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  • Pre-requisites
  • Overview
  • Definition, and statement of key property
  • Example: A normal markov chain
    • The normal markov chain as a multivariate normal
  • Addendum: interpretation of Ω in terms of conditional mean of Xi
    • Application
    • Example

Last updated: 2019-03-31

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Pre-requisites

You should be familiar with the multivariate normal distribution and the idea of conditional independence, particularly as illustrated by a Markov Chain.

Overview

This vignette introduces the precision matrix of a multivariate normal. It also illustrates its key property: the zeros of the precision matrix correspond to conditional independencies of the variables.

Definition, and statement of key property

Let X be multivariate normal with covariance matrix Σ.

The precision matrix, Ω, is simply defined to be the inverse of the covariance matrix: Ω:=Σ1.

The key property of the precision matrix is that its zeros tell you about conditional independence. Specifically: Ωij=0 if and only if Xi and Xj are conditionally independent given all other coordinates of X.

It may help to compare and contrast this with the analogous property of the covariance matrix: Σij=0 if and only if Xi and Xj are independent.

That is, whereas zeros of the covariance matrix tell you about independence, zeros of the precision matrix tell you about conditional independence.

Example: A normal markov chain

Consider a Markov chain X1,X2,X3, where the transitions are given by Xt+1|XtN(Xt,1). You might think of this Markov chain as corresponding to a type of “random walk”: given the current state, the next state is obtained by adding a random normal with mean 0 and variance 1.

The following code simulates a realization of this Markov chain, starting from an initial state X1N(0,1), and plots it.

set.seed(100)
sim_normal_MC=function(length=1000){
  X = rep(0,length)
  X[1] = rnorm(1)
  for(t in 2:length){
    X[t]= X[t-1] + rnorm(1)  
  }
  return(X)
}
plot(sim_normal_MC())

Version Author Date
02d2d36 stephens999 2017-02-20
c3b365a John Blischak 2017-01-02

The normal markov chain as a multivariate normal

If you think a little you should be able to see that the above random walk simulation is actually simulating from a 1000-dimensional multivariate normal distribution!

Why?

Well, let’s write each of the N(0,1) variables we generate using rnorm() in that code as Z1,Z2,. Then: X1=Z1 X2=X1+Z2=Z1+Z2 X3=X2+Z3=Z1+Z2+Z3 etc.

So we can write X=AZ where A is the 1000 by 1000 matrix A=(100011001110).

Let’s take a look at what the covariance matrix Sigma looks like. (We get a good idea from just looking at the top left corner of the matrix what the pattern is)

A = matrix(0,nrow=1000,ncol=1000)
for(i in 1:1000){
    A[i,]=c(rep(1,i),rep(0,1000-i))
}
Sigma = A %*% t(A)
Sigma[1:10,1:10]
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    1    1    1    1    1    1    1    1    1     1
 [2,]    1    2    2    2    2    2    2    2    2     2
 [3,]    1    2    3    3    3    3    3    3    3     3
 [4,]    1    2    3    4    4    4    4    4    4     4
 [5,]    1    2    3    4    5    5    5    5    5     5
 [6,]    1    2    3    4    5    6    6    6    6     6
 [7,]    1    2    3    4    5    6    7    7    7     7
 [8,]    1    2    3    4    5    6    7    8    8     8
 [9,]    1    2    3    4    5    6    7    8    9     9
[10,]    1    2    3    4    5    6    7    8    9    10

Now let us examine the precision matrix, Ω, which recall is the inverse of Σ. Again we just show the top left corner of the precision matrix here.

Omega = chol2inv(chol(Sigma))
Omega[1:10,1:10]
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    2   -1    0    0    0    0    0    0    0     0
 [2,]   -1    2   -1    0    0    0    0    0    0     0
 [3,]    0   -1    2   -1    0    0    0    0    0     0
 [4,]    0    0   -1    2   -1    0    0    0    0     0
 [5,]    0    0    0   -1    2   -1    0    0    0     0
 [6,]    0    0    0    0   -1    2   -1    0    0     0
 [7,]    0    0    0    0    0   -1    2   -1    0     0
 [8,]    0    0    0    0    0    0   -1    2   -1     0
 [9,]    0    0    0    0    0    0    0   -1    2    -1
[10,]    0    0    0    0    0    0    0    0   -1     2

Notice all the 0s in the precision matrix. This is because of the conditional independencies that occur in a Markov chain. In a Markov chain (any Markov chain) the conditional distribution of Xt given the other Xs (st) depends only on its neighbors Xt1 and Xt+1. That is, Xt is conditionally independent of all other Xs given Xt1 and Xt+1. This is exactly what we are seeing in the precision matrix above: the non-zero elements of the tth row are at coordinates t1,t and t+1.

Addendum: interpretation of Ω in terms of conditional mean of Xi

The following fact is also useful, both in practice and for intuition.

Suppose XNr(0,Ω1), where the subscript r indicates that X is r-variate.

Let Y1 denote the first coordinate of X and Y2 denote the remaining coordinates (so Y2:=(X2,,Xr)). Further let Ω12 denote the 1×(r1) sub matrix of Ω that consists of row 1 and columns 2 to r.

The conditional distribution of Y1|Y2 is (univariate) normal, with mean E(Y1|Y2)=(1/Ω11)Ω12Y2
and variance 1/Ω11.

Of course there is nothing special about X1: a similar result applies for any Xp. You just have to replace Ω11 with Ωpp and define Ω12 to be the pth row of Ω with all columns except p.

Application

An application of this is imputation of missing values: suppose one of the X values is missing, say Xp is missing, but you know the covariance matrix and all the other X values. Then you could impute Xp by its conditional mean, which is a simple linear combination of the other values that can be read directly off the pth row of the precision matrix. This idea is the essence of Kriging.

Example

Consider the Markov chain above. The conditional distribution of X1 given all other X values is given by X1|X2,X3,N(X2/2,1/2).

And the conditional distribution of X2 given all other X values is X2|X1,X3,N((X1+X3)/2,1/2). Similarly for Xp for p>2. The intuition is that, if one wanted to guess what the value of Xp were given all other Xs, the best guess would be the average of its neighbours.



sessionInfo()
R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.1

Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
 [1] workflowr_1.2.0 Rcpp_1.0.0      digest_0.6.18   rprojroot_1.3-2
 [5] backports_1.1.3 git2r_0.24.0    magrittr_1.5    evaluate_0.12  
 [9] stringi_1.2.4   fs_1.2.6        whisker_0.3-2   rmarkdown_1.11 
[13] tools_3.5.2     stringr_1.3.1   glue_1.3.0      xfun_0.4       
[17] yaml_2.2.0      compiler_3.5.2  htmltools_0.3.6 knitr_1.21     

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