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Elliptical and Skew Distributions · Tutorial 697 of 975

Conditional Laws of Multivariate Skew-Normal Vectors

Use Gaussian conditioning and truncation to derive a partitioned skew-normal conditional law and calculate its updated parameters and moments.

Intermediate 9 min read

What You'll Learn

  • Express a multivariate skew-normal vector through a jointly Gaussian latent variable conditioned to be positive.
  • Derive the conditional Gaussian and latent-variable parameters after observing one block.
  • Identify the conditional extended skew-normal kernel, shape and extension parameters.
  • Calculate the conditional mean and covariance without confusing them with kernel parameters.
  • Check parameter updates in independent and correlated worked examples.

Why Conditioning Changes the Skew-Normal Law

For a Gaussian vector, conditioning on one block changes the location and covariance of the remaining block but leaves the family Gaussian. A skew-normal vector is different: its asymmetry comes from selecting a joint Gaussian vector according to the sign of a latent variable. Observing part of the vector updates the distribution of that latent variable as well as the Gaussian block. The resulting conditional law is generally an extended skew-normal, with an additional extension parameter.

The derivation below makes this update explicit. It also separates the parameters of the conditional normal kernel from the actual conditional mean and covariance, which need not equal those kernel parameters.

Latent Gaussian Representation

Definition: Let \(X\) be a \(p\)-vector and \(H\) a scalar such that \[ \begin{pmatrix}X\\H\end{pmatrix} \sim N_{p+1}\!\left( \begin{pmatrix}\xi\\0\end{pmatrix}, \begin{pmatrix}\Omega&q\\q^\mathsf{T}&1\end{pmatrix} \right), \] where \(\Omega\) is positive definite and \(q^\mathsf{T}\Omega^{-1}q<1\). The skew-normal vector is the conditional distribution of \(X\) given \(H>0\). Its density is \[ f_X(x\mid H>0)=2\phi_p(x;\xi,\Omega) \Phi\!\left(\frac{q^\mathsf{T}\Omega^{-1}(x-\xi)} {\sqrt{1-q^\mathsf{T}\Omega^{-1}q}}\right). \] Here \(\phi_p(\cdot;\xi,\Omega)\) is the \(p\)-variate normal density and \(\Phi\) is the standard normal distribution function.

To obtain the density, condition the joint Gaussian law on \(X=x\). Then \(H\mid X=x\) is normal with mean \(q^\mathsf{T}\Omega^{-1}(x-\xi)\) and variance \(1-q^\mathsf{T}\Omega^{-1}q\). Thus the probability of selection at \(x\) is the displayed \(\Phi\) term. Since \(P(H>0)=1/2\), conditioning on selection multiplies the joint marginal density of \(X\) by 2.

For comparison with the usual shape-vector notation, let \(\omega\) be diagonal with entries \(\sqrt{\Omega_{jj}}\). The shape vector in the density’s standardized argument is \[ \alpha=\frac{\omega\Omega^{-1}q}{\sqrt{1-q^\mathsf{T}\Omega^{-1}q}}. \] Indeed, \(\alpha^\mathsf{T}\omega^{-1}(x-\xi)\) equals the argument of \(\Phi\) above. The latent covariance vector \(q\) is often more convenient for conditioning.

Conditional Law After Observing a Block

Partition \(X=(X_1^\mathsf{T},X_2^\mathsf{T})^\mathsf{T}\) and conformably partition \(\xi,\Omega,q\). Write \(X_2=x_2\) for the observed block. Define the Gaussian conditional parameters and the updated latent quantities by

$$ \begin{aligned} \mu&=\xi_1+\Omega_{12}\Omega_{22}^{-1}(x_2-\xi_2),\\ \Sigma&=\Omega_{11}-\Omega_{12}\Omega_{22}^{-1}\Omega_{21},\\ m&=q_2^\mathsf{T}\Omega_{22}^{-1}(x_2-\xi_2),\\ s^2&=1-q_2^\mathsf{T}\Omega_{22}^{-1}q_2,\\ c&=q_1-\Omega_{12}\Omega_{22}^{-1}q_2,\qquad v=s^2-c^\mathsf{T}\Sigma^{-1}c. \end{aligned} $$

These are obtained by conditioning the joint Gaussian vector \((X_1,X_2,H)\) on \(X_2=x_2\). In particular, \(X_1\mid X_2=x_2\) has mean \(\mu\) and covariance \(\Sigma\); \(H\mid X_2=x_2\) has mean \(m\) and variance \(s^2\); and their conditional cross-covariance is \(c\). The positive definiteness of the original joint covariance ensures \(s^2>0\) and \(v>0\).

