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Conditional law of a jointly Gaussian block with positive-definite observed covariance

Statement

Assume AC. Let $X \in R^{d}$ and $Y \in R^{m}$ , with $d, m \geq 1$ , be jointly multivariate normal with mean $(a, b)$ and positive-semidefinite block covariance $Σ = (\begin{array}{cc} A & C \\ C^{T} & B \end{array}),$ where $B$ is positive definite. Then $S = A - C B^{- 1} C^{T}$ is positive semidefinite, and for every $y \in R^{m}$ the probability law $K (y, \cdot) = N_{d} ⁣ (a + C B^{- 1} (y - b), S)$ is defined. The map $K$ is a probability kernel and is a conditional law of $X$ given $Y$ : for all Borel $E \subseteq R^{d}$ and $F \subseteq R^{m}$ , $P (X \in E, Y \in F) = \int_{F} K (y, E) P_{Y} (d y) .$ The covariance $A$ and the conditional covariance $S$ may be singular. Positive definiteness of $B$ is required; no pseudoinverse formula for singular $B$ is asserted.

Facts & Assumptions

Given: AC and a jointly Gaussian pair with the mean and block covariance in the Statement.

[A1]

AC is assumed by the multivariate-normal realization and zero-cross-covariance independence suppliers. The Axiom of Choice.

[F1]

A multivariate-normal law may have positive-semidefinite covariance and is characterized by having the stated scalar-normal linear projections; its mean and covariance are the stated $m$ and $Σ$ . Multivariate normal law, including singular covariance.

[F2]

Jointly Gaussian blocks with zero cross-covariance are independent, including when either covariance block is singular. Jointly Gaussian blocks with zero cross-covariance are independent, including singular blocks.

[F3]

A law is the pushforward probability measure of its random element. Law or distribution of a random element.

[F4]

Integrating a nonnegative product-measurable function against a finite kernel gives a measurable function of the kernel parameter. Measurability of integration against a kernel.

[F5]

Independent random elements have their product joint law. Independent random elements have product joint law.

[F6]

Tonelli evaluates nonnegative integrals on sigma-finite product spaces. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product.

[F7]

Kernel evaluations are measurable and each probability-kernel section is a probability measure. Measure kernel and probability kernel.

Proof

Given: AC and the jointly Gaussian pair $(X, Y)$ .

1.1givenalgebra

Since $B$ is positive definite, it is invertible and symmetric. For any $u \in R^{d}$ , put $v = - B^{- 1} C^{T} u$ . Positive semidefiniteness of $Σ$ gives $0 \leq {(\begin{array}{c} u \\ v \end{array})}^{T} Σ (\begin{array}{c} u \\ v \end{array}) = u^{T} (A - C B^{- 1} C^{T}) u = u^{T} S u .$ Thus $S$ is positive semidefinite. This calculation uses no inverse of $A$ .

2.1A1F1F2givenstep 1.1algebra

Put $D = C B^{- 1}$ and $R = X - a - D (Y - b)$ . Every linear projection of $(R, Y)$ is an affine scalar projection of the jointly Gaussian pair $(X, Y)$ , hence is normal by the projection characterization in [F1]; therefore $(R, Y)$ is jointly Gaussian. Its mean is $(0, b)$ , and block covariance calculation gives $Cov (R, Y) = C - D B = 0, Cov (R) = A - D C^{T} - C D^{T} + D B D^{T} = S .$ The block-independence result [F2] therefore makes $R$ independent of $Y$ , even if $S$ is singular.

3.1A1F1F3F4F7step 2.1

Let $ρ$ be the law of $R$ , and define $K (y, E) = \int_{R^{d}} 1_{E} ⁣ (a + D (y - b) + r) ρ (d r) .$ For each fixed $y$ , this is the pushforward of the probability $ρ$ under the continuous affine map $r \mapsto a + D (y - b) + r$ , so it is a probability measure by [F3]. The map $(y, r) \mapsto a + D (y - b) + r$ is continuous, hence the integrand is product-measurable for Borel $E$ . The finite-kernel measurability theorem [F4] makes $y \mapsto K (y, E)$ Borel; therefore $K$ is a probability kernel by [F7]. Since $R \sim N_{d} (0, S)$ by [F1], its pushforward is $N_{d} (a + D (y - b), S)$ for every $y$ .

4.1A1F2F5F6step 3.1algebra∎

By [F2] and [F5], the joint law of $(R, Y)$ is $ρ \otimes P_{Y}$ . The identity $X = a + D (Y - b) + R$ and Tonelli [F6] then give, for Borel $E, F$ , $P (X \in E, Y \in F) = \int_{F} \int_{R^{d}} 1_{E} (a + D (y - b) + r) ρ (d r) P_{Y} (d y) = \int_{F} K (y, E) P_{Y} (d y) .$ This is the defining conditional-law identity. It holds for every Borel observation set $F$ , so the same kernel is a conditional law of $X$ given $Y$ . The proof requires $B^{- 1}$ only; singularity of $A$ or $S$ causes no change.

Depends on

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Sources

Gelman et al., Bayesian Data Analysis, 3rd ed., §3.5 (standard reference, not scraped)
Simo Särkkä, Bayesian Filtering and Smoothing, Appendix A.1 (standard reference, not scraped)