The sign of COV( f(x,y) , g(x,y) )

S

Stole

Guest
#1
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.
:wave:
 
Last edited by a moderator:

Dragan

Super Moderator
#2
Do we need to impose more conditions to get the result?
Well, no, we don't.

The author (based on what you have provided) is just simply implying that both f(x,y) and g(x,y) are monotone functions in both x (increasing) and y (decreasing).

And, therefore, the covariance between these two functions will be positive.
 

vinux

Dark Knight
#3
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.
:wave:
I think you need a small correction here.
It should be COV( f(x,y), g(x,y) )>=0.

Consider the following situation.
Let
f(x,y) = x-y , g(x,y) = x-2y and E(x-y) =0

it holds 2&3 conditions

But f & g are independent( we can make this combination). So this lead Cov(f,g) = 0
 
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vinux

Dark Knight
#5
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?
 
#6
Re:

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.
 
S

Stole

Guest
#7
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.
 

vinux

Dark Knight
#8
Well, no, we don't.

The author (based on what you have provided) is just simply implying that both f(x,y) and g(x,y) are monotone functions in both x (increasing) and y (decreasing).

And, therefore, the covariance between these two functions will be positive.

I think you need a condition like x and y are independent.
 
S

Stole

Guest
#10
I think you need a small correction here.
It should be COV( f(x,y), g(x,y) )>=0.

Consider the following situation.
Let
f(x,y) = x-y , g(x,y) = x-2y and E(x-y) =0

it holds 2&3 conditions

But f & g are independent( we can make this combination). So this lead Cov(f,g) = 0
But the covariance can never be negative, right?
 

vinux

Dark Knight
#11
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
 
S

Stole

Guest
#12
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
That is right!
Then we need more assumptions. Except independence between x and y, is there any other assumption can guarantee this?
 
S

Stole

Guest
#14
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
May be this .
Cov(f, g ) >= 0
( Otherwise you have conditions like strictly increasing /decreasing )
Unfortunately, this assumption is exactly the result.
 
S

Stole

Guest
#16
Sorry for the confusion. I was talking about the Then part



It is not the condition
But just as you posted, let f=x-y, and g=x-2y, then
COV(f, g)=V(x)+2V(y)-3COV(x,y).
If COV(x,y) is positive and large enough, COV(f,g) can be negative.
So COV(f, g)>=0 is not guaranteed.:rolleyes:
 

vinux

Dark Knight
#17
But just as you posted, let f=x-y, and g=x-2y, then
COV(f, g)=V(x)+2V(y)-3COV(x,y).
If COV(x,y) is positive and large enough, COV(f,g) can be negative.
So COV(f, g)>=0 is not guaranteed.:rolleyes:
If X and Y are independent then Cov(X,Y)=0. :p
 

Dragan

Super Moderator
#19
All right........................:eek:
Hi Stole (and Richie): :)

Is the article that you're referring to available on-line?...or can you (Stole) give the citation of the article.

I think that would help because it would provide more context in terms of what the author(s) is(are) trying to do.
 
S

Stole

Guest
#20
Hi Stole (and Richie): :)

Is the article that you're referring to available on-line?...or can you (Stole) give the citation of the article.

I think that would help because it would provide more context in terms of what the author(s) is(are) trying to do.
I would like to, but that is my friend's working paper, and he wont let me do this. But I will let him know about this problem in his paper. Thanks!