# Trouble computing the VaR for Student's t-distribution for a minimum-variance portfolio composed of four cryptocurrencies (BTC, ETH, LTC, and XMR)

I have modelled the time-series of daily log-returns from August 2015 to October 2017 of a minimum-variance portfolio composed of four cryptocurrencies (BTC, ETH, LTC, XMR) by fitting the data to four different distributions: the Cauchy distribution, the Normal distribution, the exponential distribution, and Student's t distribution. I've then subsequently yielded the value-at-risk (VaR) and expected shortfall (ES) for all of the aforementioned distributions except for the Cauchy one, as this one has an undefined mean and variance. However, I've particularly run into a problem when computing the VaR for Student's t distribution. This is because when fitting my data I obtain a degrees of freedom parameter of $$\nu=1.73$$ and since the VaR formula for Student's t distribution is defined as $$\sqrt{\frac{\nu-2}{\nu}}t_{\nu}^{-1}(1-\alpha)\sigma-\mu$$ where $t_{\nu}^{-1}(1-\alpha)$ is the quantile function of the Student's t distribution, $\alpha$ is a probability level such that $0<\alpha<1$, $\sigma$ is the standard deviation parameter, and $\mu$ is the mean (or location) parameter, this then obviously implies that $$\nu-2<0$$ which means that the square root "does not exist" in this context.

Interestingly enough, I can nevertheless yield the expected shortfall for Student's t distribution, since its closed-form expression doesn't involve taking the square root of any number. This is all illustrated in the figures below

Therefore, does anybody have any idea as to why I obtain a degrees of freedom parameter $\nu<2$, and in which specific circumstances this tends to occur? Typically the opposite result should be obtained, and as it can be seen above, the distribution is somewhat heavy-tailed. In the past I have modelled and have performed stress testing (i.e. computing the parametric VaR and ES) on a minimum-variance portfolio of stocks, and I have never encountered this issue of yielding $\nu<2$. Any input would be highly appreciated. Thank you very much in advance.

• I believe the form of the Student t distribution you are using is only valid for $\nu>2$. For $1 \le \nu \le 2$ the Student t distribution exists (although its variance $\frac{\nu-2}{\nu}$does not) but it does not follow this equation, it follows another. Unfortunately I don't know it, and I don't have my copy of Johnson, N. L., Kotz, S., Balakrishnan, N. (1995) Continuous Univariate Distributions with me... – Alex C Oct 7 '17 at 1:02
• Do you think so? I've searched countless hours for an alternative expression without any success, and everybody seems to ignore (or neglect) the case for which $1<\nu<2$. – Jayjay95 Oct 7 '17 at 1:26