# Distribution of maximum over minimum of Weibull(alpha, 1)

Let the PDF of Weibull $(\alpha, 1)$ be

$$f_X (x) = \alpha x^{\alpha-1} \exp\left( - x^{\alpha}\right)$$

I know from probability integral transfrom that

$$X_{(1)}^{\alpha} \overset{d}{=} Z_{(1)} \\ \cdots \\ X_{(n)}^{\alpha} \overset{d}{=} Z_{(n)}$$

where $X_{(n)}$ denotes $n^{th}$ order statistic from Weibull samples, and $Z_{(n)}$ denotes $n^{th}$ order statistic from samples of $\operatorname{Exp}(1)$.

Can anybody derive the PDF of $X_{(n)}/X_{(1)}$, without using the change-of-variables technique, with which I keep failing?

• Your formula for $f(x)$ is incorrect; the last term should be $\exp\{-x^{\alpha}\}$. Would that fix the problems with your derivation for you? – jbowman Jul 18 '18 at 0:53
• @jbowman oh thanks, I simply mistyped the PDF. Still failing to solve it. – moreblue Jul 18 '18 at 1:05
• Maybe this post would be of help: stats.stackexchange.com/q/306294/119261. – StubbornAtom Jul 18 '18 at 6:09
• Using change of variables, I am faced with the integral $\int_0^\infty s^{2\alpha-1}(e^{-s^\alpha}-e^{-r^\alpha s^\alpha})^{n-2}e^{-s^\alpha(1+r^\alpha)}\,ds$ where $r>1, \alpha>0$ to get the marginal pdf of $R=X_{(n)}/X_{(1)}$. Plugging the integral directly into Mathematica (additionally setting $n\ge 2$) gives a rather ugly answer in terms of non-elementary functions. – StubbornAtom Jul 18 '18 at 14:56

I am proceeding with the change of variables technique because it is straightforward. The only catch is the integral to find the required distribution as a marginal density, but that can be taken care of. If this is what you tried to do on your first attempt, I think you can finish the derivation now if you stick with it.

Restating known facts:

The population density is

$$f(x)=\alpha x^{\alpha-1}e^{-x^{\alpha}}\mathbf1_{x>0}\,,\quad\alpha>0$$

Corresponding distribution function is $$F(x)=1-e^{-x^{\alpha}}\mathbf1_{x>0}$$

Joint pdf of $(X_{(1)},X_{(n)})=(U,V)$ is

$$f_{U,V}(u,v)=n(n-1)\alpha^2\left(e^{-u^\alpha}-e^{-v^\alpha}\right)^{n-2}(uv)^{\alpha-1}e^{-u^{\alpha}-v^{\alpha}}\mathbf1_{0<u<v}$$

Changing variables $(U,V)\to(R,S)$ such that $R=V/U$ and $S=U$.

So joint density of $(R,S)$ is

$$f_{R,S}(r,s)=n(n-1)\alpha^2r^{\alpha-1}s^{2\alpha-1}\left(e^{-s^\alpha}-e^{-r^\alpha s^\alpha}\right)^{n-2}e^{-s^{\alpha}(1+r^{\alpha})}\mathbf1_{r>1,s>0}$$

At this point, I took some help from our maths site to simplify the ensuing integral in terms of elementary functions. Now that I see it, the simplification is obvious.

For $r>1$, marginal density of $R$ is

\begin{align} f_{R}(r)&=n(n-1)\alpha^2r^{\alpha-1}\int_0^\infty s^{2\alpha-1}\left(e^{-s^\alpha}-e^{-r^\alpha s^\alpha}\right)^{n-2}e^{-s^{\alpha}(1+r^{\alpha})}\,ds \\&=n(n-1)\alpha r^{\alpha-1}\int_0^\infty y\left(e^{-y}-e^{-r^\alpha y}\right)^{n-2}e^{-(1+r^\alpha)y}\,dy\qquad[\,s\mapsto s^\alpha=y\,] \\&=n(n-1)\alpha r^{\alpha-1}\sum_{j=0}^{n-2}\binom{n-2}{j}(-1)^j\int_0^\infty ye^{-(r^\alpha (1+j)+n-j-1)y}\,dy \end{align}

That last step was just expanding the integrand using the binomial theorem.

Simplifying further we have the required density:

$$f_R(r)=n(n-1)\alpha r^{\alpha-1}\sum_{j=0}^{n-2}\binom{n-2}{j}\frac{(-1)^j}{\left(n+(r^\alpha -1)(1+j)\right)^2}\mathbf1_{r>1}$$

I think this is about the best we can do without going into special functions. This is also pretty much the same as done in this answer to a similar question.

Maybe someone could give a verification of the above pdf.