s*****i
发帖数: 93 | 1 2 independent random variables uniformly distributed in [0,1]. How do you
transform them, so that they stay uniformly distributed in [0,1], but the
correlation between them becomes \rho. | p**********g
发帖数: 9385 | 2 http://www.wilmott.com/messageview.cfm?catid=26&threadid=61763&FTVAR_MSGDBTABLE=&STARTPAGE=1
【在 s*****i 的大作中提到】
: 2 independent random variables uniformly distributed in [0,1]. How do you
: transform them, so that they stay uniformly distributed in [0,1], but the
: correlation between them becomes \rho.
| f***a
发帖数: 329 | | m*****n
发帖数: 2152 | 4 Z = X when X <=\rho
Z = (1-\rho)Y + \rho when X > \rho
(X , Z) is the answer? | b***k
发帖数: 2673 | 5 Is this a standard problem from heard on the street?
I think I have seen it somewhere, can't recall though.
【在 s*****i 的大作中提到】
: 2 independent random variables uniformly distributed in [0,1]. How do you
: transform them, so that they stay uniformly distributed in [0,1], but the
: correlation between them becomes \rho.
| b***k
发帖数: 2673 | 6 One question on the following solution I copied from wilmott.
let Z be an independant uniform variable
and
Y2 = X if Z< rho
Y2 = Y if Z > rho
Cov(X,Y2) = cov(X,X) * rho
and Y2 is uniform
(X,Y2) is our solution.
对于上面构造的Y2，如何计算E(Y2),Var(Y2),
为什么Cov(X,Y2)=cov(X,X)*rho?
是从公式Cov(X,Y2)=E(X*Y2)-E(X)*E(Y2)得到的吗？我推不出来啊。 | m***s
发帖数: 605 | 7 Y2= X*1(Zrho)
【在 b***k 的大作中提到】
: One question on the following solution I copied from wilmott.
: let Z be an independant uniform variable
: and
: Y2 = X if Z< rho
: Y2 = Y if Z > rho
: Cov(X,Y2) = cov(X,X) * rho
: and Y2 is uniform
: (X,Y2) is our solution.
: 对于上面构造的Y2，如何计算E(Y2),Var(Y2),
: 为什么Cov(X,Y2)=cov(X,X)*rho?
| b***k
发帖数: 2673 | 8 Thanks,miles. From this equation I know
E(Y2)=E(X)*rho + E(Y)*(1-rho),and
Cov(X,Y2)=Cov(X,X)*rho+Cov(X,Y)*(1-rho)=Cov(X,X)*rho
But how to evaluate Var(Y2), since I need to know it because
Corr(X,Y2)=Cov(X,Y2)/[sqrt(Var(X))*sqrt(Var(Y2))]
【在 m***s 的大作中提到】
: Y2= X*1(Zrho)
| j*****4
发帖数: 292 | 9 Y2 is U[0,1] by definition.
【在 b***k 的大作中提到】
: Thanks,miles. From this equation I know
: E(Y2)=E(X)*rho + E(Y)*(1-rho),and
: Cov(X,Y2)=Cov(X,X)*rho+Cov(X,Y)*(1-rho)=Cov(X,X)*rho
: But how to evaluate Var(Y2), since I need to know it because
: Corr(X,Y2)=Cov(X,Y2)/[sqrt(Var(X))*sqrt(Var(Y2))]
| j*****4
发帖数: 292 | | m***s
发帖数: 605 | 11 我提供一个硬算的笨办法。
E(y2)=E(E(y2|x,y))=E(rho*x+(1-rho)*y))=1/2
var(y2)=var(E(y2|x,y))+E(var(y2|x,y))
note r=rho
let u=r*x+(1-r)*y, then x-u=(1-r)*(x-y), y-u=r*(y-x)
=var(u)+E(r*[x-u]^2+(1-r)*(y-u)^2)
=[r^2+(1-r)^2]*1/12 +r*E[(x-u)^2]+(1-r)*E[(y-u)^2]
=[r^2+(1-r)^2]*1/12 + r* E[(1-r)^2*(x-y)^2]+(1-r)*E[r^2*(x-y)^2]
=[r^2+(1-r)^2]*1/12 + r*(1-r)*E[(x-y)^2]
=[r^2+(1-r)^2]*1/12 + r*(1-r)*1/6
=1/12
【在 b***k 的大作中提到】
: Thanks,miles. From this equation I know
: E(Y2)=E(X)*rho + E(Y)*(1-rho),and
: Cov(X,Y2)=Cov(X,X)*rho+Cov(X,Y)*(1-rho)=Cov(X,X)*rho
: But how to evaluate Var(Y2), since I need to know it because
: Corr(X,Y2)=Cov(X,Y2)/[sqrt(Var(X))*sqrt(Var(Y2))]
| m*****n
发帖数: 2152 | 12 only 2 sets of random number are available, where could you get Z？
【在 b***k 的大作中提到】
: One question on the following solution I copied from wilmott.
: let Z be an independant uniform variable
: and
: Y2 = X if Z< rho
: Y2 = Y if Z > rho
: Cov(X,Y2) = cov(X,X) * rho
: and Y2 is uniform
: (X,Y2) is our solution.
: 对于上面构造的Y2，如何计算E(Y2),Var(Y2),
: 为什么Cov(X,Y2)=cov(X,X)*rho?
| p*****k
发帖数: 318 | 13 maxthon, there are ways to do it just using two r.v.s.
e.g., set Z=X if Y<(rho+1)/2 and 1-X otherwise,
then take (X,Z);
or use Gaussian copula:
http://www.mitbbs.com/article_t/Quant/31176232.html | b***k
发帖数: 2673 | 14 你搜索能力总是很强，你这个链接应该就是我上次看到的地方。多谢了。
【在 p*****k 的大作中提到】
: maxthon, there are ways to do it just using two r.v.s.
: e.g., set Z=X if Y<(rho+1)/2 and 1-X otherwise,
: then take (X,Z);
: or use Gaussian copula:
: http://www.mitbbs.com/article_t/Quant/31176232.html
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