共分散行列の反転が確率変数間の部分相関をもたらすのはなぜですか?


32

ランダム変数間の偏相関は、共分散行列を反転し、そのような結果の精度行列から適切なセルを取得することで見つけることができると聞きました(この事実は http://en.wikipedia.org/wiki/Partial_correlationにいますが、証拠はありません) 。

これはなぜですか?


1
他のすべての変数について制御されたセルで偏相関を取得する場合、ここの最後の段落が明らかになる可能性があります。
ttnphns

回答:


34

場合、多変量確率変数持つ非縮退共分散行列C = γ I 、J= CovをX IXのJ、すべての実際の線形結合の組X iは、基底E = X 1X 2n次元の実ベクトル空間を形成します(X1,X2,,Xn)C=(γij)=(Cov(Xi,Xj))Xinおよび非縮退内積E=(X1,X2,,Xn)

Xi,Xj=γij .

その双対基底この内積に対して、一意の関係によって定義されますE=(X1,X2,,Xn)

Xi,Xj=δij ,

the Kronecker delta (equal to 1 when i=j and 0 otherwise).

The dual basis is of interest here because the partial correlation of Xi and Xj is obtained as the correlation between the part of Xi that is left after projecting it into the space spanned by all the other vectors (let's simply call it its "residual", Xi) and the comparable part of Xj, its residual Xj. Yet Xi is a vector that is orthogonal to all vectors besides Xi and has positive inner product with Xi whence Xi must be some non-negative multiple of Xi, and likewise for Xj. Let us therefore write

Xi=λiXi, Xj=λjXj

for positive real numbers λi and λj.

The partial correlation is the normalized dot product of the residuals, which is unchanged by rescaling:

ρij=Xi,XjXi,XiXj,Xj=λiλjXi,Xjλi2Xi,Xiλj2Xj,Xj=Xi,XjXi,XiXj,Xj .

(In either case the partial correlation will be zero whenever the residuals are orthogonal, whether or not they are nonzero.)

We need to find the inner products of dual basis elements. To this end, expand the dual basis elements in terms of the original basis E:

Xi=j=1nβijXj .

Then by definition

δik=Xi,Xk=j=1nβijXj,Xk=j=1nβijγjk .

In matrix notation with I=(δij) the identity matrix and B=(βij) the change-of-basis matrix, this states

I=BC .

That is, B=C1, which is exactly what the Wikipedia article is asserting. The previous formula for the partial correlation gives

ρij=βijβiiβjj=Cij1Cii1Cjj1 .

3
+1, great answer. But why do you call this dual basis "dual basis with respect to this inner product" -- what does "with respect to this inner product" exactly mean? It seems that you use the term "dual basis" as defined here mathworld.wolfram.com/DualVectorSpace.html in the second paragraph ("Given a vector space basis v1,...,vn for V there exists a dual basis...") or here en.wikipedia.org/wiki/Dual_basis, and it's independent of any scalar product.
amoeba says Reinstate Monica

3
@amoeba There are two kinds of duals. The (natural) dual of any vector space V over a field R is the set of linear functions ϕ:VR, called V. There is no canonical way to identify V with V, even though they have the same dimension when V is finite-dimensional. Any inner product γ corresponds to such a map g:VV, and vice versa, via
g(v)(w)=γ(v,w).
(Nondegeneracy of γ ensures g is a vector space isomorphism.) This gives a way to view elements of V as if they were elements of the dual V--but it depends on γ.
whuber

3
@mpettis Those dots were hard to notice. I have replaced them with small open circles to make the notation easier to read. Thanks for pointing this out.
whuber

4
@Andy Ron Christensen's Plane Answers to Complex Questions might be the sort of thing you are looking for. Unfortunately, his approach makes (IMHO) undue reliance on coordinate arguments and calculations. In the original introduction (see p. xiii), Christensen explains that's for pedagogical reasons.
whuber

3
@whuber, Your proof is awesome. I wonder whether any book or article contains such a proof so that I can cite.
Harry

12

Here is a proof with just matrix calculations.

I appreciate the answer by whuber. It is very insightful on the math behind the scene. However, it is still not so trivial how to use his answer to obtain the minus sign in the formula stated in the wikipediaPartial_correlation#Using_matrix_inversion.

ρXiXjV{Xi,Xj}=pijpiipjj

To get this minus sign, here is a different proof I found in "Graphical Models Lauriten 1995 Page 130". It is simply done by some matrix calculations.

The key is the following matrix identity:

(ABCD)1=(E1E1GFE1D1+FE1G)
where E=ABD1C, F=D1C and G=BD1.

Write down the covariance matrix as

Ω=(Ω11Ω12Ω21Ω22)
where Ω11 is covariance matrix of (Xi,Xj) and Ω22 is covariance matrix of V{Xi,Xj}.

Let P=Ω1. Similarly, write down P as

P=(P11P12P21P22)

By the key matrix identity,

P111=Ω11Ω12Ω221Ω21

We also know that Ω11Ω12Ω221Ω21 is the covariance matrix of (Xi,Xj)|V{Xi,Xj} (from Multivariate_normal_distribution#Conditional_distributions). The partial correlation is therefore

ρXiXjV{Xi,Xj}=[P111]12[P111]11[P111]22.
I use the notation that the (k,l)th entry of the matrix M is denoted by [M]kl.

Just simple inversion formula of 2-by-2 matrix,

([P111]11[P111]12[P111]21[P111]22)=P111=1detP11([P11]22[P11]12[P11]21[P11]11)

Therefore,

ρXiXjV{Xi,Xj}=[P111]12[P111]11[P111]22=1detP11[P11]121detP11[P11]221detP11[P11]11=[P11]12[P11]22[P11]11
which is exactly what the Wikipedia article is asserting.

If we let i=j, then rho_ii V\{X_i, X_i} = -1, How do we interpret those diagonal elements in the precision matrix?
Jason

Good point. The formula should be only valid for i=/=j. From the proof, the minus sign comes from the 2-by-2 matrix inversion. It would not happen if i=j.
Po C.

So the diagonal numbers can't be associated with partial correlation. What do they represent? They are not just inverses of the variances, are they?
Jason

This formula is valid for i=/=j. It is meaningless for i=j.
Po C.

4

Note that the sign of the answer actually depends on how you define partial correlation. There is a difference between regressing Xi and Xj on the other n1 variables separately vs. regressing Xi and Xj on the other n2 variables together. Under the second definition, let the correlation between residuals ϵi and ϵj be ρ. Then the partial correlation of the two (regressing ϵi on ϵj and vice versa) is ρ.

This explains the confusion in the comments above, as well as on Wikipedia. The second definition is used universally from what I can tell, so there should be a negative sign.

I originally posted an edit to the other answer, but made a mistake - sorry about that!

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