Power Laws & Pareto Distributions
Power Laws and the Pareto Distribution
Extremistan is the home of power laws: distributions whose tail probability scales as $P(X > x) \sim x^{-\alpha}$. The exponent $\alpha$ (the tail index) controls how catastrophic the tail is. Small $\alpha$ means a fat tail; very small $\alpha$ means the theoretical mean or variance does not even exist.
The Pareto distribution is the cleanest power law: $P(X > x) = (x_m/x)^\alpha$ for $x \ge x_m$, with density $f(x) = \alpha x_m^\alpha / x^{\alpha+1}$. Its signature log-log plot is a straight line of slope $-\alpha$, which makes visual diagnosis easy.
The famous 80/20 rule is the Pareto with $\alpha = \log_4 5 \approx 1.16$. For such tails the mean is finite but the variance is infinite, so standard errors are meaningless and the sample mean converges excruciatingly slowly. Taleb's advice: always plot the log-log survival function and estimate $\alpha$ before trusting any moment-based statistic.
$$P(X > x) = \left(\frac{x_m}{x}\right)^\alpha,\quad x \ge x_m,\ \alpha > 0$$
Mean: $\mathbb{E}[X] = \dfrac{\alpha x_m}{\alpha - 1}$ (requires $\alpha > 1$).
Variance: $\mathrm{Var}(X) = \dfrac{\alpha x_m^2}{(\alpha-1)^2(\alpha-2)}$ (requires $\alpha > 2$).
The $k$-th moment $\mathbb{E}[X^k]$ is finite iff $k < \alpha$. In particular:
• $\alpha \le 1$: mean is infinite; sample averages diverge.
• $1 < \alpha \le 2$: mean exists, variance infinite; CLT fails.
• $\alpha > 2$: classical statistics apply, but convergence is slow until $\alpha > 4$.
Show that for a Pareto with $\alpha \approx 1.16$, the top 20% of outcomes hold about 80% of the total.
The cumulative share of the top $q$-fraction is $q^{(\alpha-1)/\alpha}$. With $\alpha = 1.16$, exponent $= 0.16/1.16 = 0.138$. Top 20%: $0.2^{0.138} = e^{0.138\ln 0.2} = e^{-0.222} \approx 0.80$. So the top 20% holds 80% of the mass.
For $x_m = 1$ and $\alpha = 1.5$, compute the mean and decide whether the variance exists.
Mean: $\alpha x_m/(\alpha-1) = 1.5/0.5 = 3$.
Variance requires $\alpha > 2$; here $\alpha = 1.5$, so variance is infinite. Confidence intervals based on $s/\sqrt n$ are meaningless.
Given the largest $k = 50$ observations $X_{(1)} \ge \dots \ge X_{(50)}$ from $n = 5000$, we observe $\sum \ln(X_{(i)}/X_{(k)}) = 30$. Estimate $\alpha$.
Hill's estimator: $\hat\alpha = k / \sum_{i=1}^{k-1}\ln(X_{(i)}/X_{(k)}) = 50/30 \approx 1.67$. Borderline infinite variance — treat moment-based statistics with suspicion.
Practice Problems
Show Answer Key
1. Mean $= 2.5\cdot 2/1.5 = 3.33$; var $= 2.5\cdot 4/(1.5^2 \cdot 0.5) = 8.89$.
2. Power-law tails appear as straight lines on a log-log survival plot; slope estimates $-\alpha$.
3. CLT holds (variance finite) but convergence is slow — Gaussian approximation requires very large n.
4. $(1/100)^{1.2} = 10^{-2.4} \approx 0.004 = 0.4\%$.
5. $0.01^{0.5/1.5} = 0.01^{1/3} \approx 0.215 = 21.5\%$ of total.
6. Book sales, wealth, city populations, company sizes, insurance claims, pandemic death tolls.