Kovaryans matrisinin inversiyonu neden rastgele değişkenler arasında kısmi korelasyonlar sağlıyor?

Kovaryans matrisini ters çevirerek ve bu gibi sonuçlanan hassas matristen uygun hücreler alarak, rastgele değişkenler arasındaki kısmi korelasyonların bulunabileceğini duydum (bu gerçek http://en.wikipedia.org/wiki/Partial_correlation adresinde , ancak bir kanıt olmadan). .

Bu neden böyle?

— michal
kaynak

Diğer tüm değişkenler için kontrol edilen bir hücrede kısmi korelasyon elde etmek demek, o son paragraf burada ışık tutabilir.

— ttnphns,

Yanıtlar:

Ne zaman bir çok değişkenli rasgele değişken sahip bir dejenere olmayan bir kovaryans matrisi , tüm gerçek lineer kombinasyonları kümesi , bir oluşturur baz ile boyutlu gerçek vektör alanı $(X_1,X_2,\ldots,X_n)$ $\mathbb{C} = (\gamma_{ij}) = (\text{Cov}(X_i,X_j))$ $X_i$ $n$ $E=(X_1,X_2,\ldots, X_n)$ ve tarafından verilen dejenere olmayan bir iç ürün

⟨ X_{i}, X_{j} ⟩ = γ_{i j} .

$\langle X_i,X_j \rangle = \gamma_{ij}\ .$

Bu çift baz , bu iç ürüne göre , , benzersiz ilişkileri ile tanımlanır $E^{*} = (X_1^{*},X_2^{*}, \ldots, X_n^{*})$

⟨ X_{i}^{*}, X_{j} ⟩ = δ_{i j},

$\langle X_i^{*}, X_j \rangle = \delta_{ij}\ ,$

Kronecker deltası ( ve aksi halde olduğunda ). $1$ $i=j$ $0$

Kısmi korelasyon nedeniyle ikili temeli burada ilgi olduğunu ve parçası arasındaki korelasyon olarak elde edilir uzaya yansıtırken diğer bütün vektörler tarafından yayılmış sonra bırakılır (sade diyelim onun " "kalıntı, ) ve karşılaştırılabilir kısmı , kalıntı . Yine dışındaki tüm vektörler ortogonal olan bir vektör olan ve pozitif iç çarpım sahip nereden $X_i$ $X_j$ $X_i$ $X_{i\circ}$ $X_j$ $X_{j\circ}$ $X_i^{*}$ $X_i$ $X_i$ $X_{i\circ}$ must be some non-negative multiple of $X_i^{*}$ , and likewise for $X_j$ . Let us therefore write

X_{i \circ} = λ_{i} X_{i}^{*}, X_{j \circ} = λ_{j} X_{j}^{*}

$X_{i\circ} = \lambda_i X_i^{*},\ X_{j\circ} = \lambda_j X_j^{*}$

for positive real numbers $\lambda_i$ and $\lambda_j$ .

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

ρ_{i j \circ} = \frac{⟨ X_{i \circ}, X_{j \circ} ⟩}{\sqrt{⟨ X_{i \circ}, X_{i \circ} ⟩ ⟨ X_{j \circ}, X_{j \circ} ⟩}} = \frac{λ_{i} λ_{j} ⟨ X_{i}^{*}, X_{j}^{*} ⟩}{\sqrt{λ_{i}^{2} ⟨ X_{i}^{*}, X_{i}^{*} ⟩ λ_{j}^{2} ⟨ X_{j}^{*}, X_{j}^{*} ⟩}} = \frac{⟨ X_{i}^{*}, X_{j}^{*} ⟩}{\sqrt{⟨ X_{i}^{*}, X_{i}^{*} ⟩ ⟨ X_{j}^{*}, X_{j}^{*} ⟩}} .

$\rho_{ij\circ} = \frac{\langle X_{i\circ}, X_{j\circ} \rangle}{\sqrt{\langle X_{i\circ}, X_{i\circ} \rangle\langle X_{j\circ}, X_{j\circ} \rangle}} = \frac{\lambda_i\lambda_j\langle X_{i}^{*}, X_{j}^{*} \rangle}{\sqrt{\lambda_i^2\langle X_{i}^{*}, X_{i}^{*} \rangle\lambda_j^2\langle X_{j}^{*}, X_{j}^{*} \rangle}} = \frac{\langle X_{i}^{*}, X_{j}^{*} \rangle}{\sqrt{\langle X_{i}^{*}, X_{i}^{*} \rangle\langle X_{j}^{*}, X_{j}^{*} \rangle}}\ .$

(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 $E$ :

X_{i}^{*} = \sum_{j = 1}^{n} β_{i j} X_{j} .

