I am a PhD student in Economics, I am reading a paper on statistics, and the author states the following proposition without proof. My question is is it correct? Do we need to impose more conditions to get the result？ Thanks!!!
Suppose x and y are random variables. If
(1) E[f(x,y)]=0
(2) f(x,y) and g(x,y) are increasing in x.
(3) f(x,y) and g(x,y) are decreasing in y.
Then
COV( f(x,y), g(x,y) )>0.
Last edited by Stole; 11-01-2008 at 02:11 PM.
Last edited by vinux; 11-02-2008 at 12:29 PM.
In the long run, we're all dead.
How are f ang g independent? g = f - y.
Ok. Think about this..
(X,Y) are follows bivariate normal rv. mean =(0,0)
Var(X) = 6 Cov(X,Y) = 4 Var(Y) = 3
So what will be the distribution for f and g?
In the long run, we're all dead.
hrm, I'll consider that, but what will you do with conditional probabilities like P(f < 0 | g > 0) = P(x - y < 0 | x - 2y > 0)? This is a different event than f < 0.
Thank you for your reply.
Is there a way to prove it? or is there a textbook for reference to this proposition?
I know if f(x) and g(x) are monotonic in x in the same direction, then COV(f(x),g(x)) are positive. But I have doubt on this when it comes to two variables, x and y.
It can.. Cov(f,g) is between f & g. Not between X & Y
The same example.
(X,Y) are follows bivariate normal rv. mean =(0,0)
Var(X) = 6 Cov(X,Y) = sqrt(17) Var(Y) = 3
Here you will get Cov(f,g) <0
In the long run, we're all dead.
May be this .
Cov(f, g ) >= 0
( Otherwise you have conditions like strictly increasing /decreasing )
In the long run, we're all dead.
Sorry for the confusion. I was talking about the Then part
It is not the conditionThen
COV( f(x,y), g(x,y) )>=0.
In the long run, we're all dead.
Tweet |