Proof. The map from \((x_1,\dots ,x_n)\) to its sorted version is \(n!\)-to-one: each of the \(n!\) orderings of a sample with distinct values gives the same ordered vector. On each of the \(n!\) regions \(x_{\pi (1)}<\cdots <x_{\pi (n)}\) the map is a permutation of coordinates, with Jacobian \(\pm 1\), and the joint density of the sample, \(\prod f(x_i)\), takes the same value \(\prod f(y_i)\) at every preimage. Adding the \(n!\) contributions gives the result. □

Solution.

(i)

By Theorem 1.2.1, \[g(y_1, y_2, y_3) = 3!\, e^{-y_1}e^{-y_2}e^{-y_3} = 6 e^{-(y_1 + y_2 + y_3)}, \qquad 0 < y_1 < y_2 < y_3 < \infty .\]

(ii)

Integrate out \(y_3\) from \(y_2\) to \(\infty \), then \(y_2\) from \(y_1\) to \(\infty \): \begin {align*} g_1(y_1) &= 6e^{-y_1} \int _{y_1}^{\infty } e^{-y_2}\int _{y_2}^{\infty } e^{-y_3}\, dy_3 \, dy_2 = 6e^{-y_1} \int _{y_1}^{\infty } e^{-2y_2}\, dy_2\\ &= 6e^{-y_1}\cdot \frac {1}{2}e^{-2y_1} = 3e^{-3y_1},\qquad y_1 > 0 . \end {align*}

So the minimum of three exponential lifetimes with mean \(1\) is exponential with mean \(\frac 13\).

(iii)

Integrate out \(y_1\) from \(0\) to \(y_2\), then \(y_2\) from \(0\) to \(y_3\): \begin {align*} g_3(y_3) & = 6e^{-y_3} \int _{0}^{y_3} e^{-y_2}\int _0^{y_2}e^{-y_1}\,dy_1\,dy_2 = 6e^{-y_3} \int _{0}^{y_3} (e^{-y_2} - e^{-2y_2})\, dy_2\\ & = 6e^{-y_3} \left ( \frac {1}{2} - e^{-y_3} + \frac {1}{2}e^{-2y_3} \right ) = 3e^{-y_3} - 6e^{-2y_3} + 3e^{-3y_3},\qquad y_3 > 0 . \end {align*}

Proof. The maximum is at most \(y\) if and only if every observation is at most \(y\), so by independence \[F_{Y_n}(y) = P(X_1 \leq y,\ \dots ,\ X_n \leq y) = \prod ^n_{i = 1}P(X_i\leq y) = [F(y)]^n .\] The minimum exceeds \(y\) if and only if every observation exceeds \(y\), so \[F_{Y_1}(y) = 1 - P(X_1 > y,\ \dots ,\ X_n > y) = 1 - [1 - F(y)]^n .\] Differentiating gives the p.d.f.s. □

Solution. Here \(F(y) = \frac {y}{\theta }\) on \((0,\theta )\), so \(g_n(y) = \frac {ny^{n-1}}{\theta ^n}\) for \(0<y<\theta \), and \[E(Y_n) = \int _0^\theta y\,\frac {ny^{n-1}}{\theta ^n}\,dy = \frac {n}{n + 1}\theta .\] So \(Y_n\) underestimates \(\theta \) on average, as it must since \(Y_n<\theta \), and \(\frac {n+1}{n}Y_n\) has mean exactly \(\theta \); it is used as an estimator in the chapter on estimation.

Proof. By the multinomial argument above, \begin {align*} P(y<Y_k\leq y + h) ={}& \frac {n!}{(k - 1)!(n - k)!}[F(y)]^{k-1}[1 - F(y + h)]^{n-k}\\ &\times \left (F(y + h) - F(y)\right ) + o(h), \end {align*}

the remainder accounting for two or more observations in the interval. Dividing by \(h\) and letting \(h\to 0\) gives \(g_k(y)\). □

Proof. \(Y_k\leq y\) exactly when at least \(k\) of the \(n\) observations are at most \(y\). The number of observations at most \(y\) is binomial with parameters \(n\) and \(F(y)\), so \(P(Y_k\leq y)\) is the probability that this binomial variable is at least \(k\). □

Solution. The population c.d.f. is \(F(x) = x^2\) for \(0<x<1\) (with \(F = 0\) for \(x\leq 0\) and \(F = 1\) for \(x\geq 1\)). By Theorem 1.2.7, for \(0<y<1\), \[G_2(y) = \binom {3}{2}(y^2)^2(1 - y^2) + \binom {3}{3}(y^2)^3 = 3y^4(1 - y^2) + y^6 = y^4(3 - 2y^2) ,\] with \(G_2 = 0\) for \(y\leq 0\) and \(G_2 = 1\) for \(y\geq 1\). The p.d.f. can be found in two ways:

• differentiating the c.d.f., \(g_2(y) = \frac {d}{dy}(3y^4 - 2y^6) = 12y^3 - 12y^5 = 12y^3(1 - y^2)\);
• from Theorem 1.2.5 with \(n = 3\), \(k = 2\), \(g_2(y) = \frac {3!}{1!\,1!} (y^2)(1 - y^2)(2y) = 12y^3(1 - y^2)\),

both for \(0 < y < 1\).

Solution. By Theorem 1.2.5 with \(n = 5\), \(k = 3\) and \(F(y) = y\), \(g_3(y) = 30y^2(1 - y)^2\), a beta density. So \[P(0.4<Y_3<0.6) = 30\int _{0.4}^{0.6}y^2(1 - y)^2\,dy = 30\left [\frac {y^3}{3} - \frac {y^4}{2} + \frac {y^5}{5}\right ]_{0.4}^{0.6} \approx 0.365 .\]

Proof. One observation must be near \(y_1\), one near \(y_n\), and the other \(n - 2\) between them, each with probability \(F(y_n) - F(y_1)\). There are \(n(n - 1)\) ways to choose which observations are the minimum and the maximum. □

Solution. By Theorem 1.2.10, \(g(y_1,y_n) = n(n - 1)(y_n - y_1)^{n-2}\) for \(0<y_1<y_n<1\). Change variables to \(R = Y_n - Y_1\) and \(S = Y_1\), with Jacobian \(1\) (the method is justified in the next section). For fixed \(r\), \(s\) runs from \(0\) to \(1 - r\), so \[f_R(r) = \int _0^{1 - r}n(n - 1)r^{n-2}\,ds = n(n - 1)r^{n-2}(1 - r),\qquad 0<r<1 ,\] a beta distribution with parameters \(n - 1\) and \(2\), and \(E(R) = \frac {n - 1}{n + 1}\).