Expected Tail Exposure
When 97.5% fails
Counterparty credit risk on derivative transactions is measured against a 97.5 percentile on a distribution of losses calibrated off historical data. At this percentile, and assuming future market movements would maintain its historical bearings, means that a derivative's value would only surpass our indications 2.5% of the time, or ~6 days in a trading year. All seemed well until...
- Covid 19 - excessive losses
- Archegos default - excessive losses
- EU power supply strain - excessive losses
The increasing occurence of excessive losses over the 97.5 percentile highlights a missing statistic in the measurement of credit risk. That is, while one may understand they won't lose more than 10M (for example), 97.5% of the time, there is no indication of how much will be lost in the remaining 2.5%. Hence the need of an alternative statistic to measure tailed losses. The tailed loss measure analysed here will be the *Expected Tail Exposure* (ETE).
Expected Tail Exposure
Let $X_{t} \sim N(\mu_{t}, \sigma_{t}^{2})$ be a stochastic process over random variables representing a portfolio's value at some time $t$. The ETE is the expected value of $X_{t}$, conditioned on being past a value of $x(\alpha)_{t}$ with $\alpha$% confidence is given by \begin{align*} \text{ETE}_\alpha(X_t) = \mathbb{E}(X_t|X_t > x(\alpha)_t) \end{align*} $X_{t}$ is assumed to be Normal (Gaussian) and may be represented with respect to a standard normal random variable $Z$. That is \begin{align*} X_{t} = \mu_{t} + \sigma_{t}Z_{t}. \end{align*} From the last two equations \begin{align*} \text{ETE}_\alpha (X_t) &= \mathbb{E}(\mu_t + \sigma_t Z_t \vert \mu_t + \sigma_t Z_t > x(\alpha)_t)\\\\ &= \mathbb{E}\left(\mu_t + \sigma_t Z_t \big \vert Z_t > \frac{x(\alpha)_t - \mu_t}{\sigma_t}\right). \end{align*} For clarity, let \begin{align} q = \frac{x(\alpha)_t - \mu_t}{\sigma_t}. \end{align} Expressing conditional expectations with respect to probabilities \begin{align*} \text{ETE}_\alpha (X_t) &= \frac{\int_q^\infty (\mu_t + \sigma_t z)f(z)dz}{P(Z>q)} \\\\ &= \frac{\int_q^\infty (\mu_t + \sigma_t z)f(z)dz}{1-\alpha} \end{align*} where $f$ is the standard normal probability density function. The numerator of this last equation is evalauted below. \begin{align*} \int_q^\infty(\mu_t +\sigma_t z)f(z)dz &= \mu_t\int_q^\infty f(z)dz+\sigma_t\int_q^\infty zf(z)dz \\\\ &= \mu_t\left(1-\Phi(q)\right) + \frac{\sigma_t}{\sqrt{2\pi}}\int_q^\infty z\exp\left(\frac{-z^2}{2}\right)dz \\\\ &= \mu_t\left(1-\Phi(q)\right) + \frac{\sigma_t}{2\sqrt{2\pi}}\int_q^\infty\exp\left(\frac{-z^2}{2}\right)dz^2\\\\ &= \mu_t\left(1-\Phi(q)\right) - \frac{\sigma_t}{\sqrt{2\pi}}\left[\exp\left(\frac{-z^2}{2}\right)\right]_q^\infty\\\\ &= \mu_t\left(1-\Phi(q)\right) - \frac{\sigma_t}{\sqrt{2\pi}}\left[0-\exp\left(\frac{-q^2}{2}\right)\right]\\\\ &= \mu_t\left(1-\Phi(q)\right) + \sigma_t f(q). \end{align*} where $\Phi$ is the standard normal cumulative distribution function. With the integral evaluated, the conclusion is \begin{align*} \text{ETE}_\alpha(X_t) = \frac{\mu_t\left(1-\Phi(q)\right) + \sigma_tf(q)}{1-\alpha}. \end{align*} This expression can serve as a complementary statistic to the 97.5 percentile to gauge how much may be lost in the event prices move beyond the 97.5 percentile.