Is there a moment-generating function or a characteristic function for a univariate skew-t distribution $y\sim ST\left(\xi,\omega^2,\alpha,\nu\right)$ as defined by Azzalini?

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    $\begingroup$ What did Azzalini have to say about it, when he defined it? He is a pretty thorough sort of chap. $\endgroup$
    – wolfies
    Dec 16, 2014 at 5:30

1 Answer 1


$\newcommand{\mi}{\mathrm{i}}$ This may not be as pleasant as you were hoping for, but for both the univariate and multivariate cases see:

Hyoung-Moon Kim and Marc G.Genton, "Characteristic functions of scale mixtures of multivariate skew-normal distributions", Journal of Multivariate Analysis 102 (2011) 1105–1117

In equation (13) they introduce an earlier result due to Pewsey, that if $X \sim SN(\xi, \, \omega^2, \, \alpha)$ is skew-normal and we define $\delta = \alpha / \sqrt{1 + \alpha^2}$, then $X$ has characteristic function:

$$\Psi_X(t) = \exp(\mi \xi t - \omega^2 t^2 /2)\{1 + \mi \tau(\delta \omega t)\} \text{, for } t \in \mathbb{R}$$

Here $\tau(x) = \int_0^x \sqrt{2/\pi} \exp(u^2/2) \, \mathrm{d}u$ for positive $x$, with $\tau(-x) = -\tau(x)$.

They proceed to prove their Theorem 1, which gives an alternative form for the characteristic function of $X$, which equals the Pewsey's result.

$$\Psi_X(t) = 2 \exp(\mi \xi t - \omega^2 t^2 /2) \, \Phi(\mi \delta \omega t)$$

They note that if $X \sim SN(0, \, \omega^2, \, \alpha)$ is skew-normal and $\eta \sim IG(\nu/2, \nu/2)$ is inverse-Gamma with CDF $H(\eta)$, then $Y = \xi + \eta^{1/2}X$ has Azzalini's skew-t distribution. They don't state as such, but $\eta$ should be independent of $X$ - see e.g. equation 25 of this full-length version of A. Azzalini and A. Capitanio (2003), "Distributions generated by perturbation of symmetry with emphasis on a multivariate skew t distribution", J. Roy. Statist. Soc., B 65, 367-389. For some intuition, note that multiplying by the square root of this inverse-gamma variable is equivalent to dividing by the square root of a $\chi^2_{\nu}$ variable that has been scaled by dividing through by its degrees of freedom. They use this to show (equation 20) that the skew-t has characteristic function:

$$\Psi_Y(t) = \exp(\mi \xi t) \left( \Psi_T(\omega t) + \mi \tau^*(\delta, \, \omega t) \right)$$


$\Psi_T(t) = \frac{K_{\nu/2}(\sqrt{\nu} |t|)(\sqrt{\nu} |t|)^{\nu/2}}{\Gamma(\nu/2) 2^{\nu/2 - 1}} \text{, for } t \in \mathbb{R}, \, \nu>0$

$\tau^{*}(\delta, \, \omega t) = \int_0^\infty \exp (-\eta \omega^2 t^2 /2) \, \tau(\delta \sqrt{\eta} \omega t) \, \mathrm{d}H(\eta) \text{, for } \delta t > 0 $

$\tau^{*}(\delta, \, -\omega t) = -\tau^{*}(\delta, \, \omega t)$

$K_{\lambda}(w) = \frac{1}{2} \int_0^\infty x^{\lambda - 1} \exp \left( -\frac{1}{2}w (x + x^{-1}) \right) \, \mathrm{d}x \text{, for } w>0, \, \lambda \in \mathbb{R}$

They state that $K_{\lambda}$ is a modified Bessel function of the third kind, but I think it is more common to call it a modified Bessel function of the second kind (see Wikipedia and Wolfram MathWorld).

For completeness I will include their results for the multivariate skew-normal and skew-t, but beware that different authors define these distributions in different ways (see the skew-normal tag wiki entry for for citations)!

The $k$-dimensional multivariate skew-normal is introduced in Section 2.2, firstly as $Z \sim SN_k(\Omega_Z, \alpha)$ where $\Omega_Z$ is a correlation (not just covariance) matrix and $\alpha$ is a vector (of length $k$) for skewness. This has pdf $f_Z(z) = 2 \phi_k(z; \, \Omega_Z) \Phi(\alpha^T z)$ where $\phi(z; \, \Omega_Z)$ is the $k$-dimensional normal pdf with mean (vector) zero and covariance matrix $\Omega_Z$.

