# Thread: The sign of COV( f(x,y) , g(x,y) )

1. ## The sign of COV( f(x,y) , g(x,y) )

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.

2. Originally Posted by Stole
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.

3. Originally Posted by Stole
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.
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

4. ## O.o

How are f ang g independent? g = f - y.

(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.

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.

8. Originally Posted by Dragan
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.

9. Originally Posted by zmogggggg
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.
Did you calculated this? what is the answer

10. Originally Posted by vinux
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?

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

12. Originally Posted by vinux
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?

13. May be this .
Cov(f, g ) >= 0
( Otherwise you have conditions like strictly increasing /decreasing )

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

15. Sorry for the confusion. I was talking about the Then part

Then
COV( f(x,y), g(x,y) )>=0.
It is not the condition