Why Mix a Skew-Normal Vector Over a Scale?
A skew-normal model can represent asymmetry, while a Student \(t\) model represents heavier tails than a Gaussian model. A positive scale mixture combines these features: conditional on the scale, the vector has a skew-normal distribution; variation in the scale allows occasional observations to be much more dispersed. The resulting model is a multivariate skew-\(t\) distribution.
The construction is useful because it separates two roles. The skew-normal shape parameter controls the skewing mechanism, while the random scale controls tail thickness. The derivation also shows why the skew-\(t\) density contains a Student \(t\) distribution function rather than a Gaussian distribution function.
Definition of the Scale Mixture
Let \(p\) be the dimension, let \(\xi\in\mathbb{R}^p\), and let \(\Omega\) be a positive definite \(p\times p\) matrix. Write \(\omega=\operatorname{diag}(\sqrt{\Omega_{11}},\ldots,\sqrt{\Omega_{pp}})\), and let \(\alpha\in\mathbb{R}^p\) be a shape vector. The multivariate skew-normal density in the parameterization used here is
where \(\phi_p(\,\cdot\,;\xi,\Omega)\) is the \(N_p(\xi,\Omega)\) density and \(\Phi\) is the standard normal distribution function. In particular, \(\Omega\) is the scale matrix of the underlying Gaussian kernel; when \(\alpha\ne 0\), it need not equal the covariance matrix of the skew-normal vector.
For the conditional density, the diagonal scale matrix associated with \(\Omega/w\) is \(\omega/\sqrt w\). Thus the conditional skewing argument is \(\sqrt w\,\alpha^\top\omega^{-1}(x-\xi)\). Define
Worked Example: Generating a Bivariate Skew-\(t\) Vector
Take \(p=2\), \(\xi=(0,0)^\top\), \(\Omega=I_2\), \(\alpha=(1,1)^\top\), and \(\nu=4\). The construction is implemented in two stages.
First draw \(U\sim\chi^2_4\) and set \(W=U/4\). Since \(U>0\) with probability one, \(W\) is a positive scale variable. Its mean is \(1\), and its variance is \(2/4=1/2\).
Then, conditional on the realized \(W=w\), draw \(X\) from \(\mathrm{SN}_2(0,I_2/w,(1,1)^\top)\). For example, if \(w=1/2\), the Gaussian kernel scale matrix is \(2I_2\), and the skewing argument at \(x=(1,2)^\top\) is \(\sqrt{1/2}(1+2)=3/\sqrt 2\).
A small value of \(W\) therefore broadens the conditional kernel, while a large value narrows it. The same scale \(W\) acts on both coordinates, so unusually large scales affect the vector jointly rather than independently coordinate by coordinate.
Deriving the Marginal Density
The key calculation is an expectation of a normal distribution function over a Gamma variable. It converts the conditional skewing factor into a Student \(t\) distribution function.
Proof. Let \(U=\lambda W\), so \(U\sim\chi^2_m\), and let \(N\sim N(0,1)\) be independent of \(U\). By conditioning on \(W\),
The ratio \(N/\sqrt{U/m}\) has a standard \(t_m\) distribution by the defining normal-over-independent-chi-squared representation. The displayed probability is therefore the stated \(t_m\) distribution function. The division is valid because \(U>0\) with probability one. This argument applies to negative as well as positive \(c\), since the denominator is positive. \(\square\)
We now integrate the conditional skew-normal density. The density of \(X\) is the integral of that conditional density against the Gamma density of \(W\). In the integrand, separate the Gaussian kernel from the skewing factor. The product of the prior density of \(W\) and the Gaussian kernel is proportional to
Consequently, after normalizing this factor, the conditional distribution of \(W\) given \(x\) under the Gaussian-kernel part of the mixture is \(\operatorname{Gamma}((\nu+p)/2,(\nu+Q)/2)\). This is a calculation within the density integral; it does not assert that the full skew-\(t\) posterior distribution of \(W\) has this form.
Derivation. The Gaussian-kernel part of the mixture integral is the multivariate \(t\) density displayed in the theorem. This follows by integrating the Gaussian density against the Gamma density: the determinant contributes \(w^{p/2}\), the Gaussian exponential contributes \(\exp(-wQ/2)\), and the Gamma density contributes \(w^{\nu/2-1}\exp(-\nu w/2)\). The resulting Gamma integral has shape \((\nu+p)/2\) and rate \((\nu+Q)/2\), giving the stated normalizing constant and power.
