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Let $U_{1}, \, ... \, ,U_{n}$ be a random sample of uniform random variables $U_i \sim \mathrm{Uniform}(0,1)$. Let $U_{(1)}, \, ... \, , U_{(n)}$ be the order statistics of the sample. The goal is to prove that:

$$ W = U_{(s)}-U_{(r)} \sim \textrm{Beta}(s-r, \, n - s + r +1) \qquad 1 \leq r < s \leq r $$

My "proof"

The joint pdf of $U_{(r)}, U_{(s)}$ is:

$$ f_{U_{(r)}, U_{(s)}} (u, v) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} u^{r-1} (v-u)^{s-1-r} (1 -v)^{n-s} \cdot \textbf{1}_{\{u < v\}} $$

Consider the transformation:

$$ \begin{Bmatrix} W = U_{(s)} - U{(r)} & U_{(r)} = Z \\ Z = U_{(r)} & U_{(s)} = W + Z \\ \end{Bmatrix} $$

The absolute value of the Jacobian determinant is 1. Therefore, the joint pdf of the transformation is:

$$ f_{W,Z}(w,z) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} z^{r-1} w^{s-1-r} (1 - w - z)^{n-s} $$

$$ f_W(w) = \int_{S} f_{W,Z}(w,z) \, \textrm{d}z $$

For fixed $w$, we have $S = \{z \, : \, 0 \leq z \leq 1 - w\}$. Thus:

$$ f_W(w) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} w^{s-1-r} \int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z $$

The problem is that I do not know how to evaluate the integral, but Wolfram Alpha does:

$$ \int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z= \frac{\Gamma(r)\Gamma(n-s+1)}{\Gamma(n+r-s+1)} (1- w)^{n+r-s} $$

Now it is really easy to figure out the distribution of $W$. Indeed, the only thing that has to be done is to rewrite the factorials using gamma functions.

At first, I was going to ask how to evaluate the integral without the help of Wolfram Alpha, but then I realized that maybe there is a "better" proof that avoids it. I have tried to find an alternative proof by using multiple transformations which involved $W$ and another transformation, but it was useless.

Edit

I have corrected some mistakes I think I made:

1) The joint pdf I wrote was incorrect.

2) I wrote that I the pdf of $W$ could be found as a convolution, but $U_{(r)}, U_{(s)}$ are clearly dependent. What I actually did was an implicit change of variables.

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1 Answer 1

up vote 2 down vote accepted

One does not need to evaluate the integral. The change of variables $u=(1-w)v$ yields that $f_W(w)$ is a constant factor $C$ times $w^{s-1-r}$ times $(1-w)^{n+r-s}$. This also yields the value of the constant factor $C$ since $f_W$ must integrate to $1$.

(Or, after the change of variables, one can recognize the integral over $v$ from $0$ to $1$ as a Beta.)

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Thank you (yet) again! –  Shiwen Yao Sep 30 '11 at 15:51
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