Can portfolio Value-at-Risk be calculated analytically for multivariate t-distributed returns?

Question

It is widely known that VaR is generally not sub-additive in all but the most restrictive cases (typically when a Gaussian return distribution is assumed, which fails when it matters the most).

Values-at-Risk based on simulations violate subadditivity because the simulated VaR on a portfolio can be higher than the sum of the asset VaRs, which contradicts diversification.

Is it possible to estimate the sample covariance matrix $\hat\Sigma$ and vector of expected returns $\hat\mu$ for a portfolio of assets – as one would for a multivariate Gaussian – but then additionally to estimate (or simply assume) a value $\hat\nu$ for the degrees of freedom of a multivariate $t$-distribution and to infer from this that the portfolio return will follow a $t$-distribution with $\theta = [\hat{\mu}_p, \hat{\sigma}_p, \hat{\nu}_p]$, where $\hat{\nu}_p$ is the "aggregated" degrees of freedom parameter?

It is clear that the estimated portfolio mean return is $\hat{\mu}_p = \omega' \hat{\mu}$ and the portfolio volatility is $\hat{\sigma}_p = \omega'\hat{\Sigma}\omega$, where $\omega$ is the $N\times1$ column vector of portfolio weights, so the Gaussian distribution "aggregates" easily from multivariate to univariate. What is not clear to me is whether the degrees of freedom parameter carries over from the multivariate distribution to the univariate distribution as described above, such that $\nu_p = \nu$.

Answer 1

Let the $n-$dimensional vector of returns $\mathbf{r}$ have a multivariate t distribution with $\nu$ degrees of freedom. The marginal distribution of any component $r_i$ has a univariate t distribution also with $\nu$ degrees of freedom.

To see this, assuming mean returns have been subtracted, the multivariate t distribution decomposes as the distribution of $\mathbf{r} = s^{-1} \mathbf{z}$ where $\mathbf{z}$ has a multivariate normal distribution with some covariance matrix $\Sigma$ and the independent random variable $s$ where $\nu s^2$ has a chi-squred distribution with $\nu$ degrees of freedom.

Writing $\{\mathbf{r} \leqslant \mathbf{x}\}$ as the event $\bigcap_{i=1}^n\{r_i \leqslant x_i\}$ we see that $$P(\mathbf{r} \leqslant \mathbf{x}) = P(s^{-1}\mathbf{z} \leqslant \mathbf{x}) = P( \mathbf{z} \leqslant s\mathbf{x}),$$

and the joint distribution function (for independent s and $\mathbf{z}$) is

$$F(\mathbf{x}) = \int_0^\infty\int_{-\infty}^{s\mathbf{x}} \frac{2 (\nu/2)^{\nu/2}}{\Gamma(\nu/2)}s^{\nu-1} e^{-\nu s^2/2}(2\pi)^{-n/2} |\Sigma|^{-1/2} e^{{-\frac{1}{2}\mathbf{\xi}' \Sigma^{-1}}\mathbf{\xi}}\, d \mathbf{\xi} \, ds.$$

Since marginal distributions of a multivariate normal distribution are normal, we can let the upper integration limits $x_j \to \infty$ for all $j \neq i$ and obtain the univariate t distribution with $\nu$ degrees of freedom as the marginal distribution of $r_i$:

$$F_i(x_i) = \int_0^\infty\int_{-\infty}^{sx_i} \frac{2 (\nu/2)^{\nu/2}}{\Gamma(\nu/2)}s^{\nu-1} e^{-\nu s^2/2}(2\pi)^{-1/2} \sigma_{i}^{-2} e^{{-\frac{\xi_i^2}{2\sigma_i^2}}}\, d \xi_i \, ds.$$

The question now becomes how is a linear combination $\mathbf{\omega}'\mathbf{r}$ distributed given that the components of $\mathbf{r}$ have t distributions with $\nu$ degrees of freedom. Unlike normal variates the combination will generally not preserve the t distribution.

How the combination is distributed is discussed here.