Identifiability in Multivariate Skew-Normal Parameterizations

Why Parameterization Matters

A parameter can describe a feature of a distribution without being that feature itself. In the multivariate skew-normal model, the parameter called “location” is generally not the mean, and the matrix used in the density’s Gaussian kernel is generally not the covariance. These distinctions matter when comparing results reported in different parameterizations.

This tutorial relates the direct density parameterization to a selection construction and to a centered parameterization. It also establishes which quantities are determined by the distribution, rather than by the choice of coordinates used to describe it.

Direct and Selection Parameterizations

Definition: For a $d$-dimensional vector, the direct multivariate skew-normal density with parameters $(\xi,\Omega,\alpha)$ is $$ f(x)=2\phi_d(x;\xi,\Omega)\, \Phi\!\left(\alpha^\mathsf{T}\omega^{-1}(x-\xi)\right), $$ where $\Omega$ is positive definite, $\omega=\operatorname{diag}(\sqrt{\Omega_{11}},\ldots,\sqrt{\Omega_{dd}})$, and $\alpha\in\mathbb R^d$. Here $\phi_d(\,\cdot\,;\xi,\Omega)$ is the $N_d(\xi,\Omega)$ density and $\Phi$ is the standard normal distribution function. The matrix $\Omega$ is a scale matrix in the Gaussian kernel, not the covariance of $X$.

Set $R=\omega^{-1}\Omega\omega^{-1}$, the correlation matrix associated with $\Omega$. A useful alternative shape vector is $$ \delta=\frac{R\alpha}{\sqrt{1+\alpha^\mathsf{T}R\alpha}}, \qquad \eta=\omega\delta. $$ The vector $\delta$ is dimensionless, while $\eta$ is on the scale of $X$. For finite $\alpha$, $\delta^\mathsf{T}R^{-1}\delta<1$, equivalently $\eta^\mathsf{T}\Omega^{-1}\eta<1$.

These quantities express the selection construction. Let $U_0\sim N(0,1)$, let $\varepsilon\sim N_d(0,\Omega-\eta\eta^\mathsf{T})$ be independent of $U_0$, and define $X=\xi+\eta|U_0|+\varepsilon$. The constraint above ensures that $\Omega-\eta\eta^\mathsf{T}$ is positive definite. This construction has the direct density just given: it is the law of the Gaussian vector selected by the event $U_0>0$. In particular, $\xi$ and $\Omega$ are the location and covariance of the unselected Gaussian vector; selection changes the mean and covariance of the observed vector.

Worked Example: Converting a Direct Shape Vector

Take $d=2$, $\xi=(0,0)^\mathsf{T}$, $\Omega=I_2$, and $\alpha=(1,0)^\mathsf{T}$. Then $\omega=I_2$, $R=I_2$, and $$ \delta=\frac{(1,0)^\mathsf{T}}{\sqrt{1+1}} =\left(\frac{1}{\sqrt 2},0\right)^\mathsf{T}, \qquad \eta=\left(\frac{1}{\sqrt 2},0\right)^\mathsf{T}. $$ The constraint holds because $\eta^\mathsf{T}\Omega^{-1}\eta=1/2<1$. The selection residual covariance is $$ \Omega-\eta\eta^\mathsf{T} = \begin{pmatrix}1/2&0\\0&1\end{pmatrix}, $$ which is positive definite. Thus the same model can be generated as $X=\eta|U_0|+\varepsilon$, with this residual covariance and independent standard normal $U_0$.

Mean, Covariance, and Centered Shape

Write $c=E|U_0|=\sqrt{2/\pi}$. Since $\operatorname{Var}(|U_0|)=1-c^2$, the selection construction gives the mean and covariance directly. It motivates the centered parameterization $(\mu,\Sigma,\eta)$, where $\mu=E[X]$ and $\Sigma=\operatorname{Var}(X)$.

Theorem: Under the direct skew-normal model, $$ \mu=\xi+c\eta, \qquad \Sigma=\Omega-c^2\eta\eta^\mathsf{T}. $$ Conversely, the centered parameters determine the direct location and scale by $\xi=\mu-c\eta$ and $\Omega=\Sigma+c^2\eta\eta^\mathsf{T}$. The centered parameter domain is $\Sigma$ positive definite and $(1-c^2)\eta^\mathsf{T}\Sigma^{-1}\eta<1$.

