Extreme Value Theory

Every risk number you will be asked for is about the tail. Nobody cares what a typical Tuesday looks like; they want to know how bad the worst day of the next decade could be, and whether the firm survives it. The standard toolkit points the wrong way. Fit a distribution to twenty years of returns and the procedure spends all of its effort on the middle, because that is where all the data is; the tail is a dozen points out of five thousand and barely registers. Then you use that curve to extrapolate to a loss you have never seen. The The Central Limit Theorem does not help either — it is a theorem about averages, and averages say nothing about maxima.

Extreme value theory fills that hole: limit theorems about the largest observation rather than the sum, whose content is that a tail has a universal shape with only a handful of knobs, whatever generated the data.

The analogy: how high to build the levee

A town on a river does not size its flood wall against the average water level. It asks a different question: over the next hundred years, how high will the river get at its worst? The engineers do not model the whole catchment. They take the highest reading from each year, notice that those annual maxima follow a regular, well-behaved pattern, fit that pattern, and read off the height exceeded once a century.

That is the entire method. Throw away the ordinary days, keep only the extremes, fit a distribution to those, and let the theory tell you which distributions the extremes are even allowed to follow.

The first theorem: what maxima converge to

Chop your data into blocks — one year of trading days, say — and take the largest loss in each block. Call the largest of $n$ observations $M_n$. The Fisher–Tippett–Gnedenko theorem says that if $M_n$, after being recentred and rescaled, converges to anything at all, it can only converge to one family, the generalised extreme value distribution:

$$G(x) = \exp\left(-\Big(1 + \xi\,\frac{x-\mu}{\sigma}\Big)^{-1/\xi}\right).$$

Three symbols. $\mu$ (mu) is the location, where the typical maximum sits. $\sigma$ (sigma) is the scale, how much the maximum varies block to block. And $\xi$ (xi) is the shape, or tail index — the one that matters. In words: whatever the original data, the block maxima end up in one of exactly three shapes, and $\xi$ picks which.

• $\xi = 0$, Gumbel: the parent tail decays exponentially — normal, exponential, lognormal. Maxima grow, but agonisingly slowly.
• $\xi > 0$, Fréchet: the parent tail is a power law. Maxima grow fast and there is no ceiling. Financial returns live here, typically with $\xi$ between 0.2 and 0.4.
• $\xi < 0$, Weibull: the parent has a hard upper bound and the maxima pile up against it.

Worked example 1: how bad can the worst day get?

If daily returns are standard normal, what loss do you expect to be breached exactly once in $n$ days? Solve $P(Z > z) = 1/n$ and read $z$ off a normal table.

• One year, $n = 250$: $P = 0.004$, so $z = 2.65$.
• Ten years, $n = 2{,}500$: $P = 0.0004$, so $z = 3.35$.
• Forty years, $n = 10{,}000$: $P = 0.0001$, so $z = 3.72$.
• Four thousand years, $n = 1{,}000{,}000$: $P = 10^{-6}$, so $z = 4.75$.

Sit with those numbers. Multiplying the sample by four thousand moves the worst day from 2.65 standard deviations only as far as 4.75. Under a normal, the worst-of-$n$ grows like $\sqrt{2\ln n}$ — a logarithm inside a square root, about as slow as growth gets.

Now redo it for a power-law tail, $P(X > x) = x^{-\alpha}$ with $\alpha = 3$ (equivalently $\xi = 1/3$). Solve $n \times x^{-3} = 1$, so $x = n^{1/3}$:

sample	normal worst	power-tail worst ($\alpha = 3$)
250 days	2.65	6.3
2,500 days	3.35	13.6
10,000 days	3.72	21.5
1,000,000 days	4.75	100.0

Across that table the normal's worst case grew by a factor of 1.8; the power tail's grew by 16. This is the arithmetic behind every "we were told it was a twenty-sigma event" story: on 19 October 1987 the S&P fell about 20 percent on a day when its volatility was near 1 percent. In the normal column that never happens. In the right-hand column it is simply a bad draw.

The second theorem: peaks over threshold

Block maxima waste data — you keep one number per year and discard the rest, including the year that had five terrible days. The practical version keeps every observation above a high threshold $u$. The Pickands–Balkema–de Haan theorem says that for a high enough $u$, the excess over the threshold follows a generalised Pareto distribution:

$$P(X - u > y \mid X > u) = \left(1 + \frac{\xi y}{\beta}\right)^{-1/\xi}.$$

Here $y$ is how far past the threshold you went, $\beta$ (beta) is a scale parameter, and $\xi$ is the same shape as before. In words: given that today was already bad, how much worse than the threshold it was follows one two-parameter curve.

Worked example 2: turning exceedances into a risk limit

Twenty years of daily returns, $n = 5{,}000$ days. Set the threshold at a 2 percent loss, $u = 2$; suppose 100 days breached it, so $N_u = 100$. Fitting a generalised Pareto to those 100 excesses returns $\xi = 0.30$ and $\beta = 0.80$. The 99.9 percent loss quantile is

$$\text{VaR}_q = u + \frac{\beta}{\xi}\left[\left(\frac{n}{N_u}(1-q)\right)^{-\xi} - 1\right].$$

Work it through step by step:

1. $n/N_u = 5000/100 = 50$, and $1 - q = 0.001$, so the bracket's base is $50 \times 0.001 = 0.05$.
2. Raise it to the power $-\xi = -0.30$: $\ln 0.05 = -2.996$, times $-0.30$ is $0.899$, and $e^{0.899} = 2.456$.
3. Subtract one: $2.456 - 1 = 1.456$.
4. Multiply by $\beta/\xi = 0.80/0.30 = 2.667$: that gives $3.88$.
5. Add the threshold: $2 + 3.88 = 5.88$.

So EVT puts the one-in-a-thousand-day loss at 5.88 percent. A normal fitted to the same series, with a 1 percent daily standard deviation, puts it at 3.09 percent — nearly half, from the same data.

Now ask what the loss looks like given that the limit is breached. For a generalised Pareto that has a closed form too:

$$\text{ES}_q = \frac{\text{VaR}_q}{1-\xi} + \frac{\beta - \xi u}{1-\xi}.$$

Substituting: $5.88/0.70 = 8.41$, and $(0.80 - 0.30 \times 2)/0.70 = 0.29$. Adding gives 8.69 percent. The average breach overshoots the limit by half again. That gap between VaR and expected shortfall is the number that kills firms, and a normal model makes it look far smaller than it is.

The explorer below is a normal curve with a coverage readout. Drag the band past three standard deviations and watch how fast the remaining probability collapses — that collapse is exactly what EVT says not to trust.

What this means in practice

• Limits and margin. Clearing houses size initial margin off tail quantiles. An EVT-based limit is materially higher than a normal one, and that difference is the survival buffer.
• $\xi$ is the headline. If $\xi \ge 0.5$ the loss variance is infinite, and any risk report built on standard deviations is meaningless for that book.
• Choosing the threshold. Too low and ordinary data contaminates the fit; too high and you have twelve points and a wildly noisy $\xi$. Use the top 2 to 10 percent and check the estimate is stable across that range.
• Standardise first. The theorems assume independent, identically distributed data; returns cluster. Fit a volatility model first, then apply EVT to the residuals.