The remaining factor is the expectation of \(\Phi(z\sqrt W)\) with respect to that normalized Gamma density. Apply the lemma with \(m=\nu+p\), \(\lambda=\nu+Q\), and \(c=z\). This gives \(T_{\nu+p}(z\sqrt{(\nu+p)/(\nu+Q)})\). Multiplying by the factor \(2\) in the conditional skew-normal density proves the theorem. Since the result was obtained by integrating conditional probability densities, it is normalized and nonnegative. \(\square\)
Worked Example: Evaluating a Bivariate Skew-\(t\) Density
Let \(\xi=(0,0)^\top\), \(\Omega=I_2\), \(\alpha=(1,1/2)^\top\), and \(\nu=3\). Evaluate the density at \(x=(1,2)^\top\).
Here \(d=(1,2)^\top\), so \(Q=1^2+2^2=5\). Also, \(\omega=I_2\), giving \(z=1+(1/2)2=2\). The Student \(t\) distribution function in the skewing factor has \(\nu+p=5\) degrees of freedom, and its argument is $$ 2\sqrt{\frac{5}{3+5}}=2\sqrt{\frac{5}{8}}=\sqrt{\frac52}. $$
The base density is $$ t_2(x;0,I_2,3) =\frac{\Gamma(5/2)}{\Gamma(3/2)(3\pi)} \left(1+\frac53\right)^{-5/2} =\frac{1}{2\pi}\left(\frac83\right)^{-5/2}, $$ because \(\Gamma(5/2)/\Gamma(3/2)=3/2\). Therefore the skew-\(t\) density at \(x=(1,2)^\top\) is $$ \frac{1}{\pi}\left(\frac38\right)^{5/2}T_5\!\left(\sqrt{\frac52}\right). $$ The factor \(2\) doubles the base \(t\) density before the skewing probability is applied; it does not mean the final density must exceed the base density at every point, since the distribution-function factor is at most \(1\).
Worked Example: The Symmetric Student \(t\) Special Case
Set \(\alpha=0\), with any valid \(p,\xi,\Omega,\nu\). Then \(z=0\), and symmetry of the standard \(t\) distribution gives \(T_{\nu+p}(0)=1/2\). Substitution into the derived density yields $$ f_X(x)=2\,t_p(x;\xi,\Omega,\nu)\,\frac12 =t_p(x;\xi,\Omega,\nu). $$
Thus the scale mixture reduces exactly to the multivariate Student \(t\) distribution when there is no skewing. This check also verifies the normalizing factor \(2\) in the skew-\(t\) density: omitting it would make this special case integrate to \(1/2\), not \(1\).
How the Mixing Variable Controls Tail Thickness
The Gamma mixing variable has mean \(E[W]=1\) and variance \(\operatorname{Var}(W)=2/\nu\), as follows from the mean and variance formulas for a Gamma distribution with shape and rate \(\nu/2\). Hence increasing \(\nu\) concentrates the scale more tightly around \(1\), while smaller \(\nu\) permits greater scale variation. In the limiting regime as \(\nu\) grows, the scale becomes increasingly concentrated near one; at finite \(\nu\), small values of \(W\) can produce much broader conditional kernels.
The density derivation makes the tail mechanism precise at the level of the base kernel: mixing a Gaussian kernel over \(W\) gives a multivariate \(t\) kernel with polynomial rather than Gaussian decay. The skewing multiplier is \(2T_{\nu+p}(\cdot)\), which lies between \(0\) and \(2\). Along any fixed direction away from \(\xi\), its argument has a finite limit, so the multiplier approaches a positive finite value; it changes directional density weights but does not remove the polynomial tail generated by the scale mixture. The degrees of freedom \(\nu\) determine the polynomial order of the \(t\) kernel.
A common mistake is to confuse the Gamma variable used in the mixture with a chi-squared variable. For the prior mixing law, \(U=\nu W\sim\chi^2_\nu\), not \(2\nu W\). In the density calculation, the relevant posterior-kernel law is different: if \(W\sim\operatorname{Gamma}(m/2,\lambda/2)\), then \(\lambda W\sim\chi^2_m\). Keeping the rate parameter explicit prevents incorrect rescaling in the normal-over-chi-squared identity.
Check Your Understanding
Use the density derivation and the scale-mixture definition to answer the following.
- In the construction \(W=U/\nu\), what is the distribution of \(U\), and what are the mean and variance of \(W\)?
- For a point \(x\), define \(Q\) and \(z\). Which one enters the Student \(t\) kernel, and which one enters the skewing factor?
- Why does the posterior-kernel Gamma distribution in the density calculation have shape \((\nu+p)/2\) and rate \((\nu+Q)/2\)?
- Show that setting \(\alpha=0\) reduces the density to the ordinary multivariate Student \(t\) density.
- Explain why using a small value of \(W\) broadens the conditional kernel, and how this connects to heavier marginal tails.