s***c
发帖数: 50 | 1 两个独立随机变量X,Y服从高斯分布，求X^2 . Y的期望值。
哪位大大给个办法吧。多谢了先！ | m*********a
发帖数: 2000 | 2 not so hard, think about it
【在 s***c 的大作中提到】
: 两个独立随机变量X,Y服从高斯分布，求X^2 . Y的期望值。
: 哪位大大给个办法吧。多谢了先！
| f********9
发帖数: 65 | 3 E[X^2*Y] = E[X^2]*E[Y] on condition that X independent of Y.
E[X^2] = (Variance of X)^2 + (E[X])^2
Assume E[X] = 0, Variance of X = 1,
then E[X^2] = 1^2 + 0.0 = 1
E[Y] = 0;
Finally, E[X^2*Y] = E[X^2]*E[Y] = 1*0 = 0
哪位达人给改改作业吧! | R********n
发帖数: 519 | 4 if E[Y]=0, hehe, it is easy, X is independent with Y, so what ever the f(.)
you choose, E(f(X)Y) = E(f(X)) * E(Y) = 0
【在 f********9 的大作中提到】
: E[X^2*Y] = E[X^2]*E[Y] on condition that X independent of Y.
: E[X^2] = (Variance of X)^2 + (E[X])^2
: Assume E[X] = 0, Variance of X = 1,
: then E[X^2] = 1^2 + 0.0 = 1
: E[Y] = 0;
: Finally, E[X^2*Y] = E[X^2]*E[Y] = 1*0 = 0
: 哪位达人给改改作业吧!
| s***c
发帖数: 50 | 5 你做的对。高斯变量X的二阶矩 X^2 = mu^2 + sigma^2。如果X和Y独立，X^2也和Y独
立(?) 剩下的就好做了。
如果在一般的情况(X和Y不独立),该如何做啊？ 协方差矩阵？雅可比行列式？。。。。
哪位老大给个一般性的方法？多谢了！！
【在 f********9 的大作中提到】
: E[X^2*Y] = E[X^2]*E[Y] on condition that X independent of Y.
: E[X^2] = (Variance of X)^2 + (E[X])^2
: Assume E[X] = 0, Variance of X = 1,
: then E[X^2] = 1^2 + 0.0 = 1
: E[Y] = 0;
: Finally, E[X^2*Y] = E[X^2]*E[Y] = 1*0 = 0
: 哪位达人给改改作业吧!
| B********4
发帖数: 7156 | 6 E[X^2] = (Variance of X)^2 + (E[X])^2
Should be:
E[X^2] = (Variance of X) + (E[X])^2 | f********9
发帖数: 65 | 7 3x.
【在 B********4 的大作中提到】
: E[X^2] = (Variance of X)^2 + (E[X])^2
: Should be:
: E[X^2] = (Variance of X) + (E[X])^2
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