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Pareto distribution

A heavy-tailed positive loss where large claims stay meaningfully likely.

Parameters and support

$\alpha$

shape - larger means a lighter tail - $\alpha > 0$

$\theta$

scale - $\theta > 0$

Support

$x > 0$

The formulas

f(x)

$$f(x) = \dfrac{\alpha\theta^{\alpha}}{(x+\theta)^{\alpha+1}}$$

F(x)

$$F(x) = 1 - \left(\dfrac{\theta}{x+\theta}\right)^{\alpha}$$

Mean

$$E[X] = \dfrac{\theta}{\alpha - 1}, \quad \alpha > 1$$

Variance

$$\operatorname{Var}(X) = \dfrac{\alpha\theta^{2}}{(\alpha-1)^{2}(\alpha-2)}, \quad \alpha > 2$$

MGF

Not examinable for this distribution.

No MGF exists - the tail is heavier than any exponential.

Where the moments come from

1. $S(x) = \left(\frac{\theta}{x+\theta}\right)^{\alpha}$ decays polynomially, not exponentially.
2. $E[X] = \int_0^\infty S(x)\,dx = \theta^{\alpha}\int_0^{\infty}(x+\theta)^{-\alpha}dx$, which converges only when $\alpha > 1$, giving $\frac{\theta}{\alpha-1}$.
3. $E[X^2] = \int_0^\infty 2x\,S(x)\,dx = \frac{2\theta^2}{(\alpha-1)(\alpha-2)}$ for $\alpha > 2$.
4. Subtracting the squared mean gives $\frac{\alpha\theta^2}{(\alpha-1)^2(\alpha-2)}$.

Worked example

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run - summed over the support for a discrete distribution, integrated by quadrature for a continuous one - and compared with the closed forms printed here. A typo on this page fails that test run.

Traps

• Quoting a mean when α ≤ 1 or a variance when α ≤ 2 - they are infinite, and 'does not exist' is the answer.
• Confusing this two-parameter form with the single-parameter Pareto whose support starts at θ.

Related

• Exponential distribution
• Lognormal distribution

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.