Theorem: Under the conditions above, the conditional density of \(X_1\) given \(X_2=x_2\) and selection \(H>0\) is \[ f(y\mid x_2,H>0)= \phi_{p_1}(y;\mu,\Sigma) \frac{\Phi\!\left(\dfrac{m+c^\mathsf{T}\Sigma^{-1}(y-\mu)}{\sqrt v}\right)} {\Phi(m/s)}. \] Set \(\omega_c=\operatorname{diag}(\sqrt{\Sigma_{jj}})\). This is an extended skew-normal density with kernel location \(\mu\), kernel covariance \(\Sigma\), shape \[ \alpha_c=\frac{\omega_c\Sigma^{-1}c}{\sqrt v}, \] and extension parameter \[ \tau_c=\frac{m}{\sqrt v}. \] Its normalizing denominator can equivalently be written as \(\Phi\!\left(\tau_c/\sqrt{1+\alpha_c^\mathsf{T}R_c\alpha_c}\right)\), where \(R_c=\omega_c^{-1}\Sigma\omega_c^{-1}\).

Proof. Conditional on \(X_2=x_2\), the pair \((X_1,H)\) is jointly Gaussian with the means, covariance and cross-covariance just given. Conditional again on \(X_1=y\), the latent variable \(H\) has mean \(m+c^\mathsf{T}\Sigma^{-1}(y-\mu)\) and variance \(v=s^2-c^\mathsf{T}\Sigma^{-1}c\). Therefore its probability of being positive is the numerator’s \(\Phi\) term. Dividing the joint density of \(X_1=y\) and \(H>0\), conditional on \(X_2=x_2\), by \(P(H>0\mid X_2=x_2)=\Phi(m/s)\) gives the stated conditional density.

For the parameter form, put \(z=\omega_c^{-1}(y-\mu)\). Then \(c^\mathsf{T}\Sigma^{-1}(y-\mu)/\sqrt v=\alpha_c^\mathsf{T}z\). Also, \[ 1+\alpha_c^\mathsf{T}R_c\alpha_c =1+\frac{c^\mathsf{T}\Sigma^{-1}c}{v} =\frac{s^2}{v}, \] so \(\tau_c/\sqrt{1+\alpha_c^\mathsf{T}R_c\alpha_c}=m/s\), as required. \(\square\)

Conditional Moments Are Not Kernel Parameters

The kernel parameters \(\mu\) and \(\Sigma\) describe the normal density multiplied by the selection factor; they are not usually the conditional mean and covariance. Let \(a=m/s\) and \(r=\phi(a)/\Phi(a)\), where \(\phi\) is the standard normal density. The latent variable \(H\), conditional on \(X_2=x_2\) and \(H>0\), is a normal variable with mean \(m\) and variance \(s^2\), truncated below at zero. Its mean is \(m+sr\), and its variance is \(s^2(1-ar-r^2)\).

Result: The conditional moments are \[ E[X_1\mid X_2=x_2,H>0]=\mu+\frac{c}{s}r, \qquad \operatorname{Var}(X_1\mid X_2=x_2,H>0) =\Sigma-\frac{cc^\mathsf{T}}{s^2}(ar+r^2). \]

Derivation. Given \(X_2=x_2\), Gaussian regression writes \(X_1=\mu+(c/s^2)(H-m)+\varepsilon\), where \(\varepsilon\) is independent of \(H\) and has covariance \(\Sigma-cc^\mathsf{T}/s^2\). Taking the expectation after truncating \(H\) gives \(\mu+(c/s^2)sr=\mu+(c/s)r\). For the covariance, add the covariance of the regression term to that of \(\varepsilon\): \[ \Sigma-\frac{cc^\mathsf{T}}{s^2} +\frac{cc^\mathsf{T}}{s^4}s^2(1-ar-r^2) =\Sigma-\frac{cc^\mathsf{T}}{s^2}(ar+r^2). \] This proves both formulas. \(\square\)

Worked Example: A Two-Block Conditional Law at the Center

Let \(\Omega=I_2\), \(\xi=0\), and \(q=(1/2,1/4)^\mathsf{T}\). The joint Gaussian construction is valid because \(q^\mathsf{T}q=1/4+1/16=5/16<1\). Take \(X_1\) to be the first coordinate and observe \(X_2=0\).

Here \(\mu=0\), \(\Sigma=1\), \(m=(1/4)0=0\), \(s^2=1-(1/4)^2=15/16\), \(c=1/2\), and \(v=15/16-1/4=11/16\). Thus the conditional law has kernel location 0 and kernel variance 1, with shape \(\alpha_c=(1/2)/(\sqrt{11}/4)=2/\sqrt{11}\) and extension \(\tau_c=0\). Its density is \(2\phi(y;0,1)\Phi(2y/\sqrt{11})\); the denominator is \(\Phi(0)=1/2\).

For the moments, \(a=0\) and \(r=\phi(0)/\Phi(0)=\sqrt{2/\pi}\). Hence \[ E[X_1\mid X_2=0,H>0] =\frac{1/2}{\sqrt{15}/4}\sqrt{\frac{2}{\pi}} =\sqrt{\frac{8}{15\pi}}\doteq 0.412. \] The variance is \[ 1-\frac{(1/2)^2}{15/16}\frac{2}{\pi} =1-\frac{4}{15}\frac{2}{\pi} \doteq 0.830. \] The conditional mean is not the kernel location, and the conditional variance is not the kernel variance.

