D 理论证明补充(*)

D.1 对立变量的补充证明

定理D.1 设\(f(x_1, \dots, x_n)\)和\(g(x_1, \dots, x_n)\)是关于每个自变量单调不减的函数， 随机变量\(X_1, \dots, X_n\)相互独立， 记\(\boldsymbol X = (X_1, \dots, X_n)\)， 则下面的不等式当其中期望存在有限时成立： \[\begin{align} E[f(\boldsymbol X) g(\boldsymbol X)] \geq E[f(\boldsymbol X)]\, E[g(\boldsymbol X)] . \tag{D.1} \end{align}\]

证明： 用数学归纳法。 当\(n=1\)时，对任意实数\(x, y\)有 \[ [f(x) - f(y)][g(x) - g(y)] \geq 0 . \] 于是对任意随机变量\(X, Y\)有 \[ [f(X) - f(Y)][g(X) - g(Y)] \geq 0 . \] 于是 \[ E \left\{ [f(X) - f(Y)][g(X) - g(Y)] \right\} \geq 0 . \] 当期望存在有限时有 \[ E[f(X) g(X)] + E[f(Y) g(Y)] \geq E[f(X) g(Y)] + E[f(Y) g(X)] . \] 设\(X, Y\)独立同分布，且涉及的期望存在有限， 则 \[\begin{aligned} E[f(X) g(X)] =& E[f(Y) g(Y)], \\ E[f(X) g(Y)] =& E[f(Y) g(X)] = E[f(X)] E[g(X)], \end{aligned}\] 从而有 \[\begin{aligned} E[f(X) g(X)] \geq E[f(X)] E[g(X)] . \end{aligned}\]

设定理结论对\(n-1\)个自变量的情形成立, 设\(f(x_1, \dots, x_n)\)和\(g(x_1, \dots, x_n)\)是关于每个自变量单调不减的函数， 则 \[\begin{aligned} & E[f(\boldsymbol X) g(\boldsymbol X) | X_n = x] \\ =& E[f(X_1, \dots, X_{n-1}, x) g(X_1, \dots, X_{n-1}, x) | X_n = x ] \\ =& E[f(X_1, \dots, X_{n-1}, x) g(X_1, \dots, X_{n-1}, x)] \quad(\text{由}X_n\text{与}X_1,\dots, X_n\text{的独立性}) \\ \geq& E[f(X_1, \dots, X_{n-1}, x)] E[g(X_1, \dots, X_{n-1}, x)] \\ =& E[f(X_1, \dots, X_{n-1}, x) | X_n = x] E[g(X_1, \dots, X_{n-1}, x) | X_n = x] \end{aligned}\] 所以有 \[\begin{aligned} E[f(\boldsymbol X) g(\boldsymbol X) | X_n] \geq E[f(\boldsymbol X) | X_n] E[g(\boldsymbol X) | X_n] . \end{aligned}\]

两边取期望得 \[\begin{aligned} E[f(\boldsymbol X) g(\boldsymbol X)] \geq E\left\{ E[f(\boldsymbol X) | X_n] E[g(\boldsymbol X) | X_n] \right\} . \end{aligned}\] 注意到\(E[f(\boldsymbol X) | X_n]\)和\(E[g(\boldsymbol X) | X_n]\)分别是关于\(X_n\)的一元函数， 且关于\(X_n\)是增函数， 由\(n=1\)时已证明的结论可知 \[\begin{aligned} & E\left\{ E[f(\boldsymbol X) | X_n] E[g(\boldsymbol X) | X_n] \right\} \\ \geq& E\left\{ E[f(\boldsymbol X) | X_n] \right\} E\left\{ E[g(\boldsymbol X) | X_n] \right\} \\ =& E[f(\boldsymbol X)] E[g(\boldsymbol X)] . \end{aligned}\] 证毕。

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定理D.2 设\(h(x_1, \dots, x_n)\)是关于每个自变量分别单调的函数， \(U_1, \dots, U_n\)独立同U(0,1)分布， 则当\(h(U_1, \dots, U_n)\)二阶矩有限时 \[\begin{aligned} \text{Cov}\left[ h(U_1, \dots, U_n), h(1 - U_1, \dots, 1 - U_n) \right] \leq 0. \end{aligned}\]

证明： 因为自变量次序调整不影响结论， 所以不妨设\(h\)关于前\(r\)个分量分别是单调增的， 关于后\(n-r\)个分量分别是单调减的。 令 \[\begin{aligned} f(x_1, \dots, x_n) =& h(x_1, \dots, x_r, 1 - x_{r+1}, \dots, 1 - x_n), \\ g(x_1, \dots, x_n) =& -h(1 - x_1, \dots, 1 - x_r, x_{r+1}, \dots, x_n), \end{aligned}\] 则\(f(x_1, \dots, x_n)\)和\(g(x_1, \dots, x_n)\)是关于每个自变量单调不减的函数， 设\(V_1, \dots, V_n\)为独立同U(0,1)分布随机变量， 由定理D.1可知 \[\begin{align} & \text{Cov}[f(V_1, \dots, V_n), g(V_1, \dots, V_n)] \\ =& - \text{Cov}[h(V_1, \dots, V_r, 1 - V_{r+1}, \dots, 1 - V_n), \\ & \qquad \quad h(1 - V_1, \dots, 1 - V_r, V_{r+1}, \dots, V_n) ] \geq 0 , \tag{D.2} \end{align}\] 设\(U_1, \dots, U_n\)独立同U(0,1)分布， 令\((V_1, \dots, V_n) = (U_1, \dots, U_r, 1 - U_{r+1}, \dots, 1 - U_n)\)， 则\(V_1, \dots, V_n\)独立同U(0,1)分布，从而由(D.2)式可知 \[\begin{aligned} & \text{Cov}[h(U_1, \dots, U_r, U_{r+1}, \dots, U_n, \\ & \qquad \quad h(1-U_1, \dots, 1-U_r, 1-U_{r+1}, \dots, 1 - U_n) ] \leq 0 . \end{aligned}\] 证毕。

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