Azzalini’s Skew-Normal Density and Latent Construction

From Gaussian Symmetry to Skewness

A multivariate Gaussian density is symmetric about its center: points in opposite directions from that center have equal density. A simple way to introduce asymmetry is to multiply the Gaussian density by a probit factor that gives different weights to opposite sides of a linear boundary. Azzalini’s skew-normal distribution uses exactly this construction. The same density also arises by taking a Gaussian vector and retaining it only when a related latent Gaussian variable is positive.

The connection between these descriptions is useful because it gives the density a probability-based interpretation and identifies which parameter controls its asymmetry. It also prevents a common parameterization error: the shape parameter is not, in general, the correlation between the observed vector and the latent variable.

Density Definition and Normalization

Let $d$ be a positive integer, let $\xi\in\mathbb{R}^d$, and let $\Omega$ be a positive-definite $d\times d$ scale matrix. Write $\omega=\operatorname{diag}(\sqrt{\Omega_{11}},\ldots,\sqrt{\Omega_{dd}})$ and $R=\omega^{-1}\Omega\omega^{-1}$, the corresponding correlation matrix. For a vector $x$, the Gaussian density with location $\xi$ and scale $\Omega$ is denoted by $\phi_d(x;\xi,\Omega)$. Let $\Phi$ denote the standard univariate normal distribution function.

Definition: The $d$-dimensional Azzalini skew-normal distribution with location parameter $\xi$, scale matrix $\Omega$, and shape parameter $\alpha\in\mathbb{R}^d$ has density $$ f(x)=2\phi_d(x;\xi,\Omega)\Phi\!\left(\alpha^{\mathsf T}\omega^{-1}(x-\xi)\right),\qquad x\in\mathbb{R}^d. $$ Here $\Omega$ is positive definite. It is the scale matrix in the Gaussian kernel; in general it is not the covariance matrix of the skew-normal vector.

The standardized coordinate $u=\omega^{-1}(x-\xi)$ has Gaussian kernel with correlation matrix $R$. Thus the argument of $\Phi$ is the linear function $\alpha^{\mathsf T}u$. This standardization matters when the components have different scales: it is $\alpha$ acting on standardized coordinates, not directly on the unstandardized differences $x-\xi$.

The factor of 2 makes the expression a density. To verify this, change variables to $u=\omega^{-1}(x-\xi)$. The Gaussian part becomes the standard $d$-variate normal density with correlation matrix $R$, denoted $\phi_d(u;0,R)$, and the integral of $f$ becomes

2\int_{\mathbb{R}^d}\phi_d(u;0,R)\Phi(\alpha^{\mathsf T}u)\,du.

The centered Gaussian density is unchanged by $u\mapsto -u$. Therefore the integral without the leading 2 is equal both to the integral with $\Phi(\alpha^{\mathsf T}u)$ and to the one with $\Phi(-\alpha^{\mathsf T}u)$. Adding these two equal integrals and using $\Phi(t)+\Phi(-t)=1$ gives

2\int_{\mathbb{R}^d}\phi_d(u;0,R)\Phi(\alpha^{\mathsf T}u)\,du =\int_{\mathbb{R}^d}\phi_d(u;0,R)\,du=1.

All integrals exist because $0\leq \Phi\leq 1$ and the Gaussian density integrates to one. This proves the normalization without imposing any restriction on $\alpha$ beyond being a finite vector.

Worked Example: The One-Dimensional Shape Parameter

In one dimension, take location $\xi=0$, scale $\Omega=1$, and shape $\alpha=2$. The density is

f(u)=2\phi(u)\Phi(2u),

where $\phi$ is the standard normal density. The multiplier at $u=1$ is $2\Phi(2)$, approximately $1.9545$, whereas at $u=-1$ it is $2\Phi(-2)$, approximately $0.0455$. The Gaussian kernel is symmetric, but the probit multiplier weights these two locations very differently. If $\alpha=0$, the multiplier is $2\Phi(0)=1$ everywhere, and the density reduces exactly to $\phi(u)$.