$X_i^{*} = \sum_{j=1}^n \beta_{ij} X_j\ .$

Then by definition

δ_{i k} = ⟨ X_{i}^{*}, X_{k} ⟩ = \sum_{j = 1}^{n} β_{i j} ⟨ X_{j}, X_{k} ⟩ = \sum_{j = 1}^{n} β_{i j} γ_{j k} .

$\delta_{ik} = \langle X_i^{*}, X_k \rangle = \sum_{j=1}^n \beta_{ij}\langle X_j, X_k \rangle = \sum_{j=1}^n \beta_{ij}\gamma_{jk}\ .$

In matrix notation with $\mathbb{I} = (\delta_{ij})$ the identity matrix and $\mathbb{B} = (\beta_{ij})$ the change-of-basis matrix, this states

I = B C .

$\mathbb{I} = \mathbb{BC}\ .$

That is, $\mathbb{B} = \mathbb{C}^{-1}$ , which is exactly what the Wikipedia article is asserting. The previous formula for the partial correlation gives

ρ_{i j \cdot} = \frac{β_{i j}}{\sqrt{β_{i i} β_{j j}}} = \frac{C_{i j}^{- 1}}{\sqrt{C_{i i}^{- 1} C_{j j}^{- 1}}} .

$\rho_{ij\cdot} = \frac{\beta_{ij}}{\sqrt{\beta_{ii} \beta_{jj}}} = \frac{\mathbb{C}^{-1}_{ij}}{\sqrt{\mathbb{C}^{-1}_{ii} \mathbb{C}^{-1}_{jj}}}\ .$

— whuber
kaynak

+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

v_{1}, . . ., v_{n}

$v_1, ..., v_n$ for

V

$V$ 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

@amoeba There are two kinds of duals. The (natural) dual of any vector space

V

$V$ over a field

R

$R$ is the set of linear functions

ϕ : V \to R

$\phi:V\to R$ , called

V^{*}

$V^*$ . There is no canonical way to identify

V^{*}

$V^*$ with

V

$V$ , even though they have the same dimension when

V

$V$ is finite-dimensional. Any inner product

γ

$\gamma$ corresponds to such a map

g : V \to V^{*}

$g:V\to V^*$ , and vice versa, via

g (v) (w) = γ (v, w) .

$g(v)(w)=\gamma(v,w).$ (Nondegeneracy of

γ

$\gamma$ ensures

g

$g$ is a vector space isomorphism.) This gives a way to view elements of

V

$V$ as if they were elements of the dual

V^{*}

$V^*$ --but it depends on

γ

$\gamma$ .

— whuber

@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

@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

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

— Harry

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.

ρ_{X_{i} X_{j} \cdot V ∖ {X_{i}, X_{j}}} = - \frac{p_{i j}}{\sqrt{p_{i i} p_{j j}}}

$\rho_{X_iX_j\cdot \mathbf{V} \setminus \{X_i,X_j\}} = - \frac{p_{ij}}{\sqrt{p_{ii}p_{jj}}}$

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:

{(\begin{matrix} A & B \\ C & D \end{matrix})}^{- 1} = (\begin{matrix} E^{- 1} & - E^{- 1} G \\ - F E^{- 1} & D^{- 1} + F E^{- 1} G \end{matrix})

$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^{-1} = \begin{pmatrix} E^{-1} & -E^{-1}G \\ -FE^{-1} & D^{-1}+FE^{-1}G \end{pmatrix}$ where

E = A - B D^{- 1} C

$E = A - BD^{-1}C$ ,

F = D^{- 1} C

$F = D^{-1}C$ and

G = B D^{- 1}

$G = BD^{-1}$ .

Write down the covariance matrix as

Ω = (\begin{matrix} Ω_{11} & Ω_{12} \\ Ω_{21} & Ω_{22} \end{matrix})

$\Omega = \begin{pmatrix} \Omega_{11} & \Omega_{12} \\ \Omega_{21} & \Omega_{22} \end{pmatrix}$ where

Ω_{11}

$\Omega_{11}$ is covariance matrix of

(X_{i}, X_{j})

$(X_i, X_j)$ and

Ω_{22}

$\Omega_{22}$ is covariance matrix of

V ∖ {X_{i}, X_{j}}

$\mathbf{V} \setminus \{X_i, X_j \}$ .