To verify the formulas, write $X=\xi+\eta|U_0|+\varepsilon$, with $\operatorname{Var}(\varepsilon)=\Omega-\eta\eta^\mathsf{T}$. Independence gives

E[X]=\xi+c\eta,\qquad \operatorname{Var}(X) =\Omega-\eta\eta^\mathsf{T}+(1-c^2)\eta\eta^\mathsf{T} =\Omega-c^2\eta\eta^\mathsf{T}.

The inverse formulas follow by rearranging. To check the centered domain, put $s=\eta^\mathsf{T}\Sigma^{-1}\eta$. The rank-one matrix $\Sigma-(1-c^2)\eta\eta^\mathsf{T}$ is positive definite exactly when $(1-c^2)s<1$. This matrix equals $\Omega-\eta\eta^\mathsf{T}$ after substituting $\Omega=\Sigma+c^2\eta\eta^\mathsf{T}$, so the condition ensures a valid selection construction. It also ensures that the recovered direct shape is finite.

For completeness, recover that direct shape by setting $\omega=\operatorname{diag}(\sqrt{\Omega_{11}},\ldots,\sqrt{\Omega_{dd}})$, $R=\omega^{-1}\Omega\omega^{-1}$, and $\delta=\omega^{-1}\eta$. Then $$ \alpha= \frac{R^{-1}\delta}{\sqrt{1-\delta^\mathsf{T}R^{-1}\delta}}. $$ Indeed, if $t=\alpha^\mathsf{T}R\alpha$, the direct-to-selection formula gives $\delta=R\alpha/\sqrt{1+t}$, so $\delta^\mathsf{T}R^{-1}\delta=t/(1+t)$. Solving these relations yields the displayed inverse.

Worked Example: Computing the Centered Parameters

Continue with $\xi=(0,0)^\mathsf{T}$, $\Omega=I_2$, and $\alpha=(1,0)^\mathsf{T}$. From the previous calculation, $\eta=(1/\sqrt2,0)^\mathsf{T}$. Therefore $$ \mu=\left(\frac{c}{\sqrt2},0\right)^\mathsf{T} =\left(\frac{1}{\sqrt\pi},0\right)^\mathsf{T}, \qquad \Sigma= \begin{pmatrix}1-1/\pi&0\\0&1\end{pmatrix}. $$ The first variance is positive because $1-1/\pi>0$. Reconstructing the direct parameters gives $$ \mu-c\eta=(0,0)^\mathsf{T}=\xi, \qquad \Sigma+c^2\eta\eta^\mathsf{T} = \begin{pmatrix}1-1/\pi+1/\pi&0\\0&1\end{pmatrix} =I_2=\Omega. $$ Thus the mean and covariance differ from the direct location and scale, even though the two parameter sets describe the same distribution.

What Is Identifiable?

A parameterization is identifiable if no two distinct allowed parameter values give the same probability distribution. For this model, the centered quantities are identifiable; the direct parameters are then identifiable as well because the conversion between the two parameterizations is one-to-one on their valid domains.

Theorem: Within the multivariate skew-normal family with the stated parameter constraints, the distribution uniquely determines $\mu$, $\Sigma$, and $\eta$. Consequently, it uniquely determines $\xi$, $\Omega$, and $\alpha$ in the direct parameterization.

Here is a proof using the third cumulant. Let $H=|U_0|$. The centered construction is $X-\mu=\eta(H-c)+\varepsilon$, with independent Gaussian $\varepsilon$. The centered third cumulant tensor, whose $i,j,k$ entry is $E[(X_i-\mu_i)(X_j-\mu_j)(X_k-\mu_k)]$, is $$ \mathcal K_3(X)=\kappa_3(H)\,\eta^{\otimes 3}, \qquad \kappa_3(H)=E[(H-c)^3]=c(4/\pi-1)>0. $$ For the scalar calculation, $E[H]=c$, $E[H^2]=1$, and $E[H^3]=2c$, so $E[(H-c)^3]=2c-3c+3c^3-c^3=-c+2c^3=c(4/\pi-1)$. The Gaussian residual has zero third cumulants, and independence makes the mixed third cumulants vanish; this proves the tensor formula.