Latent-Normal Selection Construction

The selection construction starts with a jointly Gaussian pair consisting of a scalar latent variable and a $d$-vector. Work first in standardized coordinates. Let $U$ have distribution $N_d(0,R)$, where $R$ is positive definite, and let $T$ be standard normal. Specify their joint covariance matrix as

\operatorname{Cov}\begin{pmatrix}T\\U\end{pmatrix} = \begin{pmatrix} 1 & \delta^{\mathsf T}\\ \delta & R \end{pmatrix},

where $\delta=\operatorname{Cov}(U,T)\in\mathbb{R}^d$. For this to be a nondegenerate joint Gaussian distribution, the block covariance matrix must be positive definite. Since $R$ is positive definite, the Schur complement criterion says this is equivalent to

1-\delta^{\mathsf T}R^{-1}\delta>0.

Now retain $U$ only when $T>0$. Since $T$ is standard normal, $P(T>0)=1/2$. The conditional density of $U$ after selection can be derived directly.

Theorem: Under the joint Gaussian model above and the condition $\delta^{\mathsf T}R^{-1}\delta<1$, the conditional density of $U$ given $T>0$ is $$ f_{U\mid T>0}(u)=2\phi_d(u;0,R) \Phi\!\left(\frac{\delta^{\mathsf T}R^{-1}u}{\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}}\right). $$ It is an Azzalini skew-normal density with standardized shape parameter $$ \alpha=\frac{R^{-1}\delta}{\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}}. $$

Proof. The Gaussian conditioning formula gives $T\mid U=u\sim N(\delta^{\mathsf T}R^{-1}u,\,1-\delta^{\mathsf T}R^{-1}\delta)$. Hence

P(T>0\mid U=u) = \Phi\!\left(\frac{\delta^{\mathsf T}R^{-1}u}{\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}}\right).

For any measurable set $A\subseteq\mathbb{R}^d$, conditioning first on $U$ gives

P(U\in A\mid T>0) =\frac{P(U\in A,\ T>0)}{P(T>0)} =2\int_A \phi_d(u;0,R)P(T>0\mid U=u)\,du.

The integrand therefore gives the displayed conditional density. Its shape vector is the coefficient of $u$ in the probit argument, namely $R^{-1}\delta/\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}$. Finally, if $X=\xi+\omega U$, its density is the definition above with scale $\Omega=\omega R\omega$ and the same $\alpha$, because $u=\omega^{-1}(x-\xi)$. $\square$

Converting Between Shape and Latent Correlation

The theorem expresses $\alpha$ in terms of $\delta$. The inverse relation is also useful. For any finite $\alpha\in\mathbb{R}^d$, define

\delta=\frac{R\alpha}{\sqrt{1+\alpha^{\mathsf T}R\alpha}}.

This gives an admissible latent correlation vector and recovers the original shape. To check both claims, let $c=\alpha^{\mathsf T}R\alpha$, which is nonnegative since $R$ is positive definite. Then

\delta^{\mathsf T}R^{-1}\delta =\frac{\alpha^{\mathsf T}R\alpha}{1+\alpha^{\mathsf T}R\alpha} =\frac{c}{1+c}<1, \qquad \frac{R^{-1}\delta}{\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}} =\frac{\alpha/\sqrt{1+c}}{\sqrt{1/(1+c)}}=\alpha.

Thus every finite shape vector corresponds to a valid latent-normal selection model. Conversely, the formula in the theorem gives a finite shape vector for every admissible $\delta$. The vector $\delta$ records latent correlations, while $\alpha$ records the strength and direction of the probit weighting after accounting for $R$ and the residual variance of $T$ given $U$.

Worked Example: A Two-Dimensional Latent Construction

Let $R=I_2$, the $2\times2$ identity matrix, and choose $\alpha=(1,2)^{\mathsf T}$. Then $\alpha^{\mathsf T}R\alpha=1^2+2^2=5$, so the corresponding latent correlation vector is

\delta=\frac{\alpha}{\sqrt{6}} =\begin{pmatrix}1/\sqrt{6}\\2/\sqrt{6}\end{pmatrix}.

