$\newcommand{\al}{\alpha}$ 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 $$R_*:=\max_{1\le i\le n}R_i. $$ Fisher 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.

  • $\begingroup$ It seems to me that the "differential properties" for $f$ are just convenient ways of expressing certain algebraic equations for the $\alpha_i$ that Fisher is claiming must hold. $f=0$ says that $-1 + \alpha_1 + \dots + \alpha_n = 0$. $(t \frac{d}{dt})f = 0$ says that $\alpha_1 + 2\alpha_2 + \dots + n \alpha_n = 0$, and so on. How we know these are true, I don't yet see. But once they are, then it does follow that $f$ must be $f = -(1-t)^n$, and then we know exactly what its coefficients $\alpha_i$ are, from the binomial theorem. $\endgroup$ – Nate Eldredge Sep 12 at 2:05
  • $\begingroup$ @NateEldredge : Thank you for your comment. This is a part of the proof I do understand, that is, if a polynomial $f$ has these differential/algebraic properties and the free term of $f$ is $-1$, then $f$ must be $-(1-t)^n$ -- this is very easy. However, I don't see any hint by Fisher as to why $f$ has these differential properties. More importantly, I don't understand what $f$ has to do with anything else in the proof -- I don't see a word about that anywhere in Fisher's paper. $\endgroup$ – Iosif Pinelis Sep 12 at 12:22
  • $\begingroup$ Well, the goal of this step is to compute the coefficients $\alpha_i$ that appear in the function $P$. Fisher seems to be claiming that these coefficients satisfy a certain system of algebraic equations. Then $f$ is defined as it is so that this system of algebraic equations has a nice compact expression in terms of $f$ and its derivatives, and as we see, this leads to a slick way of solving the system. $\endgroup$ – Nate Eldredge Sep 12 at 13:22

$\newcommand{\al}{\alpha}$ 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.

  • $\begingroup$ $f$ is just an auxiliary polynomial used to determine the coefficients $\alpha_1,\ldots,\alpha_n$. For $k=0,\ldots,n-1$ $$\big((t\frac{d}{dt})^k f\big) (1)= (-1)^k\frac{(n-1)!}{(n-1-k)!}\big((\frac{d}{dg})^k P\big)(0)=0$$. $\endgroup$ – esg Sep 12 at 19:02
  • $\begingroup$ @esg : Thank you for solving the remaining piece of the puzzle. (Your displayed equalities hold only for $k=1,\dots,n-1$, but that is not a problem.) $\endgroup$ – Iosif Pinelis Sep 13 at 0:21

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