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Partial derivatives of f(x,y) statistically indepenedent?

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Ares Lagae

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Dec 21, 2009, 5:10:16 AM12/21/09
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Hi all,

I have a maybe somewhat strange question:

Are the partial derivatives of a continuous function f(x,y), when
interpreted as random variables, statistically independent? Under which
conditions?

Pointers into literature are appreciated.

Best regards,

Ares Lagae

Greg Heath

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Dec 21, 2009, 6:33:56 AM12/21/09
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What do you think the answer is if x and y are correlated?
What do you think the answer is if x and y are independent?

Hope this helps.

Greg

Ares Lagae

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Dec 21, 2009, 8:59:35 AM12/21/09
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Greg Heath wrote:

> What do you think the answer is if x and y are correlated?
> What do you think the answer is if x and y are independent?

Hi Greg,

Thanks for your quick response. However, I am still confused.

Are you claiming that, if X and Y are independent, the partial derivatives
of f to X and Y are also independent?

For f(X,Y)=(X+Y)^2, the partial derivatives are the same, they are both
equal to 2(X+Y), so I guess they are not dependent, even if X and Y are
dependent. Am I mis-interpreting your hint?

Best regards,
Ares Lagae

Greg Heath

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Dec 22, 2009, 11:49:38 AM12/22/09
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On Dec 21, 8:59 am, Ares Lagae <ares.la...@sophia.inria.fr> wrote:
> Greg Heath wrote:
> > What do you think the answer is if x and y are correlated?
> > What do you think the answer is if x and y are independent?
>
> Hi Greg,
>
> Thanks for your quick response. However, I am still confused.
>
> Are you claiming that, if X and Y are independent, the partial derivatives
> of f to X and Y are also independent?

No. I didn't know the answer off hand and was too lazy to think
more about it. So I gave you the first two thoughts that came to me,
hoping that you could finish and tell me the answer.

> For f(X,Y)=(X+Y)^2, the partial derivatives are the same, they are both
> equal to 2(X+Y), so I guess they are not dependent, even if X and Y are
> dependent. Am I mis-interpreting your hint?

Assume they are independent. Then

0 = <(X+Y)*(X+Y)> - <X+Y>*<X+Y>
= (<X^2>-<X>^2) + (<Y^2>-<Y>^2 )

Now, I'm sure you can find distributions for which that is
not true.

Hope that helps.

Greg

Ares Lagae

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Dec 23, 2009, 2:12:35 AM12/23/09
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Hi Greg,

Thanks again for your time.

Are you sure this is the correct formulation for the problem?

Note that I am not trying to determine whether the partial derivatives of a
function of two random variables are independent. I am trying to determine
whether the partial derivatives of a "regular" function f(x,y) at a point
(x,y) are independent random variables. But maybe that boils down to the
same?

Best regards,


Ray Koopman

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Dec 23, 2009, 3:49:16 AM12/23/09
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On Dec 22, 11:12 pm, Ares Lagae <ares.la...@sophia.inria.fr> wrote:
> [...] I am trying to determine whether the partial derivatives

> of a "regular" function f(x,y) at a point (x,y) are independent
> random variables.

You have a smooth surface, and you want to know whether the slope
in the x-direction and the slope in the y-direction are independent
random variables? Where does the randomness come from? Are you
choosing points (x,y) "at random" according to some unspecified
probability distribution over the plane, and then looking at the
pair of slopes at each point?

Ares Lagae

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Dec 23, 2009, 4:29:28 AM12/23/09
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Hi Ray,

Thanks for your interest in my problem.

Some clarifications below.

> You have a smooth surface, and you want to know whether the slope
> in the x-direction and the slope in the y-direction are independent
> random variables

Indeed.

> Where does the randomness come from? Are you
> choosing points (x,y) "at random" according to some unspecified
> probability distribution over the plane, and then looking at the
> pair of slopes at each point?

The function f(x,y) is a continuous but "random" function. For example, a
continuous / blurred / interpolated version of discrete white noise.

A proof of a property of f(x,y) relies on the fact that the partial
derivatives are statistically independent. Hence my question.

I know that this is a bit abstract, but I hope this gives you more insight
into the problem.

Best regards,
Ares Lagae

Ares Lagae

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Dec 23, 2009, 5:16:59 PM12/23/09
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Another clarification:

I guess I want to show that the partial derivatives are uncorrelated
rather than independent. (Since the proof relies on E[XY]=E[X]E[Y]).

Best regards,
Ares Lagae

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