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 \begin {align*} 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 . \end {align*}

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.

Solution.

(a)

Let \(Y = \max \{X_1,X_2\}\). The maximum is at most \(y\) exactly when both variables are, so by independence \begin {align*} P(Y>y) &= 1-P(X_1\leq y,\ X_2\leq y) = 1-P(X_1\leq y)\,P(X_2\leq y)\\ &= 1-\left (1-e^{-\lambda _1y}\right )\left (1-e^{-\lambda _2y}\right )\\ &= e^{-\lambda _1y}+e^{-\lambda _2y}-e^{-(\lambda _1+\lambda _2)y},\qquad y>0 . \end {align*}

Hence \(F_Y(y) = 1-e^{-\lambda _1y}-e^{-\lambda _2y}+e^{-(\lambda _1+\lambda _2)y}\) and \[f_Y(y) = \lambda _1e^{-\lambda _1y}+\lambda _2e^{-\lambda _2y} -(\lambda _1+\lambda _2)e^{-(\lambda _1+\lambda _2)y}.\] Each term is an exponential density, so \(E(Y)\) is the corresponding combination of exponential means. For \(X\) exponential with parameter \(\lambda \), integrating by parts, \begin {align*} E(X) &= \lambda \int _0^{\infty }xe^{-\lambda x}\,dx\\ &= \lambda \left \{\left [-\frac {x}{\lambda }e^{-\lambda x}\right ]_0^{\infty } + \frac {1}{\lambda }\int _0^{\infty }e^{-\lambda x}\,dx\right \}\\ &= \lambda \cdot \frac {1}{\lambda ^{2}} = \frac {1}{\lambda }. \end {align*}

Therefore \[E\left (\max \{X_1,X_2\}\right ) = \frac {1}{\lambda _1}+\frac {1}{\lambda _2}-\frac {1}{\lambda _1+\lambda _2}.\]

(b)

Here \(f_Y(x) = \lambda _1e^{-\lambda _1x}\) and \(F_Y(x) = 1-e^{-\lambda _1x}\), while \(Z\) has shape \(2\) and rate \(\lambda _2\) (an Erlang variable): \[f_Z(x) = \lambda _2^{2}xe^{-\lambda _2x},\qquad F_Z(x) = 1-(1+\lambda _2x)e^{-\lambda _2x}.\] Let \(X = \max \{Y,Z\}\). As in (a), \begin {align*} P(X\leq x) &= P(Y\leq x)\,P(Z\leq x) = \left (1-e^{-\lambda _1x}\right )\left (1-(1+\lambda _2x)e^{-\lambda _2x}\right )\\ &= 1-e^{-\lambda _1x}-(1+\lambda _2x)e^{-\lambda _2x} +(1+\lambda _2x)e^{-(\lambda _1+\lambda _2)x}. \end {align*}

Differentiating, \begin {align*} f_X(x) &= \lambda _1e^{-\lambda _1x}+\lambda _2^{2}xe^{-\lambda _2x}\\ &\quad -\lambda _1e^{-(\lambda _1+\lambda _2)x} -\lambda _2(\lambda _1+\lambda _2)xe^{-(\lambda _1+\lambda _2)x}. \end {align*}

Using \(\int _0^{\infty }x^{n}e^{-\lambda x}\,dx = \frac {n!}{\lambda ^{n+1}}\), \begin {align*} E(X) &= \lambda _1\cdot \frac {1!}{\lambda _1^{2}} + \lambda _2^{2}\cdot \frac {2!}{\lambda _2^{3}}\\ &\quad - \lambda _1\cdot \frac {1!}{(\lambda _1+\lambda _2)^{2}} - \lambda _2(\lambda _1+\lambda _2)\cdot \frac {2!}{(\lambda _1+\lambda _2)^{3}}\\ &= \frac {1}{\lambda _1}+\frac {2}{\lambda _2} -\frac {\lambda _1}{(\lambda _1+\lambda _2)^{2}} -\frac {2\lambda _2}{(\lambda _1+\lambda _2)^{2}}\\ &= \frac {1}{\lambda _1}+\frac {2}{\lambda _2} -\frac {\lambda _1+2\lambda _2}{(\lambda _1+\lambda _2)^{2}}. \end {align*}