# The early days of Quantitative Finance

A lot of people start studying Quantitative Finance because they see it as a great way to earn money. It seems indeed reasonable that somebody with great mathematical skills would be looking for opportunities allowing him to make a profit — sometimes at the expense of others. I can still remember like yesterday seeing a friend — clearly one of the smartest persons I’ve met — losing money at a roulette table being certain that he had a winning strategy (!) as well as countless classmates claiming they could easily make a profit out of the stock market.

Nowadays, we have books about it, computers to help us and  well-known strategies which make it always more difficult to make long-term risk-adjusted profits because all available opportunities are quickly taken away by sophisticated investors.

But how did it work in the old days? Who were the first people to study the field? When did it start? How did they put it in place? Did they make a fortune? Are they still making a fortune? All these questions are fascinating to me, and I just finished reading up a book called Fortune’s Formula by William Poundstone which addresses these topics.

Obviously, this book does not contain any fortune formulas, but it depicts the stories of the men who contributed to the field in the early days. This book was fascinating because it talks about some very important “discoveries” such as the Kelly criterion, based on Information Theory, Modern Portfolio Theory, Market Efficiency and how these techniques were put to practice at the very beginning. It also talks about how the first formula to price warrants was discovered and the struggles through which Edward Thorp went before he could really make money out of it — and how it all ended. We also learn how some brilliant minds — Noble Prize winners — got caught and lost everything in the LTCM collapse.

The good thing about this book, is that it explains all this in relatively simple words. Indeed, Poundstone focuses on the concepts rather than on the technicalities and puts everything in historical context such that any interested reader can go through the pages without being lost in details.

If you’re planning to go on holidays at the end of the year and you’re looking for a good book to read, I strongly recommend this one.

# Implied odds behind lay betting

In a precedent post,  I introduced the concept of back and lay bets on a given event, for example the fact that it will rain on a given day.  We also said that we are given some odds $o$, which express how much we can earn if we guess the right outcome between the event happening ($E$) or not ($\bar{E}$).

Usually, people betting are very familiar with the “common” back bet and the odds $o$, which generate a net cash flow of $x (o-1)$ when we guess right and $-x$ when we guess wrong on an invested amount of $x$. However, I always found lay bets less intuitive and I prefer to see them as betting on the event not to happen.  This post about seeing a lay bet on $E$ as a back bet on $\bar{E}$, for example backing that it will not rain on a given day. The difference is explained in the table below:

$E$ $\bar{E}$
Laying $E$ $-x(o-1)$ $x$
Backing $\bar{E}$ $-y$ $y(\bar{o}-1)$

We see a new term $\bar{0}$ appearing, which are the odds of backing the event not to happen. Because both lines actually imply betting on the same thing (the event not to happen), the cash flows in case of $E$ or $\bar{E}$ must be the same. Therefore, we get the following system of equations:

\begin{align} -x(o-1) &= -y\\ x &= y (\bar{o}-1) \end{align}

The fist equation tell us that $y=x(o-1)$ and we can substitute in the second equation to solve for $\bar{o}$:
\begin{align} x &= y (\bar{o}-1)\\ x &= x(o-1) (\bar{o}-1)\\ \frac{x}{x(o-1)} &= \bar{o}-1\\ \bar{o} &= 1+\frac{1}{o-1}\\ \bar{o} &= \frac{o}{o-1}\\ \end{align}

Now, we can take bets on the event not to happen much more intuitively. Let’s go back to our example and let’s say that the odds of the bet on having rain today is $o = 5$. If we want to bet on the fact that it will not rain today, we need to place a lay bet, but what are our odds $\bar{o}$? Using the formula we derived above, we can easily compute them:

$$\bar{o} = \frac{o}{o-1} = \frac{5}{5-1} = 1.25$$

Let’s say that we are happy with those odds, and we’d like to place a bet of $y=2$ at $\bar{o}=1.25$. Well, if we look at your system of equation, we see that $x = y (\bar{o}-1)$, so we need to bet:

$$x = y (\bar{o}-1) = 2 (1.25-1)=0.5$$

Let’s check:

$E$ $\bar{E}$
Laying $E$ $-x(o-1) = -0.5 (5-1) = -2$ $x = 0.5$

This is correct! An easy way of computing $x$ is to keep in mind that $x = y (\bar{o}-1)$ is simply the amount you expect to win by betting on the event not to happen.

# Expected Shortfall closed-form for Normal distribution

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.