Exercise: Risk Management

Exercise

A random variable \( L \) is called normally distributed with mean \( \mu \) in \(\mathbb{R} \) and variance \( \sigma > 0 \) if it has a probability density given by

\[ f_{\mu,\sigma}(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right) \]

and we use the notation \( L \sim \mathcal{N}(\mu, \sigma^2) \). We denote by

\[ F_{\mu,\sigma}(m) = P[L \leq m] = \int_{-\infty}^{m} f_{\mu,\sigma}(y) \, dy, \quad x \in \mathbb{R} \quad \text{and} \quad q_{\mu,\sigma}(s) = F^{-1}_{\mu,\sigma}(s), \quad s \in (0,1) \]

the CDF and quantile of the normal distribution with mean \( \mu \) and variance \( \sigma \). We use the simplified notations for the standard normal:

\[ f = f_{0,1}, \quad F = F_{0,1}, \quad q = q_{0,1}. \]

1. Show that if \( L \sim \mathcal{N}(0,1) \), it holds \( \mu + \sigma L \sim \mathcal{N}(\mu, \sigma^2) \).
2. Show that if \( L \sim \mathcal{N}(\mu, \sigma^2) \), it holds \( -L \sim \mathcal{N}(-\mu, \sigma^2) \).

3. Show that

   \[ f_{\mu,\sigma}(x) = \frac{1}{\sigma} f\left( \frac{x-\mu}{\sigma} \right), \quad \sigma^2 f^\prime_{\mu,\sigma} (x) = (\mu-x)f_{\mu,\sigma}(x) \]
   \[ F_{\mu,\sigma}(x) = F\left( \frac{x-\mu}{\sigma} \right), \quad q_{\mu,\sigma}(s) = \mu + \sigma q(s). \]

Recall that

\[ \begin{align*} V@R_{\alpha}(L) & = \inf \{m \colon P[L>m]\geq \alpha\} = q_L(1-\alpha)\\ ES_{\alpha}(L) & = \frac{1}{\alpha}\int_0^\alpha V@R_{\alpha}(s) ds = \frac{1}{\alpha}\int_{1-\alpha}^1 q_L(s)ds \end{align*} \]

and that both functionals are positive homogoneous.

Show that for \(L\sim \mathcal{N}(\mu, \sigma^2)\)

\[ V@R_{\alpha}(L) = \mu + \sigma V@R_{\alpha}(\bar{L}) \quad \text{and} \quad ES_{\alpha}(L) = \mu + \sigma ES_{\alpha}(\bar{L}) \]

where \(\bar{L} \sim \mathcal{N}(0, 1)\) (in other terms, to compute the value at risk or expected shortfal of normal distribution, you just need the V@R and ES of the standard normal distribution).

Deduce that for normally distributed random variable \(L\) with zero mean, \(V@R\) and \(ES\) are related through

\[ ES_{\alpha}(L) = C V@R_{\alpha}(L) \]

for some constant \(C\) which you provide explicitely.

Dual Representation

We already know that the expected shortfall has two possible representations:

\[ \begin{align*} ES_\alpha(L) & = \frac{1}{\alpha} \int_{1-\alpha}^{1} q_L(s) ds\\ & = \inf\left\{ m + \frac{1}{\alpha} E\left[ (L-m)^+ \right] \right\}\\ & = q_L(1-\alpha) + \frac{1}{\alpha} E\left[ \left( L - q_L(1-\alpha) \right)^+ \right] \end{align*} \]

We derive an alternative formulation in terms of duality, namely:

\[ ES_{\alpha}(L) = \sup\left\{ E^Q[L] \colon 0 \leq \frac{dQ}{dP} \leq \frac{1}{\alpha} \right\} \]

This general statement says that the expected shortfall accounts for computing the expected loss under any alternative probability model \( Q \) such that \( Q \) is not "too" far away from \( P \) in the sense that the density is bounded by \( 1/\alpha \).

1. Show that for every \( x \) and every \( y \) with \( 0 \leq y \leq 1/\alpha \) it holds:

   \[ \frac{1}{\alpha} x^+ \geq xy \]

   In other terms, \( x^+ / \alpha = \sup\{ xy \colon 0 \leq y \leq 1/\alpha \} \), which is called Fenchel-Moreau duality.

2. Using the fact that \( E[dQ/dP] = 1 \), show that if \( 0 \leq dQ/dP \leq 1/\alpha \), then it holds:

   \[ m + \frac{1}{\alpha} E\left[ (L-m)^+ \right] \geq E^Q[L] \]

   and deduce that:

   \[ ES_{\alpha}(L) \geq \sup\left\{ E^Q[L] \colon 0 \leq \frac{dQ}{dP} \leq \frac{1}{\alpha} \right\} \]

3. Assuming that \( F_L \) is continuous and strictly increasing, show that if we define the random variable:

   \[ \frac{dQ^\ast}{dP} = \frac{1}{\alpha} 1_{\{L \geq q_L(1-\alpha)\}} \]

   then it defines a probability measure \( Q^\ast \) such that \( 0 \leq dQ^\ast/dP \leq 1/\alpha \) and for which it holds:

   \[ ES_{\alpha}(L) = E^{Q^\ast}[L] \]

   and deduce the duality formula — for which you now have an explicit \( Q^\ast \) that depends on \( L \).