Trying to understand Fisher's proof

Question

$\newcommand{\al}{\alpha}$ For $i=1,\dots,n$, let \begin{equation} R_i:=\frac{X_i}{X_1+\dots+X_n}, \end{equation} where the $X_i$'s are iid standard exponential random variables. Let $$R_*:=\max_{1\le i\le n}R_i. $$ Fisher¹ gave the formula \begin{equation} P(R_*>x)=\sum_{j=1}^n(-1)^{j-1}\binom nj(1-jx)_+^{n-1} \end{equation} for $x\in(0,1)$ (using somewhat different notation), where $u_+:=\max(0,u)$. I have a proof of this result and a certain generalization of it.

My problem is that I understand almost nothing in Fisher's proof (on pages 57--58 of his paper). In particular, I don't understand the following:

1. What does (the polynomial (?)) $f$ in $t$ (introduced (?) on page 57 of Fisher's paper) have to do with the spline (?) $\text{P}$ in $g$;
2. Why does $f$ have to have the differential properties in a neighborhood of $t=1$ that Fisher says $f$ has to have?
3. How does Fisher make the jump from those properties of $f$ to the (correct) final expression for $\text{P}$? Fisher seems to provide absolutely no details on this.

I will appreciate any help in filling these huge gaps in my understanding.

¹Fisher, R. A., Tests of significance in harmonic analysis., Proceedings Royal Soc. London (A) 125, 54-59 (1929). ZBL55.0950.16, MR2079.

Answer 1

$\newcommand{\al}{\alpha}$ I seem to finally get it. Fisher denotes $P(R_*>x)$ by $\text{P}$ (also using $g$ in place of $x$) and calls this probability "the probability integral". He says:

We may therefore represent the probability integral by the form
$$\text{P}=\al_1(1-g)^{n-1}+\al_2(1-2g)^{n-1}+\ldots+\al_n(1-ng)^{n-1}, $$ in which as many terms are to be taken as have positive quantities within the brackets. The last term is therefore included for no possible value of $g$, but is written above in order to utilise the condition that when $g<1/n$ the probability integral shall be unity. This condition is sufficient to determine the $n$ coefficients by equation of the coefficients of $g^0,g^1,\ldots,g^{n-1}$.

In other words, Fisher says that $\text{P}$ must be of the form $$\text{P}=\sum_{j=1}^n \al_j(1-jg)_+^{n-1}, $$ and the coefficients $\al_j$ can be determined by the condition that $\text{P}=1$ when $g<1/n$.

This is a clever observation. Indeed, for $g<1/n$, $\text{P}$ coincides with the genuine polynomial $\sum_{j=1}^n \al_j(1-jg)^{n-1}$, and this polynomial must be identically equal to $1$ for $g\in(0,1/n)$, by the probability meaning of $\text{P}$. This will require that, in particular, all the derivatives of $\text{P}$ in $g$ of orders $\ge1$ be identically $0$, which will of course uniquely determine the $\al_i$'s.

It is still unclear to me what $f$ has to do with $\text{P}$, but this now does not seem to matter much.