this post) saying that ``McKean's expression for $f_{(n)} - f_{(n-1)}$ is (1.15)'', but the right side of (1.15) only involves the $n$-fold Wild convolution $Q^+_n(F)$. This means that if the formula for $f_{(3)}$ indeed includes the term \eqref{3} which I think the authors have somehow missed, then $f_{(3)}$ (with $n=3$) will involve $(F\circ F)\circ (F\circ F)$, which is a member of $Q^+_4(F)$. At this point I am super confused... By the way, I also need to mention that McKean's paper are usually written in the 1950s or 1960s and his papers are usually pretty short (or say "condensed") and hard to digest in my personal opinion...

Remark 2: Professor Terrence Tao's answer is very helpful, but my concern is from a different perspective. It is a classical result (see for instance Villani's monograph)) that a semi-explicit representation formula for solutions of the Boltzmann equation \eqref{1} can be written as $$f_t = \mathrm{e}^{-t}\sum_{n=1}^\infty \left(1-\mathrm{e}^{-t}\right)^{n-1} Q^+_n(f_0) \label{4}\tag{4} $$ where the $n$-fold nonlinear operator $Q^+_n$ is defined recursively by $$Q^+_1(f_0) = f_0,\quad Q^+_n(f_0) = \frac{1}{n-1}\,\sum_{k=1}^{n-1} Q^+\left(Q^+_k(f_0),Q^+_{n-k}(f_0)\right).$$ Thus, using Professor Terrence Tao's notation, we have that $$f_t = \lim_{n \to \infty} f_{[n]} = \lim_{n \to \infty} \mathrm{e}^{-t}\sum_{k=1}^n \left(1-\mathrm{e}^{-t}\right)^{k-1} Q^+_k(f_0) .$$ From a motivation to construct the solution formula \eqref{4} instead of checking directly that \eqref{4} indeed solves \eqref{1}, we would proceed as in the Wild's iterative scheme $f_{(j+1)} = \Phi(f_{(j)})$ described before. What is funny is that we have $f_{(0)} = f_{[0]} = 0$, $f_{(1)} = f_{[1]}$, $f_{(2)} = f_{[2]}$, while $f_{(3)} > f_{[3]}$. In some heuristic sense, it is not clear at all as to how to arrive at the "ansatz" (or say "guess") \eqref{4} at the first place if we just want to set/take $$f_t := \lim_{n \to \infty} f_{[n]}.$$ By the way, I tried email Professor Eric Carlen some year ago asking about a (different) question I have regarding one of his papers and I never got replied (which occurs very often in general I guess), so I would not try to bother him again.