# Why does a zero entry in the inverse covariance matrix of a joint Gaussian distribution imply conditional independence?

When $$X_1, X_2, \ldots, X_n$$ are random variables jointly following a Gaussian distribution, let $$A$$ be the inverse of the covariance matrix ($$A=\Sigma^{-1}$$). I am wondering how to prove following theorem:

If $$A_{ij}=0,$$ then $$X_i$$ and $$X_j$$ are conditionally independent on the other variables.

## 1 Answer

The aim of this answer is to demonstrate this result with minimal algebraic effort. The secret is to maintain a laser-like focus on what matters, ignoring all the rest. Let me illustrate.

By definition (many definitions anyway), the value of the density of a multivariate Gaussian with vector mean $$\mu$$ and invertible covariance matrix $$\Sigma$$ at the point $$\mathbf x = (x_1, x_2, \ldots, x_n)^\prime$$ is proportional to

$$\exp\left((\mathbf x - \mu)^\prime \Sigma^{-1} (\mathbf x - \mu)/2\right).$$

Conditioning on all the variables except $$(x_i,x_j)$$ is tantamount to viewing all those other variables as constants. Focus, then, on how this density depends on $$(x_i,x_j).$$ To do so, we must examine the argument of $$\exp.$$ Since $$\mu$$ is constant, too, let's consider how it depends on $$(x_i-\mu_i,x_j-\mu_j) = (y_i,y_j).$$ Letting $$(a_{rs}),$$ $$1\le r,$$ $$1\le s$$ be the coefficients of $$\Sigma^{-1},$$ the rules of matrix multiplication imply

\begin{aligned} (\mathbf x - \mu)^\prime \Sigma^{-1} (\mathbf x - \mu) &= a_{ii}y_i^2/2 + \text{constants}\times y_i \\&+ a_{jj}y_j^2/2 + \text{other constants}\times y_j \\ &+ a_{ij}y_iy_j \\&+ \text{yet other constants}. \end{aligned}

That's the sum of four expressions, one per line. The rules of exponentiation then tell us the conditional density is the product of five terms:

1. A normalizing constant from the original (joint) density.

2. A term that is a function only of $$y_i$$ -- which implies it's a function only of $$x_i.$$

3. Another term that is a function only of $$y_j$$ -- which implies it's a function only of $$x_j.$$

4. The $$a_{ij}y_iy_j$$ term.

5. The exponential of yet other constants.

When, as assumed in the question, $$a_{ij} = 0,$$ term (4) drops out. This leaves the product of constants (from $$(1)$$ and $$(5)$$), a function of $$x_i$$ alone (from $$(2)$$), and a function of $$x_j$$ alone (from $$(3)$$), showing explicitly how the conditional density factors as separate functions of $$x_i$$ and $$x_j.$$ This factorization implies the corresponding random variables are independent, QED.

• Thank you very much! Commented Jun 8, 2022 at 5:48