VaR and Expected Shortfall Under Student-t Returns

Worked Solution

How to Think About It: VaR and ES are the bread and butter of risk management. VaR tells you the threshold loss at a given confidence level; ES tells you the average loss in the tail beyond that threshold. Under Gaussian returns, the gap between ES and VaR is modest. The whole point of using Student-$t$ is to capture heavy tails -- and that is exactly where the ES-VaR gap blows up. Before writing any formulas, you should expect: (1) VaR is just a linear transformation of the $t$-quantile, (2) ES involves integrating the $t$-density in the tail, and (3) as $\nu$ shrinks (fatter tails), ES pulls further away from VaR because the conditional expectation in the tail grows.

Quick Sanity Checks: - As $\nu \to \infty$, the Student-$t$ becomes Gaussian, so both VaR and ES should reduce to their Gaussian forms. - For small $\nu$ (say $\nu = 3$), the tail is much fatter than Normal, so ES should be substantially larger than VaR. - ES $\geq$ VaR always (by definition, the conditional mean beyond the threshold exceeds the threshold). - The ES-VaR gap should be an increasing function as $\nu$ decreases.

Derivation:

Part 1: VaR

We define $\text{VaR}_\alpha$ as the loss (negative return) that is exceeded with probability $\alpha$. Writing $X = \mu + \sigma T$ with $T \sim t_\nu$:

$P(X \leq -\text{VaR}_\alpha) = \alpha$

$P\left(T \leq \frac{-\text{VaR}_\alpha - \mu}{\sigma}\right) = \alpha$

So $\frac{-\text{VaR}_\alpha - \mu}{\sigma} = t_\nu(\alpha)$, giving:

$\boxed{\text{VaR}_\alpha = -\mu - \sigma \, t_\nu(\alpha)}$

Since $\alpha$ is small (e.g., 0.01 or 0.05), $t_\nu(\alpha) < 0$, so $\text{VaR}_\alpha > 0$ when $\mu$ is near zero -- as expected.

Part 2: Expected Shortfall

ES is the conditional expected loss given that the loss exceeds VaR:

$\text{ES}_\alpha = -E[X \mid X \leq -\text{VaR}_\alpha]$

Substituting $X = \mu + \sigma T$:

$\text{ES}_\alpha = -\mu - \sigma \, E[T \mid T \leq t_\nu(\alpha)]$

We need $E[T \mid T \leq q]$ where $q = t_\nu(\alpha)$. By definition:

$E[T \mid T \leq q] = \frac{1}{\alpha} \int_{-\infty}^{q} t \, f_\nu(t) \, dt$

The key integral identity for the standard $t_\nu$ distribution is:

$\int_{-\infty}^{q} t \, f_\nu(t) \, dt = -\frac{\nu + q^2}{\nu - 1} \cdot f_\nu(q)$

This follows from writing $f_\nu(t) = \frac{1}{\sqrt{\nu}B(\nu/2, 1/2)}\left(1 + \frac{t^2}{\nu}\right)^{-(\nu+1)/2}$ and integrating by substitution $u = 1 + t^2/\nu$. The derivative of the density kernel gives us back a density-like term, producing the closed form above.

Therefore:

$E[T \mid T \leq q] = -\frac{1}{\alpha} \cdot \frac{\nu + q^2}{\nu - 1} \cdot f_\nu(q)$

Plugging back in (with $q = t_\nu(\alpha)$):

$\boxed{\text{ES}_\alpha = -\mu + \frac{\sigma}{\alpha} \cdot \frac{\nu + [t_\nu(\alpha)]^2}{\nu - 1} \cdot f_\nu(t_\nu(\alpha))}$

Part 3: ES-VaR Separation

The gap is:

$\text{ES}_\alpha - \text{VaR}_\alpha = \frac{\sigma}{\alpha} \cdot \frac{\nu + [t_\nu(\alpha)]^2}{\nu - 1} \cdot f_\nu(t_\nu(\alpha)) + \sigma \, t_\nu(\alpha)$

Defining $q = t_\nu(\alpha)$:

$\text{ES}_\alpha - \text{VaR}_\alpha = \sigma \left[\frac{1}{\alpha} \cdot \frac{\nu + q^2}{\nu - 1} \cdot f_\nu(q) + q\right]$

The dependence on $\nu$ enters through three channels:

• The quantile $q = t_\nu(\alpha)$: As $\nu$ decreases, the quantile moves further into the left tail (more negative), making VaR larger.
• The density $f_\nu(q)$: The $t$-density has heavier tails for smaller $\nu$, so $f_\nu(q)$ is larger at extreme quantiles.
• The factor $\frac{\nu + q^2}{\nu - 1}$: This is the tail-amplification factor. The $\frac{1}{\nu - 1}$ term diverges as $\nu \to 1^+$, not at $\nu = 2$ -- and $\nu = 1$ is precisely the boundary where the mean (hence ES itself) ceases to exist. As $\nu \to 2^+$ (the variance boundary, but mean and ES are still finite), the gap does NOT blow up: it converges to a finite limit. In standardized form ($\mu = 0$, $\sigma = 1$) that limit is $\frac{1}{\sqrt{2\alpha(1-\alpha)}}$ -- e.g. $\approx 7.107$ at $\alpha = 0.01$ and $\approx 3.244$ at $\alpha = 0.05$. So heavier tails widen the gap monotonically, but the widening is bounded all the way down to $\nu = 2$, and only becomes unbounded as $\nu \to 1^+$.

Limiting behavior:

• $\nu \to \infty$ (Gaussian limit): $t_\nu(\alpha) \to z_\alpha$ (standard Normal quantile), $f_\nu(q) \to \phi(z_\alpha)$, and $\frac{\nu + q^2}{\nu - 1} \to 1$. So ES reduces to the Gaussian formula: $\text{ES}_\alpha = -\mu + \frac{\sigma}{\alpha}\phi(z_\alpha)$.
• Small $\nu$ (heavy tails): The ratio $\text{ES}_\alpha / \text{VaR}_\alpha$ grows significantly. For example, at $\alpha = 0.01$, the Gaussian ES/VaR ratio is about 1.11, but with $\nu = 4$ it is roughly 1.35, and with $\nu = 3$ it is about 1.55. The tail conditional expectation is much worse than the threshold.

Practical Interpretation: This is why regulators moved from VaR to ES (Basel III to Basel III.1/FRTB). VaR tells you the door to the tail; ES tells you what is behind it. Under fat-tailed models, VaR severely understates the true risk in the tail. The $\frac{\nu + q^2}{\nu - 1}$ factor is the precise mechanism: it measures how much worse the average tail loss is compared to the threshold loss.

Answer: The one-day VaR is $\text{VaR}_\alpha = -\mu - \sigma\, t_\nu(\alpha)$. The one-day ES is $\text{ES}_\alpha = -\mu + \frac{\sigma}{\alpha} \cdot \frac{\nu + [t_\nu(\alpha)]^2}{\nu - 1} \cdot f_\nu(t_\nu(\alpha))$. The ES-VaR gap grows as $\nu$ decreases (heavier tails), driven by the amplification factor $\frac{\nu + q^2}{\nu - 1}$, and collapses to the Gaussian gap as $\nu \to \infty$. The gap stays finite all the way down to the variance boundary $\nu \to 2^+$ (standardized limit $\frac{1}{\sqrt{2\alpha(1-\alpha)}}$, e.g. $\approx 7.107$ at $\alpha = 0.01$) and diverges only as $\nu \to 1^+$, where the mean and ES cease to exist.