Worked Example: An Observation Produces a Nonzero Extension

Use the same \(\Omega=I_2\), \(\xi=0\), and \(q=(1/2,1/4)^\mathsf{T}\), but now observe \(X_2=1\). The values \(\Sigma=1\), \(c=1/2\), and \(s^2=15/16\) are unchanged. The updated latent mean is \(m=1/4\), so \(v=15/16-1/4=11/16\).

The conditional kernel location and variance remain 0 and 1. The shape is \(2/\sqrt{11}\), while the extension becomes \(\tau_c=(1/4)/(\sqrt{11}/4)=1/\sqrt{11}\). The density is therefore \[ \phi(y;0,1)\frac{\Phi\!\left(1/\sqrt{11}+2y/\sqrt{11}\right)} {\Phi(1/\sqrt{15})}. \] The denominator follows directly from \(m/s=(1/4)/(\sqrt{15}/4)=1/\sqrt{15}\). An observation away from the center changes the extension even though the Gaussian conditional kernel and shape in this example stay fixed.

For a moment check, \(a=1/\sqrt{15}\), \(r=\phi(1/\sqrt{15})/\Phi(1/\sqrt{15})\), and \(c/s=2/\sqrt{15}\). Thus the conditional mean is \(2r/\sqrt{15}\), and the conditional variance is \(1-(4/15)(ar+r^2)\). These expressions use the updated truncation point; replacing \(a\) by zero would incorrectly reuse the centered-case moments.

Worked Example: Conditioning with Correlated Coordinates

Let \(\xi=0\), \(\Omega=\begin{pmatrix}2&1\\1&2\end{pmatrix}\), and \(q=(0.2,0.1)^\mathsf{T}\). The matrix \(\Omega\) is positive definite, and \(q^\mathsf{T}\Omega^{-1}q=(0.08-0.04+0.02)/3=0.02<1\), so the latent construction is valid. Take \(X_1\) as the first coordinate and observe \(X_2=2\).

Since \(\Omega_{22}=2\), the Gaussian conditional location is \(\mu=(1/2)2=1\), and the conditional kernel variance is \(\Sigma=2-1(1/2)1=3/2\). The updated latent quantities are \(m=(0.1)(1/2)2=0.1\), \(s^2=1-(0.1)^2/2=0.995\), and \(c=0.2-(1/2)(0.1)=0.15\). Consequently \(v=0.995-(0.15)^2/(3/2)=0.98\).

The conditional shape, using \(\omega_c=\sqrt{3/2}\), is \[ \alpha_c=\frac{0.15}{\sqrt{3/2}\sqrt{0.98}}, \qquad \tau_c=\frac{0.1}{\sqrt{0.98}}. \] The normal-kernel location and variance are 1 and \(3/2\), and the density’s normalizing denominator is \(\Phi(0.1/\sqrt{0.995})\). This example shows that the update uses the Schur complement and the latent cross-covariance \(c\), not the original marginal parameters alone.

Common Pitfall and Practical Check

A common mistake is to treat the conditional skew-normal as an ordinary skew-normal with only a changed location and shape, setting the extension parameter to zero. That is valid in special cases, such as \(m=0\), but not generally: after observing \(X_2=x_2\), the latent Gaussian variable has conditional mean \(m\), which need not be zero. Omitting this updated truncation point changes both the density’s normalizer and its moments.

A second mistake is to report \(\mu\) and \(\Sigma\) as the actual conditional mean and covariance. They are the Gaussian-kernel parameters. The selection factor shifts the mean and reduces or otherwise modifies the covariance according to the truncated-latent moments. A reliable calculation therefore checks the updated \(m,s^2,c,v\), retains \(\Phi(m/s)\) in the normalizer, and applies the moment formulas only afterward.

1
Condition the Gaussian blocks.
Compute \(\mu\) and \(\Sigma\) for \(X_1\mid X_2=x_2\).
2
Update the latent variable.
Compute its conditional mean \(m\), variance \(s^2\), and cross-covariance \(c\) with \(X_1\).
3
Form the extended skew-normal parameters.
Set \(v=s^2-c^\mathsf{T}\Sigma^{-1}c\), then calculate \(\alpha_c\) and \(\tau_c\).
4
Calculate moments if needed.
Use \(a=m/s\) and \(r=\phi(a)/\Phi(a)\); do not substitute kernel parameters for moments.

Check Your Understanding

Use the latent Gaussian formulation and distinguish distribution parameters from moments.

  1. Why does the conditional density’s denominator involve \(P(H>0\mid X_2=x_2)\), rather than the unconditional probability \(P(H>0)\)?
  2. In the conditional-law theorem, which quantity determines whether the extension parameter is zero?
  3. Explain why \(v=s^2-c^\mathsf{T}\Sigma^{-1}c\) is the conditional variance of \(H\) given both \(X_2=x_2\) and \(X_1=y\).
  4. For the centered worked example, verify that \(q^\mathsf{T}q<1\), then state the conditional kernel variance and actual conditional variance separately.
  5. When using the moment formulas, what changes if \(m\neq 0\) compared with the centered case?
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