ทำไมการแปรผกผันของเมทริกซ์ความแปรปรวนร่วมจึงให้สหสัมพันธ์บางส่วนระหว่างตัวแปรสุ่ม

ผมได้ยินมาว่าบางส่วนความสัมพันธ์ระหว่างตัวแปรสุ่มสามารถพบได้โดย inverting เมทริกซ์ความแปรปรวนและการเซลล์ที่เหมาะสมจากที่เกิดเช่นความแม่นยำเมทริกซ์ (ความเป็นจริงนี้ถูกกล่าวถึงในhttp://en.wikipedia.org/wiki/Partial_correlationแต่ไม่มีหลักฐาน) .

เหตุใดจึงเป็นเช่นนี้

— คาล
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หากคุณหมายถึงการได้รับความสัมพันธ์บางส่วนในเซลล์ควบคุมสำหรับตัวแปรอื่นทั้งหมดย่อหน้าสุดท้ายที่นี่อาจทำให้กระจ่าง

— ttnphns

คำตอบ:

เมื่อตัวแปรสุ่มหลายตัวแปร $(X_1,X_2,\ldots,X_n)$ มีเมทริกซ์ความแปรปรวนแบบไม่เปลี่ยนแปลง $\mathbb{C} = (\gamma_{ij}) = (\text{Cov}(X_i,X_j))$ ชุดของการรวมเชิงเส้นจริงทั้งหมดของ $X_i$ สร้างปริภูมิเวกเตอร์จริง $n$ มิติที่มีพื้นฐานและผลิตภัณฑ์ภายในที่ไม่เสื่อมคุณภาพที่กำหนดโดย $E=(X_1,X_2,\ldots, X_n)$

⟨ X_{i}, X_{j} ⟩ = γ_{i j} .

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

ใช้พื้นฐานคู่ที่เกี่ยวกับผลิตภัณฑ์ด้านนี้ , ถูกกำหนดให้ไม่ซ้ำกันโดยความสัมพันธ์ $E^{*} = (X_1^{*},X_2^{*}, \ldots, X_n^{*})$

⟨ X_{i}^{*}, X_{j} ⟩ = δ_{i j},

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

Kronecker delta (เท่ากับเมื่อและอย่างอื่น) $1$ $i=j$ $0$

พื้นฐานที่น่าสนใจก็คือที่นี่เพราะความสัมพันธ์บางส่วนของและนั้นได้มาจากความสัมพันธ์ระหว่างส่วนของที่เหลือหลังจากที่ฉายมันเข้าไปในอวกาศที่เวกเตอร์อื่น ๆ ทอด (เรียกง่ายๆว่า " ที่เหลือ" ) และส่วนหนึ่งเดียวกันของ , คงเหลือ ∘แต่เป็นเวกเตอร์ที่ตั้งฉากกับเวกเตอร์ทั้งหมดนอกเหนือจากและมีผลบวกภายในกับดังนั้น $X_i$ $X_j$ $X_i$ $X_{i\circ}$ $X_j$ $X_{j\circ}$ $X_i^{*}$ $X_i$ $X_i$ $X_{i\circ}$ 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
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+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

V

$V$

R

$R$

ϕ : V \to R

$\phi:V\to R$

V^{*}

$V^*$

V^{*}

$V^*$

V

$V$

V

$V$

γ

$\gamma$

g : V \to V^{*}

$g:V\to V^*$

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.
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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
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