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知乎里思学回答：作者：思雪
链接：zhihu.com/question/614507007/answer/3140296910
来源：知乎
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设 a <= x1 <= ... <= xn <= b，由于方差平移不变，不妨设 mean(xi) = 0，又由对称性，不妨设 |a| < |b|。我们有mean(a, xi, |a|) = 0var(a, xi, |a|)= (sum (xi)^2 + 2a^2) / (n+2)>= (sum (xi)^2 * (1 + 2/n)) / (n + 2)= sum (xi)^2 / n= var(xi)当且仅当 a^2 = xi^2 取等故只需证 var(a, xi, b) >= var(a, xi, |a|)注意到 var(x) = E(x^2) - (Ex)^2做差得(n+2)^2*(var(a, xi, b) - var(a, xi, |a|))= (sum xi^2 + a^2 + b^2)*(n + 2) - (a + b)^2 - (sum xi^2 + 2a^2)*(n + 2)= (b^2 - a^2)(n + 2) - (a + b)^2= (a + b)((n + 1)(b - a) - 2a)>= 0当且仅当 a + b = 0 取等综上，var(a, xi, b) >= var(xi)，当且仅当 a = xi = b 或 n 为偶数 2k 且 a = x1 = ... = xk < x(k+1) = ... = xn = b 时取等。 |
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