It is known that &#x03C1;<!-- ρ --> , the pearson correlation, is a measure for the linear dep

Merati4tmjn

Merati4tmjn

Answered question

2022-05-09

It is known that ρ , the pearson correlation, is a measure for the linear dependence of two random variables say X, Y. But can't you say just transform X and Y such that we have,
ρ X , Y ( f ( X ) , g ( Y ) )
where f, g are non-linear functions such that it measures other kinds of dependce (take for example f ( s ) = g ( s ) = s 2 for quadratic dependence).

Answer & Explanation

Ahmed Kirby

Ahmed Kirby

Beginner2022-05-10Added 17 answers

Step 1
Consider the vector space S = L 2 ( Ω , F , P ) of square-integrable random variables. Here, you can introduce the inner-product
( X Y ) = Ω X Y d P . .
Suppose now that X and Y are two elements in S and we further assume they are standarised (a.k.a. studentised), in other words E ( X ) = 0 and V ( X ) = 1 , , similar for Y. Under these circumstances,
c o r r ( X , Y ) = ( X Y ) .
Interpretation. We know from elementary linear algebra that the inner-product between two vectors is the cosine of the angle they form (in the plane generated by both). So, c o r r ( X , Y ) = 0 means orthogonal in S and is equivalent to θ { 0 , π } , , where θ is the angle formed by X and Y. Of course, θ=0 θ = 0 means that X and Y point in the same direction while θ = π means that they point in opposite directions, in either case, X and Y are colinear, and this reduces to Y = a X + b in the general setting (i.e. not necessarily standarised).
What people often confuse is that c o r r ( X , Y ) = 0 should signify Y = f ( X ) for some measurable function f. In fact, Y = f ( X ) if and only if Y S X where S X is the subspace of S generated by all bounded measurable images of X, it can be shown that S X = L 2 ( Ω , X 1 ( B R ) , P ) . . This is not the same as the span of X, the span of X is X = R X = { a X a R } . . In this linear algebra language, the correlation is then the orthogonal projection operator onto the span of X. That is
p r X ( Y ) = c o r r ( X , Y ) X .
Furthermore, the function Z↦E(Z∣X) Z E ( Z X ) from S S X is the orthogonal projection from S onto S X ,, which is the right object when we want functional dependency. Explictly,
p r S X ( Z ) = E ( Z X ) .
As you can tell, the mathematical framework is crystal clear. If you want to show that Y is not functionally dependent of X you have to show that Y S X which is the same as E ( Y X ) = 0. . (Actually, the condition "Y is functionally dependent on X" signifies Y S X , so its negation really is Y S X and not Y S X , , this latter relation has no English translation, or not an obvious one, it is akin to "Nothing of Y can be functionally dependent on X.") If you show that c o r r ( X , Y ) = 0 , the you are just showing that Y X , , but as already mentioned above, many functions f ( X ) are orthogonal to X.
hisyhauttaq84w

hisyhauttaq84w

Beginner2022-05-11Added 4 answers

Step 1
As you note, Pearson's correlation coefficient reflects linear dependence of random variables. It is invariant to changes in scale and location, i.e.
ρ ( X , Y ) = ρ ( a + b X , c + d Y )
for constants a, b, c, d and b , d > 0 . It is not invariant to more general nonlinear transformations.
An alternative measure of dependence is mutual information. Intuitively, this captures how different a joint distribution of two random variables is from the product of their marginals, and thus reflects a broader notion of dependence (not just linear dependence). Moreover, mutual information is invariant to bijective transformations, i.e.
I ( X ; Y ) = I ( f ( X ) ; g ( Y ) )
for bijections f, g.

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