Solution. The map \(y = -2\log x\) is decreasing on \((0,1)\) with inverse \(x = e^{-y/2}\) and \[\left |\frac {dx}{dy}\right | = \frac 12 e^{-y/2}.\] Since \(f_X(x)=1\) on \((0,1)\) and \(x\in (0,1)\) corresponds to \(y>0\), \[f_Y(y) = 1\cdot \frac 12 e^{-y/2} = \frac 12 e^{-y/2},\qquad y>0 ,\] which is EXP\(\left (\tfrac 12\right )\), equivalently \(\chi ^{2}_{(2)}\). This is the step behind the Box–Muller construction of the simulation chapter.

Solution. The map is increasing with \(x = y^{1/3}\) and \(\dfrac {dx}{dy} = \tfrac 13 y^{-2/3}\). Hence for \(0<y<1\), \[f_Y(y) = 3\left (y^{1/3}\right )^{2}\cdot \frac 13 y^{-2/3} = 3y^{2/3}\cdot \frac 13 y^{-2/3} = 1 ,\] so \(Y\sim \) UNIF\((0,1)\). This is the probability integral transform in action: \(F_X(x) = x^{3}\), so \(Y = F_X(X)\), and the answer had to be uniform.

Solution.

The maximum

\(P\left (X_{(n)}\leq t\right ) = \left (t/\theta \right )^{n}\) for \(0<t<\theta \), so differentiating, \[f_{X_{(n)}}(t) = \frac {n\,t^{n-1}}{\theta ^{n}},\qquad 0<t<\theta .\]

The range

The joint density of the extremes is \[f(u,v) = n(n-1)\left [F(v)-F(u)\right ]^{n-2}f(u)f(v) = \frac {n(n-1)}{\theta ^{n}}\left (v-u\right )^{n-2},\qquad 0<u<v<\theta .\] Putting \(v = u+r\) and integrating out \(u\) over \(0<u<\theta -r\), \[f_R(r) = \frac {n(n-1)}{\theta ^{n}}\,r^{n-2}\int _0^{\theta -r}du = \frac {n(n-1)}{\theta ^{n}}\,r^{n-2}\left (\theta -r\right ),\qquad 0<r<\theta .\] As a check, \(E(R) = \theta (n-1)/(n+1)\), which is the value quoted in the completeness discussion of a related model, and it tends to \(\theta \) as \(n\) grows — the sample eventually spans the whole interval.