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If the variables $alpha_1$...$alpha_n$ are distributed uniformly in $(0,1)$,

1. How do I show that the spread $alpha_{(n)}$ - $alpha_{(1)}$ has density $n (n-1) x^{n-2} (1-x)$ and expectation $(n-1)/(n+1)$?


2. What is the probability that all $n$ points lie within an interval of length $t$?

This is a (straight forward?) exercise is order statistics.

From Wikipedia, the pdf for the joint order statistic is:

$$ f_{ X_{(i)}, X_{(j)} }(u, v) = $$
$$ frac{ n! }{ (i-1)! (j-1-i)! (n-j)! } [ F_X(u) ]^{i-1} [F_X(v) - F_X(u)]^{j-1-i} [1 - F_X(v)]^{n-j} f_X(u) f_X(v) $$

(note the change in arguments)
Where $F_X(u)$ is the cumulative distribution function (or cdf) and $f_X(u)$ is the probability density function (or pdf) of the random variable $X$.

Plugging in $i=1$ and $j=n$ with $X = alpha$, $f_{alpha}(u) = 1$ and $F_{alpha}(u) = u$, we get:

$$ f_{ alpha_{(1)}, alpha_{(n)} }(s, s+t) = frac{ n! }{ (n-2)! } [ s+t - s ]^{n-2} = n (n-1) t^{n-2} $$

for a starting point $s in [0, 1-t] $ with interval length $t$. To find the probability of finding $alpha_{(1)}$ and $alpha_{(n)}$ within some interval of $t$, we need to integrate over all (permissible) starting positions, $s$:

$$ int_0^{1-t} f_{alpha_{(1)},alpha_{(n)}}(s, s+t) ds = int_{0}^{1-t} n (n-1) t^{n-2} ds = n (n-1) t^{n-2} int_0^{1-t} ds $$
$$ = n (n-1) t^{n-2} (1-t) $$

Which gives the first part of the answer for question 1. For the second part to question 1, you could integrate the above equation for $t in [0, 1]$ or you could just take the expectation of the difference of the random variable $alpha_{(n)}$ with $alpha_{(1)}$. I will do the latter. The pdf of the $k$'th order statistic is:

$$ f_{X_{(k)}}(u) = frac{ n! }{ (k-1)! (n-k)! } [F_X(u)]^{k-1} [1-F_X(u)]^{n-k} f_X(u) $$

and plugging in $X = alpha$, $k=1$ and $k=n$ for the 1st and $n$'th order statistic (of the uniform distribution) respectively, we get:

$$ f_{alpha_{(1)}}(u) = n (1 - u)^{n-1} $$
$$ f_{alpha_{(n)}}(u) = n u^{n-1} $$

and taking the expectation of their difference:

$$ E[ alpha_{(n)} - alpha_{(1)} ] = int_{0}^{1} u (n u^{n-1} - n (1 - u)^{n-1} ) du = frac{n}{n+1} - frac{1}{n+1} = frac{n-1}{n+1} $$

As desired. (note that $alpha_{(n)}$ and $alpha_{(1)}$ are not independent but that the expectation of the sums of random variables is still the same regardless).

To answer part 2, one must find $Pr{ alpha_{(n)} - alpha_{(1)} le t } $. We just derived the pdf of the joint distribution that the difference of $alpha_{(n)}$ and $alpha_{(1)}$ is of length $t$, so now we just have to sum over all lengths less than $t$:

$$ int_0^{t} n (n-1) x^{n-2} (1-x) dx = t^{n-1} ( n - (n-1) t) $$

I must admit that I just looked up the pdf of the joint order statistic. Could anyone tell me (or give a reference) on how to derive this equation?

Here's a hint: The answer to part 2 is quite easy. And if you consider it, you might see a trick on how to get the answer to part 1.

Here's approach for computing the density function that doesn't rely on knowledge of order statistics. First let $R = alpha_{(n)} - alpha_{(1)}$ be the range. We compute the cumulative distribution function $F_R(x)$ by conditioning on the event $alpha_{(n)} leq x$.

begin{align}
F_R(x) &= P{R leq x} \
&= P{R leq x text{ and } alpha_{(n)} leq x}+P{R leq x text{ and } alpha_{(n)} > x} \
&=P{R leq x text{ | } alpha_{(n)} leq x}P{alpha_{(n)} leq x}+P{R leq x text{ and } alpha_{(n)} > x} label{a}tag{1}
end{align}

To compute the first term in $ref{a}$, observe that if the largest value is less than $x$, then the range $R$ must be less than x, so we have

begin{align}
P{R leq x text{ | } alpha_{(n)} leq x} = 1 \
end{align}

Also,

begin{align}
P{alpha_{(n)}leq x} &= P{alpha_1 leq x, alpha_2 leq x,ldots,alpha_n leq x} \
&= P{alpha_1 leq x}P{alpha_2 leq x}cdots P{alpha_n leq x} \
&= x^n
end{align}

To finish our computation of $F_R$, we must compute $P{R leq x text{ and } alpha_{(n)} > x} $. Let $A$ be the event that $R leq x text{ and } alpha_{(n)} > x$. Let $A_i$ be the event that $alpha_i > x$ and $alpha_i - x leq alpha_j leq alpha_i$ for all $j neq i$. Clearly $P{alpha_i} > x = (1-x)$. Because the interval $[alpha_i - x,alpha_i]$ is of width $x$, we have $P{alpha_i - x leq alpha_j leq alpha_i} = x$, which implies

$$P{alpha_i - x leq alpha_j leq alpha_i text{ for all } j neq i } = x^{n-1}$$

Therefore

$$P{A_i}= (1-x)x^{n-1} $$

Notice that $A = bigcuplimits_{i=1}^n A_i$, and the events $A_i$ are mutually exclusive, so we have

$$P(A) = P{R leq x text{ and } alpha_{(n)} > x} = n(1-x)x^{n-1}$$

Substituting these results into $ref{a}$, we get

$$F_R(x) = x^n + n(1-x)x^{n-1}$$

Differentiation and a little algebra gives the desired density function

begin{align}
f_R(x) &= F_R'(x) \
&= n(n-1)x^{n-2}(1-x)
end{align}