Pareto Hazard And Conditional Excess Loss

Why the Pareto Hazard and Excess Matter

A large-loss analysis often conditions on a claim already exceeding a threshold. The conditional excess describes how much larger the loss is than that threshold, while the hazard describes the local rate of failure to survive further as the loss level increases. For a Pareto Type I severity, both quantities have simple forms, but their interpretation depends on keeping the threshold and the Pareto scale distinct.

The results below use a continuous Pareto Type I loss $X$ with scale parameter $\theta>0$ and shape parameter $\alpha>0$. There is no interest or mortality assumption in these severity calculations. All dollar amounts and hypothetical loss scenarios are illustrative.

Definition: A Pareto Type I loss with scale $\theta$ and shape $\alpha$ has support $x\geq\theta$, distribution function $F_X(x)=1-(\theta/x)^\alpha$ for $x\geq\theta$, and survival function $S_X(x)=(\theta/x)^\alpha$ for $x\geq\theta$. For $x<\theta$, $F_X(x)=0$ and $S_X(x)=1$. Its density for $x\geq\theta$ is $f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}$. The distribution is continuous, so there is no point mass at $\theta$.

Deriving the Pareto Type I Hazard

For a continuous loss with density $f_X(x)$ and positive survival $S_X(x)$, the hazard at level $x$ is the density divided by the probability of surviving beyond $x$:

h_X(x)=\frac{f_X(x)}{S_X(x)}.

For the Pareto Type I model and $x\geq\theta$, substituting the density and survival function gives

h_X(x) = \frac{\alpha\theta^\alpha x^{-(\alpha+1)}}{(\theta/x)^\alpha} = \frac{\alpha}{x}.

The scale parameter cancels. The hazard decreases as the loss level increases: $h_X(x)$ is proportional to $1/x$. Below the support, $f_X(x)=0$ and $S_X(x)=1$, so the hazard is zero there. The expression $\alpha/x$ applies on the supported range $x\geq\theta$; its units are inverse dollars when $x$ is measured in dollars.

Result: For a Pareto Type I loss, $h_X(x)=\alpha/x$ for $x\geq\theta$. A decreasing hazard does not mean that the loss decreases; it describes the hazard rate as the threshold level $x$ increases.

Conditional Excess Loss Above a Threshold

Let $u\geq\theta$ be a threshold and define the excess amount, given that the loss exceeds $u$, by $Z_u=X-u\mid X>u$. This is a nonnegative random variable. For $y\geq0$, its survival function follows directly from conditional probability:

\begin{aligned} P(Z_u>y) &=P(X-u>y\mid X>u)\\ &=\frac{P(X>u+y)}{P(X>u)}\\ &=\frac{(\theta/(u+y))^\alpha}{(\theta/u)^\alpha}\\ &=\left(\frac{u}{u+y}\right)^\alpha. \end{aligned}

The event $X>u+y$ already implies $X>u$, which justifies the conditional-probability numerator. The scale $\theta$ cancels, leaving the threshold $u$ as the scale of the excess distribution. In particular, the conditional excess does not have the same support as the original Pareto Type I loss: $Z_u$ starts at zero.

Definition: Given $X>u$, the conditional excess $Z_u=X-u$ has survival function $S_{Z_u}(y)=(u/(u+y))^\alpha$ for $y\geq0$, and distribution function $F_{Z_u}(y)=1-(u/(u+y))^\alpha$. Its density is $f_{Z_u}(y)=\alpha u^\alpha/(u+y)^{\alpha+1}$ for $y\geq0$. This is a Lomax (Pareto Type II) form with scale $u$ and shape $\alpha$.

This conditional distribution has a useful interpretation: after the loss is known to exceed $u$, the probability of an additional excess larger than $y$ depends on $u$ and $\alpha$, but not separately on the original scale $\theta$. The shape parameter remains $\alpha$; the conditional excess is not exponential and does not have the exponential memoryless property.

Expected Excess and Its Finiteness

For a nonnegative random variable $Z$, its expectation can be found by integrating its survival function, provided the integral is finite. Applying this identity to the conditional excess gives

\begin{aligned} E[Z_u] &=\int_0^\infty P(Z_u>y)\,dy\\ &=\int_0^\infty \left(\frac{u}{u+y}\right)^\alpha dy\\ &=u\int_1^\infty t^{-\alpha}\,dt, \end{aligned}

where the substitution is $t=(u+y)/u$, so $dy=u\,dt$. For $\alpha>1$, the improper integral converges:

u\int_1^\infty t^{-\alpha}\,dt = u\left[\frac{t^{1-\alpha}}{1-\alpha}\right]_1^\infty = \frac{u}{\alpha-1}.

