Let $X_1,X_2,\ldots,X_n$ be i.i.d random variables having density

$$f(x)=2(1-x)\mathbf1_{0<x<1}$$

I am trying to derive the distribution of the sample range $R=X_{(n)}-X_{(1)}$.

The usual way I do these problems is to first find the joint density of $(R,S)$ taking $S=X_{(1)}$, and then find the distribution of $R$ as a marginal density. This is quite straightforward in general because we know the joint distribution of $(X_{(1)},X_{(n)})$. For this particular problem however, the integration to find the marginal pdf is pretty cumbersome to evaluate by hand.

For absolutely continuous distributions, it is easily shown via a change of variables that the joint density of $(R,S)$ is given by

$$f_{R,S}(r,s)=n(n-1)(F(r+s)-F(s))^{n-2}f(s)f(r+s)\mathbf1_{s<r+s}$$

, where $F$ is the population distribution function.

So here I have after simplification

$$f_{R,S}(r,s)=4n(n-1)(r(2-2s-r))^{n-2}(1-s)(1-r-s)\mathbf1_{0<s<r+s<1}$$

That means the pdf of $R$ for $0<r<1$ should be

\begin{align} f_R(r)&=\int_0^{1-r}f_{R,S}(r,s)\,ds \\&=4n(n-1)r^{n-2}\int_0^{1-r}(2-2s-r)^{n-2}(1-s)(1-r-s)\,ds \end{align}

Now I integrate by parts $$I=\int_0^{1-r}(2-2s-r)^{n-2}(1-s)(1-r-s)\,ds$$

noting that $$d\,[(1-s)(1-r-s)]=(2s+r-2)\,ds$$

Skipping some details, I get

\begin{align} I&=\left[(1-s)(1-r-s)\frac{(2-2s-r)^{n-1}}{2(1-n)}\right]_0^{1-r}+\int_0^{1-r}\frac{(2-2s-r)^n}{2(1-n)}\,ds \\\\&=\frac{(r-1)(2-r)^{n-1}}{2(1-n)}-\frac{1}{4(1-n)}\int_{2-r}^{r}t^n\,dt \\\\&=\frac{(r-1)(2-r)^{n-1}}{2(1-n)}+\frac{1}{4(n^2-1)}\left[r^{n+1}-(2-r)^{n+1}\right] \end{align}

It might not seem so, but doing this by hand and writing down every step took a fair amount of time.

Finally, I get the pdf of $R$ as

$$f_R(r)=4n(n-1)r^{n-2}\left[\frac{(r-1)(2-r)^{n-1}}{2(1-n)}+\frac{1}{4(n^2-1)}\left\{r^{n+1}-(2-r)^{n+1}\right\}\right]\mathbf1_{0<r<1}$$

Honestly, after the tedious computation, I don't know if I want to check that this integrates to $1$ or not (without using software that is). So I don't know if this answer even makes sense.

I would like to know of any alternative procedure to solve the problem, and perhaps a more efficient way. I think the CDF method results in almost the same complexity.

  • 2
    I can confirm the same result using mathStatica (so am confident your working is correct). – wolfies Nov 6 at 16:00
  • 3
    About the only simplification I can suggest--and it's truly a tiny one--is to recognize that the operation $X\to 1-X$ preserves the range while converting the density into $2x\mathcal{I}(0\lt x\lt 1).$ This makes the integrations ever so slightly easier. Asymptotic expressions are readily available, though. – whuber Nov 7 at 14:52

edit: please stop down voting this comment, it was a comment and should not have been accepted as the answer by the OP. Please read the comments and read the literature if you are unfamiliar with this technique. It seems that the relationship between the uniform distribution and beta distribution of it's ordered statistics is not taught/well understood.

I'm guessing you've realized the range is distributed as a form of the beta density? Convert to Uniform if you don't like working with Beta. My advice is don't integrate it out, just think about what form this resembles, it may not have a closed form if involves the incomplete beta function but is not really necessary to find the distribution.

  • 3
    I believe your advice is incorrect--perhaps it is based on misreading the question. If, after re-reading the question, you believe your approach is right, then please prove me wrong by giving us an explicit answer. – whuber Nov 7 at 17:48
  • Why did you delete the comments with the links to the answer to this question? Stop down voting my response. And OP should not have accepted this comment as the answer. This is literally the same question as the link I posted. Your advice to preserve the range is literally dividing by the scale parameter so you can invoke the relationship between Uniform and Beta densities . This is a Beta distribution...if you look at the links I posted its very obvious – Rosalie Nov 7 at 19:20
  • here's the wikipedia page for the beta distribution... "The beta distribution has an important application in the theory of order statistics. A basic result is that the distribution of the kth smallest of a sample of size n from a continuous uniform distribution has a beta distribution.[51] This result is summarized as: $U_{(k)}$~$\beta$(k, n+1-k) From this, and application of the probability integral transform, the distribution of any individual order statistic from any continuous distribution can be derived." en.wikipedia.org/wiki/Beta_distribution – Rosalie Nov 7 at 19:31
  • 3
    Thank you--but that doesn't address the question, which concerns the range (which is a linear combination of correlated order statistics) of a particular underlying distribution. As the question makes clear, an exact formula for the distribution of the range is available in this case--but the approach you describe does not appear to be a legitimate way to derive that distribution. – whuber Nov 7 at 19:56
  • 1
    You make the mistake of assuming that your many downvoters do not understand this material. I suspect they do, and regardless of how you feel, you should do them the courtesy of assuming as much. I would reiterate my invitation (in my original comment) to support your claims with a real demonstration of an explicit result rather than merely making assertions that your point of view is "easily" shown. – whuber Nov 9 at 13:31

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