Today, I would like to share a little exercise I did to compute the Expected Shortfall of a normal variable. For those of you who are not familiar with this risk measure, it evaluates the *average* of the $(1-\alpha)$-worst outcomes of a probability distribution (example and formal definition follow).

If you have enough data, the expected shortfall can be *empirically* estimated. Let’s say we want to compute the expected shortfall at 95% (denoted $ES_\text{95%}$) and that we have 1000 points. The first thing to do is to sort the 1000 points. Then, we take the $1-95\%=5\%$ worst outcomes, which is for 1000 points $1000 \cdot 5\% = 50$ points. Finally, we simply average these points to get our estimated expected shortfall. Nevertheless, this is * only an estimate* and it will be precise only if you have enough points available.

Sometimes, you can assume the distribution of the points and when these points are a set of returns, it is quite common to assume that they follow some normal distribution $\mathcal{N}(\mu,\sigma^2)$. In this case, you can find a closed form formula which gives you the *exact* solution for a given $\alpha$, given the parameters $\mu$ and $\sigma$. This result is available in several papers but usually comes without the proof, which I will provide below.

Let’s assume we have a random variable $X \sim \mathcal{N}(\mu,\sigma^2)$. The first thing to do is to re-express this variable in term of a standard normal variable $Z \sim \mathcal{N}(0,1)$. This is easily done as follows:

$$ X = \mu + \sigma Z $$

The next thing to do is to express the Value at Risk at a certain level $\alpha$, which is defined as follows:

$$\text{VaR}_\alpha (X) = \{ Y ~ | ~ \mathbb{P}(X \leq Y) = 1-\alpha \}$$

This can be easily found as well, since:

$$

\begin{align}

\mathbb{P}(X \leq Y) &= \mathbb{P}( \mu + \sigma Z \leq Y)\\

&= \mathbb{P} \left(Z \leq \frac{Y-\mu}{\sigma} \right)\\

&= \Phi \left( \frac{Y-\mu}{\sigma} \right) = 1-\alpha\\

\end{align}

$$

where $\Phi(\cdot)$ is the cumulative standard normal distribution.

We can then deduce that:

$$\text{VaR}_\alpha (X) = Y = \Phi^{-1}(1-\alpha) \sigma + \mu$$

where $\Phi^{-1}(\cdot)$ is the inverse cumulative standard normal distribution and which can be looked up online.

Now, we can actually start working on the closed-form. Let’s first express the expected shortfall in terms of the value at risk:

$$ES_\alpha(X) = \frac{1}{1-\alpha} \int_\alpha^1 \text{VaR}_u(X) du$$

Using our previous result, we can rewrite the definition above as follows:

$$

\begin{align}

ES_\alpha(X) &= \frac{1}{1-\alpha} \int_\alpha^1 ( \Phi^{-1}(1-u) \sigma + \mu ) du\\

&= \frac{1}{1-\alpha} \left( \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \int_\alpha^1 \mu du \right)\\

&= \frac{1}{1-\alpha} \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \mu

\end{align}

$$

As you can see, we managed to split the big integral into two parts, one of which is solved very easily. However, there remains an integral with $\Phi^{-1}(\cdot)$ which is a very complex function so we need to find a way around. To do this, we are going to use a change of variable introducing $u = \Phi(y)$. We have then $y = \Phi^{-1}(u)$ and $du = \phi(y)dy$ where $\phi(y)=\frac{1}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right)$ is the standard normal density function.

$$

\begin{align}

\int_\alpha^1 \Phi^{-1}(1-u) \sigma du &= \int_{\Phi^{-1} (\alpha)}^{\Phi^{-1}(1)} \Phi^{-1}\left(\underbrace{ 1-\Phi(y)}_{\Phi(-y)} \right) \sigma \phi (y) dy \\

&= \int_{\Phi^{-1} (\alpha)}^\infty -y \sigma \phi (y) dy \\

&= \int_{\Phi^{-1} (\alpha)}^\infty -\frac{y \sigma}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right) dy \\

\end{align}

$$

So, we got rid of $\Phi^{-1}(\cdot)$ and the remaining integral is relatively easy to solve:

$$

\begin{align}

\int_{\Phi^{-1} (\alpha)}^\infty -\frac{y \sigma}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right) dy &= \frac{\sigma}{\sqrt{2\pi}} \int_{\Phi^{-1} (\alpha)}^\infty -y \exp \left( – \frac{y^2}{2} \right) dy\\

&= \frac{\sigma}{\sqrt{2\pi}} \left( \exp \left( – \frac{y^2}{2} \right)\right)|_{\Phi^{-1} (\alpha)}^\infty\\

&= \frac{\sigma}{\sqrt{2\pi}} \left( 0 – \exp \left( – \frac{\Phi^{-1} (\alpha)^2}{2} \right)\right)\\

&= -\sigma \frac{1}{\sqrt{2\pi}} \exp \left( – \frac{\Phi^{-1} (\alpha)^2}{2} \right)\\

&= -\sigma \phi \left( \Phi^{-1} (\alpha) \right)\\

\end{align}

$$

Now, we just have to include this in our previous result to get:

$$

\begin{align}

ES_\alpha(X) &= \frac{1}{1-\alpha} \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \mu\\

&= \frac{1}{1-\alpha} \left[ -\sigma \phi \left( \Phi^{-1} (\alpha) \right) \right] + \mu\\

&= \mu – \sigma \cdot \frac{\phi \left( \Phi^{-1} (\alpha) \right)}{1-\alpha}

\end{align}

$$

This is a very good result, since we know that $\lambda( \alpha ) = \frac{\phi \left( \Phi^{-1} (\alpha) \right)}{1-\alpha}$ is a function of $\alpha$ that can be computed very easily, for example in Excel. Below are the results for $\alpha$ from 80% to 99%.

That’s it, with this simple formula, you can compute the expected shortfall of a normally distributed variable without needing any data point.