Let me preface this by saying that this is not a question about the Irwin-Hall distribution , at least directly. Rather, in the proof of Lemma 1 in the appendix of this paper , the author shows that the joint density of $S_n:= \sum_{I=1}^{n}X_i$ where $X_i\stackrel{\rm i.i.d}{\sim}U\left[0,1\right]$ is given by $$f_n(s) = \frac{s^{n-1}}{(n-1)!}P\left(\max_{1\leq i\leq n} L_n \leq s^{-1}\right)$$ where $\{L_i\}_{i=1}^n$ are the lengths of consecutive sub-intervals of $\left[0,1\right]$ cut out by $(n-1)$ uniform random numbers in the interval. I know that the the joint density of the first $(n-1)$ such lengths is $(n-1)!$ on its support. However, I failed to follow the argument of the author to its conclusion.
Specifically, he argues that the joint density of $S_n,X_1,\ldots,X_{n-1}$ is the same as the joint density of the $n$ independent uniform which is 1 on its support. Then, applying a transformation, the joint density of $S_n,\frac{X_1}{S_n},\ldots,\frac{X_{n-1}}{S_n}$ can be found to be $s^{n-1}$. I cannot follow the argument from this point on, where he presumably integrates out the other variables to derive $f(s)$. If anyone could clarify this step for me, it'd be great.
Best Answer
Ok so this answer basically avoids what Pittel does (maybe that is easier but it's not obvious to me), and follows Feller's approach.
We have $X_1$, $\cdots$, $X_n$ independent uniform random variables on $[0,a]$. Let $S_n$ be their sum. We want the density of $S_n$, let's call that $f_{n,a}$. Obviously $f_{1,a}$ is $\frac{1}{a}$ on $[0,a]$ and $0$ on the rest of the positive real line. And the rest of the $f_{n,a}$ obey the convolution recurrence, so for $n \geq 2$,
$$f_{n,a}(x) = \frac{1}{a} \int_0^a f_{n-1,a}(x-y) \, dy.$$
Let's set this aside and turn to the other item Pittel uses. Say we partition $[0,t]$ into n intervals using independent uniform random points $Y_1, \cdots, Y_{n-1}$. We want to track the probability that none of the subintervals are larger than $a$, i.e. $P(\max_{1 \leq i \leq n-1} \left(Y_{i+1} - Y_i\right) \leq a)$. Call this thing $\varphi_{n,a}(t)$. By this definition, $\varphi_{1,a}(t)$ is clearly $1$ for $0 \leq t \leq a$, and $0$ on the rest of the positive real line.
What about for larger $n$? We have $n-1$ choices for a random point $Y_i$. Once we choose, call its position $x$. Then the probability that the $Y_i$ we chose is the leftmost one is $\left(\frac{t-x}{t}\right)^{n-2}$.
Also, this leftmost point must be less than $a$. Now the remaining variables are distributed with the same conditions over $[x,t]$, so the conditional probability they satisfy the requisite conditions is $\varphi_{n-1,a}(t-x)$. Integrate this over the possible $x$ and get
$$\varphi_{n,a}(t) = (n-1) \int_0^a \varphi_{n-1,a}(t-x) \left(\frac{t-x}{t}\right)^{n-2} \, \frac{1}{t}dx.$$
Now set
$$u_{n,a}(t) = \frac{\varphi_{n,a}(t)t^{n-1}}{a^{n-1}(n-1)!},$$
and observe for $n \geq 2$,
$$ u_{n,a}(t) = \frac{1}{a} \int_0^a u_{n-1,a}(t-x) \, dx.$$
This matches our recurrence above and the initial condition matches as well when $a=1$. So $u_{n,1}$ is our desired density function. As the problem requires, we have $a=1$, then put $t = s$ to get what Pittel has:
$$ u_{n,1}(s) = \frac{s^{n-1}}{(n-1)!} \varphi_{n,1}(s),$$
where the $\varphi$ term is what you want: the probability that no sub-interval exceeds length $1$ when $[0,s]$ is cut up into $n$ sub-intervals, or after rescaling, the probability that no sub-interval exceeds length $\frac{1}{s}$ when $[0,1]$ is cut up into $n$ sub-intervals.