Suppose two parameter triples in the centered parameterization yield the same distribution. Their means and covariance matrices agree, since these are determined by the distribution. Their third cumulant tensors agree as well. If this tensor is zero, its displayed formula and $\kappa_3(H)>0$ imply $\eta=0$ for both triples. Otherwise, choose an index $j$ with $\eta_j\ne0$. The $j,j,j$ entry is $\kappa_3(H)\eta_j^3$, which determines the real value of $\eta_j$ uniquely. For each $i$, the $i,j,j$ entry is $\kappa_3(H)\eta_i\eta_j^2$, which then determines $\eta_i$ uniquely. Thus both triples have the same $\eta$, as well as the same $\mu$ and $\Sigma$. The inverse formulas recover the same $\xi$ and $\Omega$; the formula for $\alpha$ then recovers the same direct shape vector. This proves identifiability. $\square$

Worked Example: Reading Shape From the Third Cumulant

Suppose a two-dimensional skew-normal distribution has third cumulant entries $\mathcal K_{3,111}=2\kappa_3(H)$ and $\mathcal K_{3,211}=-\kappa_3(H)$, where $\kappa_3(H)=c(4/\pi-1)$. Since $\mathcal K_{3,111}=\kappa_3(H)\eta_1^3$, it follows that $\eta_1^3=2$, hence $\eta_1=\sqrt[3]{2}$. Next, $$ \mathcal K_{3,211}=\kappa_3(H)\eta_2\eta_1^2 \quad\Longrightarrow\quad \eta_2=\frac{-1}{(\sqrt[3]{2})^2}. $$ The signs and magnitudes follow from the tensor entries, not from a choice of sign convention. The other third cumulant entries must agree with these recovered components if the proposed tensor comes from this model.

Which Quantities Can Be Compared?

The mean $\mu$ and covariance $\Sigma$ are distributional quantities: every valid parameterization of the same distribution gives the same values. The centered shape vector $\eta$ is also uniquely determined within this family. Equivalently, its direction and magnitude are encoded by the nonzero third cumulant tensor $\kappa_3(H)\eta^{\otimes3}$. The direct parameters $\xi,\Omega,\alpha$ are identifiable too, but they have different roles: $\xi$ is the Gaussian-kernel location, $\Omega$ is its scale, and $\alpha$ is a standardized shape coefficient.

When changing coordinates by $Y=AX+b$, with $A$ nonsingular, the distributional quantities transform as $E[Y]=A\mu+b$, $\operatorname{Var}(Y)=A\Sigma A^\mathsf{T}$, and the centered shape vector as $A\eta$. Thus a numerical shape vector is meaningful only with its coordinate system and parameterization specified.

Common Mistakes

Calling $\xi$ the mean. In general, $E[X]=\xi+c\eta$. The two coincide only when the shape vector is zero.
Calling $\Omega$ the covariance. The covariance is $\Omega-c^2\eta\eta^\mathsf{T}$; $\Omega$ belongs to the Gaussian kernel and selection construction.
Comparing shape vectors without conversion. The direct vector $\alpha$, standardized selection vector $\delta$, and centered vector $\eta$ have different definitions. Convert them before comparing numerical values.
Assuming every centered triple is valid. A positive definite $\Sigma$ alone is not enough: the additional constraint $(1-c^2)\eta^\mathsf{T}\Sigma^{-1}\eta<1$ is required.

Key takeaway: For the scalar-selection multivariate skew-normal family, direct and centered parameters are different coordinates for the same identifiable distribution. The mean, covariance, and centered shape are not interchangeable with the direct location, scale, and shape parameters.

Check Your Understanding

Use the definitions and conversions above to answer these questions.

Why is the direct scale matrix $\Omega$ generally not the covariance matrix of $X$?
Given $\eta$ and $\Omega$, what condition ensures that the selection residual covariance is positive definite?
How can the third cumulant tensor distinguish two candidate centered shape vectors?
When does the direct location $\xi$ equal the mean $\mu$?
Why must shape vectors from different parameterizations be converted before their numerical values are compared?