They extend this with scale and location parameters to $X \sim SN_k(\xi, \Omega, \alpha)$. The rows and columns of the correlation matrix are rescaled to a full-rank $k$-by-$k$ covariance matrix $\Omega = (\omega_{ij})$ using a diagonal matrix $\omega$ whose diagonal entries are the square roots of the diagonal entries of the desired $\Omega$: writing $\omega_i = \omega_{ii}$, we use $\omega = \text{diag}(\omega_1,\dots,\omega_k)$ to obtain $\Omega = \omega \Omega_Z \omega$. There is then a translation by the location vector $\xi = (\xi_1, \dots, \xi_k)^T$. Overall, $X = \xi + \omega Z$ with pdf $f_X(x) = 2 \phi_k(x - \xi; \, \Omega) \Phi\left(\alpha^T \omega^{-1} (x - \xi)\right)$.

Later in the paper they return to the multivariate case. After tweaking the definition of $\delta$ to $\delta = \Omega_z \alpha / \sqrt{1 + \alpha^T \Omega_z \alpha}$ and letting $t$ now be a vector in $\mathbb{R}^k$ then the characteristic function of $X$ is given in Theorem 6, firstly by:

$$\Psi_X(t) = \exp(\mi t^T \xi - t^T \Omega t /2)\{1 + \mi \tau(\delta \omega t)\}$$

Note there seems to be a typographical error in the published paper; the $\omega$ in $\tau(\delta \omega t)$ is written as $w$. That form resembles Pewsey's, but they also give a form closer to Theorem 1:

$$\Psi_X(t) = 2 \exp(\mi t^T \xi - t^T \Omega t /2) \, \Phi(\mi \delta^T \omega t)$$

In equation (24) they move on to the multivariate skew-t distribution $Y = \xi + \eta^{1/2} X$ where $X \sim SN_k(0, \Omega, \alpha)$ and $\eta$ follows the same inverse-Gamma distribution as before, $\eta \sim IG(\nu/2, \nu/2)$. (Again, though unstated in this paper, take $\eta$ as independent of $X$.)

Equation (25) gives the characteristic function as:

$$\Psi_Y(t) = \exp(\mi t^T \xi) \left( \Psi_{T_k}(\Omega^{1/2} t) + \mi \tau^+(\delta, \, \omega t) \right)$$


$\Psi_{T_k} = \frac{\| \sqrt{\nu} t \|^{\nu / 2}}{\Gamma(\nu/2) 2^{\nu/2 -1}} K_{\nu/2}(\| \sqrt{\nu} t \|) \text{, for } t \in \mathbb{R}^k, \, \nu > 0$

$\tau^{+}(\delta, \, \omega t) = \int_0^\infty \exp (-\eta t^T \Omega t /2) \, \tau(\sqrt{\eta} \delta^T \omega t) \, \mathrm{d}H(\eta) \text{, for } \delta^T \omega t > 0 $

$\tau^{+}(\delta, \, -\omega t) = -\tau^{+}(\delta, \, \omega t)$

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    $\begingroup$ Quite possibly there are some typos in my own Latex but as indicated in my answer, I think I found a typo in the original paper too. The paper is easily found online but I don't want to link to respect the author's copyright. If someone identifies a typo, please feel free to edit my answer. Explanation of whether the $w$ was really meant to be an $\omega$ or not would also be welcome. If someone with greater familiarity with the topic agrees with me, I will change the main equation to the "correct" form rather than trying to match the paper. $\endgroup$
    – Silverfish
    Dec 22, 2014 at 23:26
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    $\begingroup$ Citeseer has a pdf of what looks to be a working paper (2010) of that here -- I believe it would be okay to point to that, at least, though as you say, the actual publication isn't hard to locate. If we're lucky, the working paper might throw a little more light on some of the issues. $\endgroup$
    – Glen_b
    Dec 31, 2014 at 0:07
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    $\begingroup$ I'm pretty sure it's necessary to specify $\eta$ independent of $X$ (that's not your fault; it seems to be an omission from the paper). [Edit: Independence is mentioned in the corresponding place in that citeseer paper (but it's a somewhat broader paper organized differently, so maybe I missed it somewhere else in the published version)] $\endgroup$
    – Glen_b
    Dec 31, 2014 at 0:19
  • $\begingroup$ Reading through it, I agree about the "$w$" vs "$\omega$" thing $\endgroup$
    – Glen_b
    Dec 31, 2014 at 0:45
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    $\begingroup$ @Glen_b Your statement about independence was correct and I have edited my answer accordingly. $\endgroup$
    – Silverfish
    Dec 31, 2014 at 2:35

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