b*******e
发帖数: 51 | 1 10 blue houses and 8 red hourse, what's the expected number of houses which
have one neighbour of different color?
dX_t = X_t dt + e^(-rt)dW_t
what's X_t? |
m***s
发帖数: 605 | 2 d(e^(-t)*x)=-e^(-t)*xdt + e^(-t)dx
=e^(-t) * (e^(-rt) dW_t
=e^(-(r+1)t) dW_t
which
【在 b*******e 的大作中提到】
: 10 blue houses and 8 red hourse, what's the expected number of houses which
: have one neighbour of different color?
: dX_t = X_t dt + e^(-rt)dW_t
: what's X_t?
|
d*j
发帖数: 13780 | 3 没看懂。。。
能解释一下吗？
【在 m***s 的大作中提到】
: d(e^(-t)*x)=-e^(-t)*xdt + e^(-t)dx
: =e^(-t) * (e^(-rt) dW_t
: =e^(-(r+1)t) dW_t
:
: which
|
m***s
发帖数: 605 | 4 就是对那个e^(-t)*x求导，然后中间用原来的等式去化简。 帮忙检查下对不对。
【在 d*j 的大作中提到】
: 没看懂。。。
: 能解释一下吗？
|
d*j
发帖数: 13780 | 5 看懂了， 多谢
【在 m***s 的大作中提到】
: 就是对那个e^(-t)*x求导，然后中间用原来的等式去化简。 帮忙检查下对不对。
|
b*******e
发帖数: 51 | 6 谢谢，谁能说说第一道呀
【在 m***s 的大作中提到】
: d(e^(-t)*x)=-e^(-t)*xdt + e^(-t)dx
: =e^(-t) * (e^(-rt) dW_t
: =e^(-(r+1)t) dW_t
:
: which
|
t*******e
发帖数: 172 | 7 I find a way, but I do not like it,
Let X_{j}=1 if the j-th ball has the property we want, or 0 in other case.
So what we want is E(\sum_{j=1..18} X_{j}). by linearity of expectation, we
only need to determin E(X_{j}), which is easy, maybe need careful on two
boundary.
which
【在 b*******e 的大作中提到】
: 10 blue houses and 8 red hourse, what's the expected number of houses which
: have one neighbour of different color?
: dX_t = X_t dt + e^(-rt)dW_t
: what's X_t?
|
d*j
发帖数: 13780 | 8 题目够难的啊 。。。
which
【在 b*******e 的大作中提到】
: 10 blue houses and 8 red hourse, what's the expected number of houses which
: have one neighbour of different color?
: dX_t = X_t dt + e^(-rt)dW_t
: what's X_t?
|
b*******e
发帖数: 51 | 9 是的啊，俺都不会
【在 d*j 的大作中提到】
: 题目够难的啊 。。。
:
: which
|
b*******e
发帖数: 51 | 10 谢谢牛人，明天看，今天我先睡觉去，脑子转不动了。。。
we
【在 t*******e 的大作中提到】
: I find a way, but I do not like it,
: Let X_{j}=1 if the j-th ball has the property we want, or 0 in other case.
: So what we want is E(\sum_{j=1..18} X_{j}). by linearity of expectation, we
: only need to determin E(X_{j}), which is easy, maybe need careful on two
: boundary.
:
: which
|
b*****t
发帖数: 10 | |
p*****k
发帖数: 318 | 12 for the 1st question, seems it's easier to consider the opposite
question, i.e., the expected # of houses with same color as its
neighbor(s).
the idea is outlined by timesgone, or check a similar thread:
http://www.mitbbs.com/article_t/Quant/31237731.html
which results:
2*[(10/18)*(9/17)+(8/18)*(7/17)] +
16*[(10/18)*(9/17)*(8*/16)+(8/18)*(7/17)*(6/16)] = 674/153 ~ 4.4
1st term considers the houses on both ends.
so the answer is ~ 18-4.4 = 13.6
combinatorial approach seems very tedious, but doabl |
d*j
发帖数: 13780 | 13 这个没有简单点的？
真麻烦 。。。
【在 p*****k 的大作中提到】
: for the 1st question, seems it's easier to consider the opposite
: question, i.e., the expected # of houses with same color as its
: neighbor(s).
: the idea is outlined by timesgone, or check a similar thread:
: http://www.mitbbs.com/article_t/Quant/31237731.html
: which results:
: 2*[(10/18)*(9/17)+(8/18)*(7/17)] +
: 16*[(10/18)*(9/17)*(8*/16)+(8/18)*(7/17)*(6/16)] = 674/153 ~ 4.4
: 1st term considers the houses on both ends.
: so the answer is ~ 18-4.4 = 13.6
|
