Single-Parameter Pareto Distribution

The Single-Parameter Pareto Distribution with shape $\alpha > 0$ and a known lower bound $\theta > 0$ is a continuous distribution on $(\theta, \infty)$ , used for losses observed only above a fixed threshold. Only $\alpha$ is a true parameter: $\theta$ is set in advance.

$f(x) = \frac{\alpha\,\theta^{\alpha}}{x^{\alpha + 1}}, \quad x > \theta$

$F(x) = 1 - \left(\frac{\theta}{x}\right)^{\alpha}, \quad x > \theta$

$E[X^k] = \frac{\alpha\,\theta^k}{\alpha - k}, \quad k < \alpha$

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

$E[X \wedge x] = \frac{\alpha\,\theta}{\alpha - 1} - \frac{\theta^{\alpha}}{(\alpha - 1)\,x^{\alpha - 1}}, \quad x \geq \theta$

This is the parameterization of the CAS MAS-I tables. The mean is $E[X] = \frac{\alpha\theta}{\alpha - 1}$ for $\alpha > 1$ ; the variance is $E[X^2] - E[X]^2$ from the moment formula. The last block is the Limited Expected Value.
Link to the exponential: $\ln(X/\theta)$ has the exponential distribution with rate $\alpha$ (mean $1/\alpha$ ). That is why the maximum likelihood estimate of $\alpha$ from a sample is $\hat{\alpha} = n \big/ \sum \ln(x_i/\theta)$ .
Link to the two-parameter Pareto Distribution: $X - \theta$ has the Pareto distribution with shape $\alpha$ and scale $\theta$ . The two are easy to confuse; the single-parameter version's support starts at $\theta$ , not 0.
As with any Pareto, the $k$ -th moment exists only for $k < \alpha$ , so the tail is heavy.

Fitting Large Losses Above a Reporting Threshold An insurer records only losses above 1,000, and models them as single-parameter Pareto with $\theta = 1{,}000$ . Five recorded losses are $1{,}200,\ 1{,}500,\ 2{,}500,\ 1{,}100,\ 4{,}000$ . Find the maximum likelihood estimate of $\alpha$ , and use it to estimate the probability that a recorded loss exceeds 3,000.
Answer The log-likelihood is $\ell(\alpha) = n\ln\alpha + n\alpha\ln\theta - (\alpha + 1)\sum \ln x_i$ . Setting $\ell'(\alpha) = n/\alpha - \sum \ln(x_i/\theta) = 0$ :
 $\begin{align*} \sum \ln\frac{x_i}{1{,}000} &= \ln 1.2 + \ln 1.5 + \ln 2.5 + \ln 1.1 + \ln 4 \\ &= 2.9857 \\ \hat{\alpha} &= \frac{5}{2.9857} \\ &= 1.6747 \end{align*}$ 
Then
 $\begin{align*} \hat{P}(X > 3{,}000) &= \left(\frac{1{,}000}{3{,}000}\right)^{1.6747} \\ &= 0.1588 \end{align*}$ 
About 16% of recorded losses are estimated to exceed 3,000. With $\hat{\alpha} < 2$ the fitted distribution has an infinite variance.

Limited Expected Value of Excess Losses Losses above a threshold follow a single-parameter Pareto with $\alpha = 3$ and $\theta = 500$ . A policy caps each payment at 2,000. Find the expected payment, and the expected amount by which losses exceed the cap.
Answer
 $\begin{align*} E[X] &= \frac{3(500)}{3 - 1} \\ &= 750 \\ E[X \wedge 2{,}000] &= 750 - \frac{500^3}{(2)(2{,}000)^2} \\ &= 750 - 15.625 \\ &= 734.375 \end{align*}$ 
The cap removes $E[X] - E[X \wedge 2{,}000] = 15.625$ per loss, even though only $(500/2{,}000)^3 = 1.56\%$ of losses exceed it.