Let $X_1,X_2, \ldots , X_n$ be independent and identically distributed Uniform random variables on the interval (0, a) for a > 0, each having a density function $f(x) = \frac{1}{a}$, $0<x<a$. Let $X_{(1)},X_{(2)}, \ldots , X_{(n)}$ denote the order statistics. The range of the data is defined as $R = X_{(n)}−X_{(1)}$ and the midrange is defined as $V = \frac{1}{2}(X_{(1)}+X_{(n)})$ Derive the joint distribution of $(R, V )$ and deduce the marginal distributions of $R$ and $V.$


First I am attempting to generate the joint distribution of $(V,X_{(1)})$ and then integrate out the $X_{(1)}$.

For the joint distribution of $(V,X_{(1)})$ I have:

$$ f_{v,x_{(1)}}(v,x_{(1)})=2n(n-1)\int_{-\infty}^v [F(2v-x_{(1)})-F(x_{(1)})]^{n-2} f(2v-x_{(1)})f(x_{(1)}) \, dx_{(1)}$$

At this point I have no idea how to integrate out the $x_{(1)}$.

Edit: I have located the joint distribution of the range and mid-range:

$$ f_{r,v}(r,v) = n(n-1) \left[F\left(v+\frac{r}{2}\right) - \left(F\left(v-\frac{r}{2}\right)\right)\right]^{n-2} f\left(v-\frac r 2 \right) f(v+\frac{r}{2})$$

For each marginal distribution, I have to integrate out:

$$ f_r(r) = \int_{-\infty}^\infty n(n-1)[F(v+\frac{r}{2})-(F(v-\frac{r}{2})]^{n-2}f(v-\frac{r}{2})f(v+\frac{r}{2}) \, dv$$

$$ f_v(v) = \int_0^{\infty}n(n-1)[F(v+\frac{r}{2})-(F(v-\frac{r}{2})]^{n-2}f(v-\frac{r}{2})f(v+\frac{r}{2}) \, dr$$

  • 1
    $\begingroup$ Since $(R,V)$ is a linear transform of $U=(X_{(1)},X_{(n)})$, it is enough to determine he distribution of $U$. And this distribution is well known... $\endgroup$
    – Did
    Sep 24, 2014 at 6:16
  • $\begingroup$ I have added the joint density. Now I just need to figure out how to integrate v and r to get the marginal densities. $\endgroup$
    – statsguyz
    Sep 25, 2014 at 0:29
  • $\begingroup$ Let $n=4.$ Then the standard deviation of the midrange of three observations is about 0.158, while the SD of their sample mean is larger: $\sqrt1/36) =0.167.$ Thus, the midrange is a better estimator of the population mean than is the sample mean. Perhaps see this Q&A, if it still exists. $\endgroup$
    – BruceET
    Aug 16, 2021 at 16:24

1 Answer 1


Solution from Casella and Berger:

Joint density of $X_{(1)},X_{(n)}$:

$\Large f_{X_{(1)}, X_{(n)}}(x_1,x_n) = \frac{n(n-1)}{a^2}(\frac{x_n}{a}-\frac{x_1}{a})^{n-2}$

We know that $X_{(1)} = V - R/2$ and $X_{(n)} = V + R/2$. The Jacobian is $|-1| = 1$.

The transformation maps {$(x_1,x_n): 0 < x_1< x_n <a$ } onto the set {$(r,v): 0 < r< a, r/2<v<a-r/2$ } .

Therefore, the joint pdf of (R,V) is:

$\Large f_{R,V}(r,v) = \frac{n(n-1)r^{n-2}}{a^n}$

The marginal pdf of R is:

$\Large f_R(r)= \int_{r/2}^{a-r/2}\frac{n(n-1)r^{n-2}}{a^n}dv$

= $\Large \frac{n(n-1)r^{n-2}(a-r)}{a^n}$, $0<r<a$.

If a = 1, then r has a beta(n-1,2) distribution.

The marginal pdf of V is:

$\Large f_V(v) = \int_0^{2v}\frac{n(n-1)r^{n-2}}{a^n}dr$

= $\Large \frac{n(2v)^{n-1}}{a^n}$, $0<v \leq a/2$,


$\Large f_V(v) = \int_0^{2(a-v)}\frac{n(n-1)r^{n-2}}{a^n}dr$

= $\Large \frac{n[2(a-v)]^{n-1}}{a^n}$, $a/2<v \leq a$.


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