Let $P = \Omega^{-1}$ . Similarly, write down $P$ as

P = (\begin{matrix} P_{11} & P_{12} \\ P_{21} & P_{22} \end{matrix})

$P = \begin{pmatrix} P_{11} & P_{12} \\ P_{21} & P_{22} \end{pmatrix}$

By the key matrix identity,

P_{11}^{- 1} = Ω_{11} - Ω_{12} Ω_{22}^{- 1} Ω_{21}

$P_{11}^{-1} = \Omega_{11} - \Omega_{12}\Omega_{22}^{-1}\Omega_{21}$

We also know that $\Omega_{11} - \Omega_{12}\Omega_{22}^{-1}\Omega_{21}$ is the covariance matrix of $(X_i, X_j) | \mathbf{V} \setminus \{X_i, X_j\}$ (from Multivariate_normal_distribution#Conditional_distributions). The partial correlation is therefore

ρ_{X_{i} X_{j} \cdot V ∖ {X_{i}, X_{j}}} = \frac{[P_{11}^{- 1}]_{12}}{\sqrt{[P_{11}^{- 1}]_{11} [P_{11}^{- 1}]_{22}}} .

$\rho_{X_iX_j\cdot \mathbf{V} \setminus \{X_i,X_j\}} = \frac{[P_{11}^{-1}]_{12}}{\sqrt{[P_{11}^{-1}]_{11}[P_{11}^{-1}]_{22}}}.$ I use the notation that the

(k, l)

$(k,l)$ th entry of the matrix

M

$M$ is denoted by

[M]_{k l}

$[M]_{kl}$ .

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

(\begin{matrix} [P_{11}^{- 1}]_{11} & [P_{11}^{- 1}]_{12} \\ [P_{11}^{- 1}]_{21} & [P_{11}^{- 1}]_{22} \end{matrix}) = P_{11}^{- 1} = \frac{1}{det P_{11}} (\begin{matrix} [P_{11}]_{22} & - [P_{11}]_{12} \\ - [P_{11}]_{21} & [P_{11}]_{11} \end{matrix})

$\begin{pmatrix} [P_{11}^{-1}]_{11} & [P_{11}^{-1}]_{12} \\ [P_{11}^{-1}]_{21} & [P_{11}^{-1}]_{22} \\ \end{pmatrix} = P_{11}^{-1} = \frac{1}{\text{det} P_{11}} \begin{pmatrix} [P_{11}]_{22} & -[P_{11}]_{12} \\ -[P_{11}]_{21} & [P_{11}]_{11} \\ \end{pmatrix}$

Therefore,

ρ_{X_{i} X_{j} \cdot V ∖ {X_{i}, X_{j}}} = \frac{[P_{11}^{- 1}]_{12}}{\sqrt{[P_{11}^{- 1}]_{11} [P_{11}^{- 1}]_{22}}} = \frac{- \frac{1}{det P_{11}} [P_{11}]_{12}}{\sqrt{\frac{1}{det P_{11}} [P_{11}]_{22} \frac{1}{det P_{11}} [P_{11}]_{11}}} = \frac{- [P_{11}]_{12}}{\sqrt{[P_{11}]_{22} [P_{11}]_{11}}}

$\rho_{X_iX_j\cdot \mathbf{V} \setminus \{X_i,X_j\}} = \frac{[P_{11}^{-1}]_{12}}{\sqrt{[P_{11}^{-1}]_{11}[P_{11}^{-1}]_{22}}} = \frac{- \frac{1}{\text{det}P_{11}}[P_{11}]_{12}}{\sqrt{\frac{1}{\text{det}P_{11}}[P_{11}]_{22}\frac{1}{\text{det}P_{11}}[P_{11}]_{11}}} = \frac{-[P_{11}]_{12}}{\sqrt{[P_{11}]_{22}[P_{11}]_{11}}}$ which is exactly what the Wikipedia article is asserting.

— Po C.
kaynak

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.

Note that the sign of the answer actually depends on how you define partial correlation. There is a difference between regressing $X_i$ and $X_j$ on the other $n - 1$ variables separately vs. regressing $X_i$ and $X_j$ on the other $n - 2$ variables together. Under the second definition, let the correlation between residuals $\epsilon_i$ and $\epsilon_j$ be $\rho$ . Then the partial correlation of the two (regressing $\epsilon_i$ on $\epsilon_j$ and vice versa) is $-\rho$ .

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!

— Johnny Ho
kaynak