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】
: 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】
: 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】
: 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】
: 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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