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

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 $\nu_p = \nu$.

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.

• Fantastic answer. The paper referenced is slightly challenging, though. Do you happen to have any insights into the conditions (e.g., parameters) under which the distribution may be preserved? – Constantin Jan 25 '18 at 10:57
• Thank you. Let me see if I can find a better reference. As far as I know, the Student's t distribution is not in the family of stable distributions that includes the normal, Cauchy and Levy distributions. Stable means the linear combination of two independent random variables with this distribution has the same distribution. Now it may be possible that combining t distributed random variables all with the same number of degrees of freedom has a similar but not identical distribution. I just don't know for sure at this point. – RRL Jan 25 '18 at 19:34
• The reason I am asking for this specifically is that I have computed the portfolio-VaR for multivariate t-distributed returns both using simulations and analytically, in both cases using the sample covariance matrix and mean returns vector as scale and location parameters and assuming a constant (and somewhat arbitrary) degrees of freedom parameter of $v=3$. Both methods appear to yield the same values (the MC VaR converges to the analytical VaR for an increasing number of runs) so there appear to be some conditions under which the t-distribution is preserved under linear combinations. – Constantin Jan 25 '18 at 20:20
• When you computed VaR from the simulations, did you get this as a multiple of an estimated standard deviation or as an estimated percentile (by actually counting the number of losses below a threshold in the tail)? Also what do you mean by analytically computed VaR using a multivariate t distribution. – RRL Jan 25 '18 at 22:15
• I did not compute the VaR as a multiple of $\sigma_p$. Simulation: I computed $N$ returns per simulation for each asset with $K$ simulations in total, drawing from a multivariate $t$-distribution with $\mathbf{r}$ and $\Sigma$ estimated from real data and $\nu=3$. I then computed portfolio returns for each run as $\omega' \mathbf{r}(k), k \in [1, \dots, K]$ and computed the VaR as the $(1-\alpha)$ percentile of those returns. Analytical: I computed the $(1-\alpha)$ percentile of the univariate $t$-distribution using $\nu = 3$ and $\mu = \omega'\mathbf{r}$ and $\sigma_p = \omega'\Sigma \omega$. – Constantin Jan 25 '18 at 22:49