a*******1
发帖数: 1554 | 1 对任意随机变量X,证明:(Median(X)-Mean(X))^2<=Var(X)
谢谢。 | m****r
发帖数: 237 | 2 By using Cantelli's inequality, assuming the mean is mu, standard deviation
is sigma, we have:
P(X-mu>=k*sigma)<=1/(1+k^2)
Take k=1, we have,
P(X-mu>=sigma)<=1/2, which is
P(X>=mu+sigma)<=1/2
Similarly, from P(mu-X>=k*sigma)<=1/(1+k^2),
we get P(mu-X>=sigma)<=1/2, which is
P(X<=mu-sigma)<=1/2
Therefore, the median of X shoule lie between mu-sigma and mu+sigma. (median
(X)-E(X))^2<=var(X) follows.
【在 a*******1 的大作中提到】
: 对任意随机变量X,证明:(Median(X)-Mean(X))^2<=Var(X)
: 谢谢。
| s*****n
发帖数: 2174 | 3 Var(X)
=E(X-Median(X) + Median(X) - Mean(X))^2
=E(X-Median(X))^2 + 2(Median(X)-Mean(X))*E(X-Median(X))+(Median(X)-Mean(X))^
2
=E(X-Median(X))^2 - (Mean(X)-Median(X))^2
于是此题等价于证明
E(X-Median(X))^2 >= 2 * (Mean(X)-Median(X))^2
这个是显而易见的, 对于固定样本可以很容易的构造性证明(两两配对).
如果要严格对于概率测度来证明, 恐怕要复杂多了, 我能想到的是以median最为概率测
度的中分点, 然后两边还有个匹配问题. 结论很直观, 不过要说严格了可能还不容易. | a*******1
发帖数: 1554 | | c******r
发帖数: 300 | 5 applying Jensen's inequality twice along with the fact that median minimizes
the l1 losss.
sqrt(E(|X-EX|^2)) >= sqrt((E|X-EX|)^2)=E|X-EX|
>=E|X-med(X)|>=|E(X-med(X))|=|E(X) - med(X)|
【在 a*******1 的大作中提到】
: 对任意随机变量X,证明:(Median(X)-Mean(X))^2<=Var(X)
: 谢谢。
| a*******1
发帖数: 1554 | 6 Thank you very much.
minimizes
【在 c******r 的大作中提到】
: applying Jensen's inequality twice along with the fact that median minimizes
: the l1 losss.
: sqrt(E(|X-EX|^2)) >= sqrt((E|X-EX|)^2)=E|X-EX|
: >=E|X-med(X)|>=|E(X-med(X))|=|E(X) - med(X)|
| d******e
发帖数: 7844 | 7 sqrt(E(|X-EX|^2)) >= E(sqrt(|X-EX|)^2))这里应该是typo
这个方法好巧妙啊
minimizes
【在 c******r 的大作中提到】
: applying Jensen's inequality twice along with the fact that median minimizes
: the l1 losss.
: sqrt(E(|X-EX|^2)) >= sqrt((E|X-EX|)^2)=E|X-EX|
: >=E|X-med(X)|>=|E(X-med(X))|=|E(X) - med(X)|
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