Ratio between Expected Shortfall and Value at Risk for $t$-distribution

Question

If $X$ is a random variable with $t$-distribution of parameter $\mathcal{v}$, how can I prove that $$ \lim_{\alpha \to 1^{-}} \frac{\mathrm{ES}_{\alpha}(X)}{\mathrm{VaR}_{\alpha}(X)} = \frac{\mathcal{v}}{\mathcal{v}-1}? $$ In the book Quantitative Risk Management: Concepts, Techniques and Tools by Alexander J. McNeil et al., I found the following formulas: $$\mathrm{ES}_{\alpha} = \frac {g_v(t_v^{-1}(\alpha))}{1- \alpha} \frac{v+(t_v^{-1}(\alpha))^2}{v-1} ,\qquad\mathrm{VaR}_{\alpha} = t_v^{-1}(\alpha)$$ where $t_v$ denotes the distribution function of the standard student and $g_v$ the density function of the same distribution. But I'm having dificulties evaluating such limit. Any suggestions?

Answer 1

Let $u=t^{-1}_v(\alpha)$ and recall that $g_v(u)=c_v(v+u^2)^{-\frac{v+1}2}$ for some constant $c_v$. By the formulas you provided, $$\begin{eqnarray*}\lim_{\alpha\to 1^-}\frac{\mathrm{ES}_\alpha(X)}{\mathrm{VaR}_\alpha(X)}&=&\lim_{\alpha\to 1^-} \frac{g_v(t^{-1}_v(\alpha))}{(1-\alpha)(v-1)\left(\frac{t^{-1}(\alpha)}{v+(t^{-1}(\alpha))^2}\right)}\\ =\lim_{u\to\infty} \frac{g_v(u)}{(1-t_v(u))(v-1)\left(\frac{u}{v+u^2}\right)}&=&\lim_{u\to\infty}\frac{(v+u^2)^{-(v-1)/2}u^{-1}c_v/(v-1)}{(1-t_v(u))}.\end{eqnarray*}$$ This is of the form $[0/0]$. L'Hopital's rule gives $$\frac{\frac{d}{du}\left[(v+u^2)^{-(v-1)/2}u^{-1}c_v/(v-1)\right]}{-g_v(u)}=\frac{c_v}{v-1}\frac{\frac{d}{du}\left[(v+u^2)^{-(v-1)/2}u^{-1}\right]}{-g_v(u)}$$ $$=\frac{c_v}{v-1}\frac{\left[-\frac{v-1}2(v+u^2)^{-(v+1)/2}2uu^{-1}+(v+u^2)^{-(v-1)/2}(-u^{-2})\right]}{-c_v(v+u^2)^{-(v+1)/2}}$$ $$=\frac{1}{v-1}\left[\frac{v-1}1+(v+u^2)(u^{-2})\right]=1+\frac{v+u^2}{(v-1)u^2}\to1+0+\frac1{v-1}=\frac{v}{v-1},$$ as desired.