D**u
发帖数: 204 | 1 【 以下文字转载自 Mathematics 讨论区 】
发信人: DuGu (火工头陀), 信区: Mathematics
标 题: Ordering a sequence (2)
发信站: BBS 未名空间站 (Wed Dec 16 12:17:39 2009, 美东)
Let x1,...xn,y1,...,yn be 2n distinct real numbers, and xi+xj is not equal
to yk+yl for any i,j,k,l.
Prove that you can properly reordering x1,...,xn to z1,...zn, such that
(zi + zj - yi - yj)*(zi - zj)*(yi - yj) > 0
for all i and j. | p*****k
发帖数: 318 | 2 DuGu, not sure if I missed anything, but cannot you just take
x_i'=exp(x_i), and apply the solution last time?
http://www.mitbbs.com/article_t/Quant/31214785.html | D**u
发帖数: 204 | 3 That's correct, x_i'=exp(x_i) will work, so this is essentially a (trivial)
generalization of the previous problem.
What puzzels me is that this new question looks simpler (or "cleaner") than
last one (because we now only compare SUMS "zi+zj vs yi+yj", instead of
PRODUCTS "zi*zj vs yi*yj"), but the solution looks more complicated than
last one(because we now need an extra transformation x_i'=exp(x_i)).
I am wondering if there is a more straight-forward solution, which does not
rely on the soluti
【在 p*****k 的大作中提到】
: DuGu, not sure if I missed anything, but cannot you just take
: x_i'=exp(x_i), and apply the solution last time?
: http://www.mitbbs.com/article_t/Quant/31214785.html
| w*****e
发帖数: 197 | 4 Notice the choice of {y_i} is arbitrary. So we can assume y_i=i.
Now we want to make ( z_i + z_j - i - j ) ( z_i - z_j ) ( i - j ) > 0
for all i and j.
Now consider this function M(P)=sum( i * z_i * z_i ) - sum( i * i * z_i )
over i,
where P denotes a permutation of {z_i}.
If we switch two numbers z_i and z_j here, the overall change to M(P) - M(P'
) is:
( i - j ) * ( z_i * z_i - z_j * z_j ) - ( i * i - j * j )( z_i - z_j )
which is
( i - j ) * ( z_i - zj ) ( z_i + z_j - i - j ).
Volla, we got t
【在 D**u 的大作中提到】
: That's correct, x_i'=exp(x_i) will work, so this is essentially a (trivial)
: generalization of the previous problem.
: What puzzels me is that this new question looks simpler (or "cleaner") than
: last one (because we now only compare SUMS "zi+zj vs yi+yj", instead of
: PRODUCTS "zi*zj vs yi*yj"), but the solution looks more complicated than
: last one(because we now need an extra transformation x_i'=exp(x_i)).
: I am wondering if there is a more straight-forward solution, which does not
: rely on the soluti
| w*****e
发帖数: 197 | 5 Never mind, I checked the original post and it indeed comes out such an
optimization problem. :)
P'
【在 w*****e 的大作中提到】
: Notice the choice of {y_i} is arbitrary. So we can assume y_i=i.
: Now we want to make ( z_i + z_j - i - j ) ( z_i - z_j ) ( i - j ) > 0
: for all i and j.
: Now consider this function M(P)=sum( i * z_i * z_i ) - sum( i * i * z_i )
: over i,
: where P denotes a permutation of {z_i}.
: If we switch two numbers z_i and z_j here, the overall change to M(P) - M(P'
: ) is:
: ( i - j ) * ( z_i * z_i - z_j * z_j ) - ( i * i - j * j )( z_i - z_j )
: which is
| D**u
发帖数: 204 | 6 This is an excellent proof, which is very clean. It is better than the one
that the problem was originally designed for.
P'
【在 w*****e 的大作中提到】
: Notice the choice of {y_i} is arbitrary. So we can assume y_i=i.
: Now we want to make ( z_i + z_j - i - j ) ( z_i - z_j ) ( i - j ) > 0
: for all i and j.
: Now consider this function M(P)=sum( i * z_i * z_i ) - sum( i * i * z_i )
: over i,
: where P denotes a permutation of {z_i}.
: If we switch two numbers z_i and z_j here, the overall change to M(P) - M(P'
: ) is:
: ( i - j ) * ( z_i * z_i - z_j * z_j ) - ( i * i - j * j )( z_i - z_j )
: which is
| s****n
发帖数: 700 | 7 我在想hedge fund的人问这些问题和他们控制orders有什么关联呢？
P'
【在 w*****e 的大作中提到】
: Notice the choice of {y_i} is arbitrary. So we can assume y_i=i.
: Now we want to make ( z_i + z_j - i - j ) ( z_i - z_j ) ( i - j ) > 0
: for all i and j.
: Now consider this function M(P)=sum( i * z_i * z_i ) - sum( i * i * z_i )
: over i,
: where P denotes a permutation of {z_i}.
: If we switch two numbers z_i and z_j here, the overall change to M(P) - M(P'
: ) is:
: ( i - j ) * ( z_i * z_i - z_j * z_j ) - ( i * i - j * j )( z_i - z_j )
: which is
| D**u
发帖数: 204 | 8 不错的想法。不是面试题。
【在 s****n 的大作中提到】
: 我在想hedge fund的人问这些问题和他们控制orders有什么关联呢？
:
: P'
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