v*c
发帖数: 42 | 1 if xi, i=1,...,n are iid distributed with mean mu. x_ba=1/n*(x1+...+xn).
So in another word, expectation of x: E(x)=mu. x_ba is the average of x1,...
,xn.
Then what is the variance(|x-x_ba|-|x-mu|)? Is there any simple form of it?
We can assume x has cumulative density function F, and pdf f. | s***n
发帖数: 9499 | 2 I think you assumed x and x_ba are indepedent, and xi were samples of x.
Then the result
is (2 + 1/n)var(x)
..
【在 v*c 的大作中提到】
: if xi, i=1,...,n are iid distributed with mean mu. x_ba=1/n*(x1+...+xn).
: So in another word, expectation of x: E(x)=mu. x_ba is the average of x1,...
: ,xn.
: Then what is the variance(|x-x_ba|-|x-mu|)? Is there any simple form of it?
: We can assume x has cumulative density function F, and pdf f.
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