B******e
发帖数: 16928 | 1 X,Y joint normally distributed with mean 0 and variance sigma^2 and
correlation rho. t=min(X,Y). What is the expression of E(tX)? Thanks a lot. | g******2
发帖数: 234 | 2 I don't think there is an analytical form. Maybe try numerical methods. | c********h
发帖数: 330 | 3 我觉得是 0.5 * sigma^2 * (1 + rho)
tX = X^2 if X < Y
tX = XY if X > Y
E(tX) = E(tX * I(X
= E(X^2 * I(X
E(XY) = E(XY * I(X
symmetry, that's where 0.5 * sigma^2 * rho comes from.
For E(X^2 * I(X
region {xy}. So
E(X^2 * I(X
You may try simulations to check the results. Correct me if I'm wrong.
.
【在 B******e 的大作中提到】
: X,Y joint normally distributed with mean 0 and variance sigma^2 and
: correlation rho. t=min(X,Y). What is the expression of E(tX)? Thanks a lot.
| u*h
发帖数: 397 | 4 I think you are right.
here is another prove:
for let X' = -X, Y'=-Y, t'=max(X',Y')=-min(X,Y)=-t;
E(max((-X),(-Y))*(-X))=E(t'X') = E((-t)*(-X)) = E(tX);
thus : E(min(X,Y) * X) = E(max(X,Y)*X);
on another hand:
E(min(X,Y)*X + max(X,Y)*X)=E((X+Y)*X)= E(X^2) + E(X*Y)
= sigma^2*(1+rho)
so: E(tX) = 1/2 * sigma^2(1+rho);
【在 c********h 的大作中提到】
: 我觉得是 0.5 * sigma^2 * (1 + rho)
: tX = X^2 if X < Y
: tX = XY if X > Y
: E(tX) = E(tX * I(X: = E(X^2 * I(X: E(XY) = E(XY * I(X: symmetry, that's where 0.5 * sigma^2 * rho comes from.
: For E(X^2 * I(X: region {xy}. So
: E(X^2 * I(X
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