n****8 发帖数: 37 | 1 X1和X2是独立的随机变量,分别服从不同的symmetric (about 0) unimodal分布。
h(.)是一个单调递增函数,输入一个正数,返回一个正数。
问题:
最小化E{h(x1^2 * k1 + x2^2 * k2)},conditioned on k1*k2 = constant, k1,k2是
只有唯一解吗?(E是expected value)这个在图像上有点象Mahalanobis distance.
如果X1,X2服从一样的分布,那显然没有唯一解,以为k1,k2可以互换。现在问题是它
们的分布是不同的(pfd没有任何部分是重合的)。
求高手指点一下思路。。。
谢谢! | y**t 发帖数: 205 | 2 不知道这样想对不对:
Let A1 and A2 be two r.v. defined as
A1=h(x1^2*k1+x2^2*k2), A2=h(2*x1*x2*sqrt(k1*k2))
so A1>=A2 since h uniform non decrease
so E(A1)>=E(A2) from property of expectation
min(E(A1)=E(A2)=E(h(C*x1*x2))
so k1, k2 can have multiple solution
【在 n****8 的大作中提到】![](/moin_static193/solenoid/img/up.png) : X1和X2是独立的随机变量,分别服从不同的symmetric (about 0) unimodal分布。 : h(.)是一个单调递增函数,输入一个正数,返回一个正数。 : 问题: : 最小化E{h(x1^2 * k1 + x2^2 * k2)},conditioned on k1*k2 = constant, k1,k2是 : 只有唯一解吗?(E是expected value)这个在图像上有点象Mahalanobis distance. : 如果X1,X2服从一样的分布,那显然没有唯一解,以为k1,k2可以互换。现在问题是它 : 们的分布是不同的(pfd没有任何部分是重合的)。 : 求高手指点一下思路。。。 : 谢谢!
| O*****y 发帖数: 222 | 3 I don't think this is correct.
Since given X1=x1, X2=x2 (X1,X2 are r.v.s, while x1,x2 are numbers),
A1 >= A2, and equality holds when k1/k2 = (x2/x1)^2
for other X1=x1', X2=X2', A1 > A2 still holds. However, the equality won't
hold any more.
My point is
min(A1) != E(A2)
【在 y**t 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 不知道这样想对不对: : Let A1 and A2 be two r.v. defined as : A1=h(x1^2*k1+x2^2*k2), A2=h(2*x1*x2*sqrt(k1*k2)) : so A1>=A2 since h uniform non decrease : so E(A1)>=E(A2) from property of expectation : min(E(A1)=E(A2)=E(h(C*x1*x2)) : so k1, k2 can have multiple solution
| y**t 发帖数: 205 | 4 I didn't get your point. The problem is asking expectation of A1 and A2, you
were considering value of A1 and A2.And I think your last line should be
min(E(A1)!=E(A2).
But by the monotonicity of expectation operator, if two variables A1 and A2,
with A1>=A2 almost sure, there is E(A1)>=E(A2). You can prove it easily.
【在 O*****y 的大作中提到】![](/moin_static193/solenoid/img/up.png) : I don't think this is correct. : Since given X1=x1, X2=x2 (X1,X2 are r.v.s, while x1,x2 are numbers), : A1 >= A2, and equality holds when k1/k2 = (x2/x1)^2 : for other X1=x1', X2=X2', A1 > A2 still holds. However, the equality won't : hold any more. : My point is : min(A1) != E(A2)
| n****8 发帖数: 37 | 5 I think OMalley is right.
My thinking is:
Suppose h(x) = x (simple example, or we can try h(x) = x^2,...)
Suppose k1 and k2 are one set of solution, and k1' and k2' are another set
of solution.
If we make E{h(x1,x2,k1,k2)} = E{h(x1,x2,k1',k2')}, it is easy to verify
that it is possible to have k1 != k1'.
But the problem is, h() can be different...
I don't know if I am on the right track.
【在 O*****y 的大作中提到】![](/moin_static193/solenoid/img/up.png) : I don't think this is correct. : Since given X1=x1, X2=x2 (X1,X2 are r.v.s, while x1,x2 are numbers), : A1 >= A2, and equality holds when k1/k2 = (x2/x1)^2 : for other X1=x1', X2=X2', A1 > A2 still holds. However, the equality won't : hold any more. : My point is : min(A1) != E(A2)
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