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

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#### Stole

##### Guest
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:

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#### Dragan

##### Super Moderator
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
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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#### zmogggggg

##### New Member

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

#### vinux

##### Dark Knight

(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?

#### zmogggggg

##### New Member
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.

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#### Stole

##### Guest
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
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.

#### vinux

##### Dark Knight
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

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#### Stole

##### Guest
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
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

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#### Stole

##### Guest
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?

#### vinux

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

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#### Stole

##### Guest
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.

#### vinux

##### Dark Knight
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

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#### Stole

##### Guest
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.

#### vinux

##### Dark Knight
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.
If X and Y are independent then Cov(X,Y)=0.

#### Dragan

##### Super Moderator
All right........................
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.

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#### Stole

##### Guest
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!