Derivatives of VaR and ES

This section derives expressions for the derivatives of Value-at-Risk and Expected Shortfall. Equation (22) is directly implied by Lemma A.2.

We consider a bivariate random variable $(X,Y)$ with continuous density $f(x,y)$ such that $X + \epsilon Y$ has for any $\epsilon\in\mathbb{R}$ a continuous density, too. Define VaR$_{\alpha}(\epsilon)$ for $0 < \alpha <1$ as

$$_{\alpha}(\epsilon) := _{\alpha}(X + \epsilon Y) .$$ (60)

Since $X + \epsilon Y$ is a continuous random variable, the infimum in (9) is actually reached (i.e. is a minimum) and the respective probability is exactly $\alpha$.

LEMMA A.1   Under certain strong assumptions on the density $f$, the Value-at-Risk VaR$_{\alpha}(\epsilon)$ can be differentiable in $\epsilon$ and

$$\frac{\partial\text{VaR}_{\alpha}(\epsilon)}{\partial\epsilon} = - \mathbf{E}[Y \vert X + \epsilon Y = -\text{VaR}_{\alpha}(\epsilon)] .$$ (61)

The following proof is analogous to Gouriéroux, Laurent and Scaillet (2000). The mentioned authors have derived the expression for the derivative if existing, but have not proven the existence (of the derivative).

Proof. [Proof (Gouriéroux, Laurent and Scaillet, 2000).] If $\partial$VaR$_{\alpha}(\epsilon)/\partial\epsilon$ exists, we have

$$\int\int_{-\infty}^{-\text{VaR}_{\alpha}(\epsilon) - \epsilon y} f(x,y) dx dy = \alpha ,$$ (62)

and hence by differentiation with respect to $\epsilon$

$$\int [- \partial _{\alpha}(\epsilon)/\partial\epsilon - y]f(- _{\alpha}(\epsilon) - \epsilon y,y) dy = 0 .$$ (63)

This implies

$$\frac{\partial\text{VaR}_{\alpha}(\epsilon)}{\partial\epsilon} = ... ...) - \epsilon y,y) dy}{\int f(-\text{VaR}_{\alpha}(\epsilon) - \epsilon y,y) dy}$$ (64)

and therefore (61). $\qedsymbol$

As already mentioned, the main problem in this reasoning is the missing proof of the differentiability of VaR. Also the strict positivity of the integral $\int f(-$VaR$_{\alpha}(\epsilon) - \epsilon y,y) dy$ should be an important ingredient in a proper proof of the lemma. In the paper of Tasche (2000), there is given a sufficient condition, named (S), for VaR-differentiation. However, condition (S) is in the most cases not easy to prove (the normal distribution excluded) and differentiation may be possible even if (S) is not fulfilled.

LEMMA A.2   Under certain strong assumptions on the density $f$, the Expected Shortfall ES$_{\alpha}(\epsilon) :=$   ES$_{\alpha}(X + \epsilon Y)$ can be differentiable in $\epsilon$ and

$$\frac{\partial\text{ES}_{\alpha}(\epsilon)}{\partial\epsilon} = - \mathbf{E}[Y \vert X + \epsilon Y \leq -\text{VaR}_{\alpha}(\epsilon)] .$$ (65)

Proof. We have

$$_{\alpha}(\epsilon) = - \frac{1}{\alpha} \int\int_{-\infty}^{-\text{VaR}_{\alpha}(\epsilon) - \epsilon y} (x+\epsilon y)f(x,y) dx dy .$$ (66)

Differentiation with respect to $\epsilon$ leads to

$$\frac{\partial\text{ES}_{\alpha}(\epsilon)}{\partial\epsilon} = - \frac{1}{\alpha} \int\int_{-\infty}^{-\text{VaR}_{\alpha} (\epsilon) - \epsilon y} y f(x,y) dy$$ (67)

$$+ \frac{1}{\alpha} \int _{\alpha}(\epsilon)[-\partial _{\alpha}(\epsilon)/\partial\epsilon - y]f(- _{\alpha}(\epsilon) - \epsilon y, y) dy .$$

Due to (64), the second summand is 0. $\qedsymbol$

In a more general context, Tasche (2000) also derives (65). Again, the most important parts of an existence proof would be the existence of the respective integrals and the proof of the correct application of all used differentiation rules.