Its admissibility is verified by $\delta^{\mathsf T}R^{-1}\delta=(1+4)/6=5/6<1$. Substituting into the shape formula gives

\frac{R^{-1}\delta}{\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}} = \frac{(1,2)^{\mathsf T}/\sqrt{6}}{\sqrt{1/6}} = \begin{pmatrix}1\\2\end{pmatrix}.

The selected density is $2\phi_2(u;0,I_2)\Phi(u_1+2u_2)$. At $u=(1/2,-1/2)^{\mathsf T}$, the probit argument is $1/2-1= -1/2$, so the density there is $2\phi_2(u;0,I_2)\Phi(-1/2)$. The argument’s sign shows why selection does not simply favor positive values of both coordinates: it favors values according to the particular linear combination $u_1+2u_2$.

Worked Example: Evaluating a Density With Nonunit Scale

Consider the univariate model with $\xi=2$, $\Omega=4$, and $\alpha=3/2$. Here $\omega=2$. At $x=3$, the standardized coordinate is $u=(3-2)/2=1/2$, so the probit argument is $(3/2)(1/2)=3/4$. The density is

f(3) =2\left[\frac{1}{2}\phi\!\left(\frac{3-2}{2}\right)\right]\Phi(3/4) =\phi(1/2)\Phi(3/4) \approx (0.3521)(0.7734) \approx 0.2723.

The factor $1/2$ inside the brackets is the Jacobian scale factor $1/\omega$ in the Gaussian density. Omitting it would give the wrong density. The same scale also appears in the standardized shape argument; replacing that argument by $\alpha(x-\xi)$ would incorrectly produce $3/2$ instead of $3/4$.

Interpretation and a Common Pitfall

The selection representation explains the source of asymmetry: observations are drawn from a joint Gaussian model, but only vectors paired with a positive latent value are retained. The resulting density remains Gaussian times a Gaussian distribution function, although selection changes its symmetry. This construction is useful for specifying and interpreting skew-normal models, while the Gaussian kernel parameters should not be mistaken for summaries of the selected distribution.

Pitfall: treating $\delta$ as the shape vector. The wrong reasoning is that the latent correlation vector directly becomes the coefficient in the probit argument. In fact, conditioning gives the coefficient $R^{-1}\delta/\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}$. Only in special cases would this happen to equal $\delta$.
Pitfall: using unstandardized coordinates with the stated shape parameter. The density uses $\alpha^{\mathsf T}\omega^{-1}(x-\xi)$. Using $\alpha^{\mathsf T}(x-\xi)$ changes the model when component scales are not one.
Pitfall: calling $\Omega$ the selected covariance matrix. It is the scale matrix of the Gaussian kernel before selection. Selection changes the distribution, so $\Omega$ should not automatically be interpreted as the covariance of $X$.

Key takeaway: In standardized coordinates, Azzalini’s density is $2\phi_d(u;0,R)\Phi(\alpha^{\mathsf T}u)$. A latent Gaussian selection model produces it with $\alpha=R^{-1}\delta/\sqrt{1-\delta^{\mathsf T}R^{-1}\delta}$, and every finite $\alpha$ has a valid inverse representation.

Check Your Understanding

Use the density and selection construction to answer these questions.

Why does the factor of 2 normalize the density for every finite shape vector $\alpha$?
In the latent model, what condition on $R$ and $\delta$ ensures a nondegenerate joint Gaussian distribution?
For $R=I_2$ and $\alpha=(2,0)^{\mathsf T}$, calculate the corresponding latent correlation vector $\delta$.
Explain why the Gaussian kernel scale matrix $\Omega$ should not automatically be called the covariance matrix after selection.
For a general scale matrix, write the probit argument in terms of $x,\xi,\omega$, and $\alpha$.