Proof. We note that there are \(n!\) elements in \(\mathcal {B}\). We need to show that \(R\) assumes each of the permutations of \((1,\, 2,\, \cdots ,\, n)\) with probability \(\dfrac {1}{n!}\).

Let \(r = (r_1, \, r_2, \, \cdots , \, r_n)\) be an arbitrary element in \(\mathcal {B}\). Then \begin {align*} P_{pro}(R = r) & = P\left ((X_1,\, X_2,\, \cdots ,\, X_n) = (X_{(r_1)}, \, X_{(r_2)},\, \cdots \,,\,X_{(r_n)}\right )\\ & = P(X_{d_1} < X_{d_2} < X_{d_3} < \cdots < X_{d_n}) \end {align*}

where \(d_i\) is the position of the number \(i\) in the permutation \(r\), for \(i = 1, \, 2, \, \cdots , \, n\).
Note that if \(Y_1, \, Y_2,\, \cdots \, ,\, Y_m\) are \(iid\) random variables and \((\alpha _1, \, \alpha _2, \, \cdots \,, \, \alpha _m)\) is any permutation of the integers \((1, \, 2,\, \cdots \, , \, m)\) \[\left (Y_1, \, Y_2, \, \cdots \cdots \, , \, Y_m\right )\, \overset {d}{=}\, \left (Y_{\alpha _1}, \, Y_{\alpha _2}, \, \cdots \cdots \, , \, Y_{\alpha _m}\right )\] Therefore \(\,\left (X_1, \, X_2, \, \cdots \cdots \, , \, X_m\right )\, \overset {d}{=}\, \left (X_{d_1}, \, X_{d_2}, \, \cdots \cdots \, , \, X_{d_m}\right )\,\) which implies that \[P(X_{d_1} < X_{d_2} < X_{d_3} < \cdots < X_{d_n}) = P(X_1 < X_2 < X_3 < \cdots < X_n).\] Thus \begin {align*} P(R = r) & = P(X_{d_1} < X_{d_2} < X_{d_3} < \cdots < X_{d_n})\\ & = P(X_1 < X_2 < X_3 < \cdots < X_n)\\ & = P(R = r_0)\hspace {0.3cm}, \hspace {0.5cm}\text {where}\hspace {0.2cm} r_0 = (1, \, 2,\, \cdots \, , \, n) \end {align*}

since there are \(n!\) elements in \(\mathcal {B}\), the result follows since \(r\) is an arbitrary element in \(\mathcal {B}\). \[\sum _{r\in \mathcal {B}} P(R = r) = 1.\hspace {1.5cm} n!\, (a) = 1 \, \implies \, a = \frac {1}{n!}\] □

Proof. Recall that under \(H_0\) the rank vector \(R\) is uniformly distributed (theorem 2.3.2), the set of permutations of integers \((1,\, 2, \, \cdots \, , N)\) where \(N= n + m\). We note that under \(H_0: c = 0,\) \(\, \left (-X_1, \,-X_2, \, \cdots \, , \, -X_m, \, -Y_1, \, -Y_2, \, \cdots \, , \, -Y_n\right )\) are \(iid\) continuous random variables and so their rank vector is also uniformly distributed on \(\mathcal {B}\). However multiplying by \(-1\) makes the largest be the smallest and so forth, and hence the reverse the order of ranking. Therefore the rank vector of \(\left (-X_1, \,-X_2, \, \cdots \, , \, -X_m, \, -Y_1, \, -Y_2, \, \cdots \, , \, -Y_n\right )\) written in terms of the components of the original \(R\) is \[\left (N + 1 - Q_1, \, N + 1 - Q_2, \, \cdots \, , \, N + 1 - Q_m\, , \, N + 1 - R_1, \, N + 1 - R_2,\, \cdots \, , \, N + 1 - R_n\right ).\] Therefore \[ \left (Q_1,\, \cdots \, , \, Q_m\, , \, R_1,\, \cdots \, , \, R_n\right ) \, \overset {d}{=}\left (N + 1 - Q_1, \, \cdots \, , \, N + 1 -Q_m\, , \, N + 1 - R_1, \, \cdots \, , \,N + 1 - R_n\right )\] computing the rank sum statistic on each side and applying the theorem which says that if \(Z\overset {d}{=}V\) and \(K(\cdot )\) a measurable function then \(K(Z) \overset {d}{=} K(V)\). \[W_y = \sum ^n_{j = 1} R_j \overset {d}{=} \sum ^n_{j = 1} (N + 1 - R_j)\] \[W_y \overset {d}{=} n(N+ 1) - \sum ^n_{j = 1} R_j = n(N + 1) - W_y\] \[W_y \overset {d}{=} n(N+1) - W_y\] \begin {align*} W_y - \frac {n(N+1)}{2}\, & \overset {d}{=}\, n(N+1)- W_y - \frac {n(N+1)}{2}\\\\ W_y - \frac {n(N+1)}{2}\, & \,\overset {d}{=} \frac {n(N+1)}{2} - W_y \end {align*}

\(\implies \hspace {0.3cm} W_y\,\) is symmetric at \(\, \dfrac {n (N + 1)}{2}\). □