কেন কোনও কোভারিয়েন্স ম্যাট্রিক্সের বিপরীতে র্যান্ডম ভেরিয়েবলের মধ্যে আংশিক পারস্পরিক সম্পর্ক রয়েছে?

Question

আমি শুনেছি যে এলোমেলো ভেরিয়েবলের মধ্যে আংশিক পারস্পরিক সম্পর্কগুলি কোভেরিয়েন্স ম্যাট্রিক্সকে উল্টিয়ে দিয়ে এবং এর ফলে প্রাপ্ত নির্ভুলতা ম্যাট্রিক্স থেকে উপযুক্ত কোষ গ্রহণ করে (এই সত্যটি http://en.wikedia.org/wiki/Partial_correlation এ উল্লেখ করা হয়েছে , তবে একটি প্রমাণ ছাড়াই) ।

কেন এই ক্ষেত্রে?

Comments

আপনি যদি একটি সেল সব অন্যান্য ভেরিয়েবল নিয়ন্ত্রণ আংশিক পারস্পরিক সম্পর্ক পেতে চাওয়ার কথা বলছেন, তাহলে শেষ অনুচ্ছেদ এখানে আলোকপাত করতে পারে।

— ttnphns

Answer 1

যখন একটি মাল্টিভিয়ারেট এলোমেলো পরিবর্তনশীল $(X_1,X_2,\ldots,X_n)$ এর একটি অবিস্মরণীয় কোভারিয়েন্স ম্যাট্রিক্স $\mathbb{C} = (\gamma_{ij}) = (\text{Cov}(X_i,X_j))$ তখন সমস্ত বাস্তব লিনিয়ার সংমিশ্রণের সেট থাকে এক্স i ভিত্তি E = ( এক্স 1 , এক্স 2 , … ,) এর সাথে $X_i$একটি $n$ মাত্রিক আসল ভেক্টর স্পেস তৈরি করে$E=(X_1,X_2,\ldots, X_n)$ and a non-degenerate inner product given by

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

Its dual basis with respect to this inner product, $E^{*} = (X_1^{*},X_2^{*}, \ldots, X_n^{*})$, is uniquely defined by the relationships

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

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

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

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

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:

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

(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 $E$:

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

Then by definition

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

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

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

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

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

Comments

+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 $v_1, ..., v_n$ 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

@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 $\phi:V\to R$, 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 $\gamma$ corresponds to such a map $g:V\to V^*$, and vice versa, via $$g(v)(w)=\gamma(v,w).$$ (Nondegeneracy of $\gamma$ 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 $\gamma$.

— 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

Answer 2

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.

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

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:

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

Write down the covariance matrix as

$$\Omega = \begin{pmatrix} \Omega_{11} & \Omega_{12} \\ \Omega_{21} & \Omega_{22} \end{pmatrix}$$ where $\Omega_{11}$ is covariance matrix of $(X_i, X_j)$ and $\Omega_{22}$ is covariance matrix of $\mathbf{V} \setminus \{X_i, X_j \}$.

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

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

By the key matrix identity,

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

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

$$\rho_{X_iX_j\cdot \mathbf{V} \setminus \{X_i,X_j\}} = \frac{[P_{11}^{-1}]_{12}}{\sqrt{[P_{11}^{-1}]_{11}[P_{11}^{-1}]_{22}}}.$$ I use the notation that the $(k,l)$th entry of the matrix $M$ is denoted by $[M]_{kl}$.

Just simple inversion formula of 2-by-2 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,

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

Comments

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.

Answer 3

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!