If $0<\alpha\leq1$, then $t^{-\alpha}$ has a divergent integral over $[1,\infty)$, so the expected conditional excess is infinite. Thus the finiteness condition is a property of the shape parameter, not of how large the threshold is.

Theorem: For $u\geq\theta$, the conditional expected excess is $E[X-u\mid X>u]=u/(\alpha-1)$ when $\alpha>1$. It is infinite when $0<\alpha\leq1$.

A related quantity is the unconditional expected payment above the threshold, often written as $E[(X-u)_+]$, where $(X-u)_+=\max(X-u,0)$. The loss contributes zero when it does not exceed $u$. Conditioning on whether $X>u$ gives

E[(X-u)_+] = P(X>u)E[X-u\mid X>u] = \left(\frac{\theta}{u}\right)^\alpha\frac{u}{\alpha-1}, \qquad \alpha>1.

This is not the same as the conditional expected excess. The conditional quantity averages only over losses that exceed $u$; the unconditional quantity also accounts for the probability that a loss exceeds $u$.

Check the support.
For the formulas here, verify $u\geq\theta$. Then $P(X>u)=(\theta/u)^\alpha$.

Identify the requested quantity.
Use $E[X-u\mid X>u]$ for the conditional mean excess, and multiply by $P(X>u)$ for $E[(X-u)_+]$.

Check finiteness.
Either expected excess is finite only when $\alpha>1$; then substitute the threshold and, for the unconditional quantity, the exceedance probability.

Worked Applications

Worked Example: Hazard and Conditional Exceedance Probability

A hypothetical portfolio uses a Pareto Type I model with $\theta=\$1{,}000$ and $\alpha=3$. Find the hazard at $x=\$4{,}000$, and find the probability that the excess above $u=\$2{,}000$ exceeds $\$1{,}000$, given that the loss exceeds $\$2{,}000$. No interest or mortality assumption is involved.

Method. Use the Pareto hazard for the first calculation and the conditional excess survival function for the second. Both thresholds are at least $\theta$, as required.

The hazard is

h_X(4{,}000)=\frac{3}{4{,}000}=0.00075

per dollar. The requested conditional probability is

P(Z_{2{,}000}>1{,}000) = \left(\frac{2{,}000}{2{,}000+1{,}000}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \approx 0.2963.

Check. The hazard is positive and has inverse-dollar units; the conditional probability is between zero and one. The calculation uses the excess $1{,}000$, not the total loss $3{,}000$, as the amount being tested.

Common mistake avoided. Substituting $3{,}000$ as the excess would compute the probability that the loss exceeds $\$5{,}000$ given it exceeds $\$2{,}000$, not the probability asked for.

Worked Example: Conditional Mean and Unconditional Excess

Suppose a hypothetical equipment-loss severity follows Pareto Type I with $\theta=\$500$ and $\alpha=4$. At threshold $u=\$2{,}000$, calculate both the expected excess conditional on exceeding the threshold and the unconditional expected amount above it. No interest or mortality assumption is involved.

Method. Since $u\geq\theta$ and $\alpha=4>1$, both expectations are finite. First calculate the conditional mean excess:

E[X-u\mid X>u] = \frac{u}{\alpha-1} = \frac{2{,}000}{4-1} = \$666.67

Next calculate the exceedance probability:

P(X>2{,}000) = \left(\frac{500}{2{,}000}\right)^4 = \left(\frac14\right)^4 = \frac{1}{256} = 0.00390625.

Therefore,

E[(X-2{,}000)_+] = \frac{1}{256}\cdot\frac{2{,}000}{3} = \frac{2{,}000}{768} \approx \$2.60.

Check. The unconditional value is much smaller than the conditional mean because only $1/256$ of losses exceed the threshold. Both dollar amounts are rounded to the nearest cent.

Common mistake avoided. Reporting $\$666.67$ as the unconditional expected amount ignores the probability of an exceedance. That figure is conditional on $X>u$.

Worked Example: Conditional Tail, Hazard, and Mean Excess

A hypothetical severity model has Pareto Type I parameters $\theta=\$200$ and $\alpha=2.5$. At $u=\$800$, calculate the hazard at $x=\$1{,}600$, the probability the excess exceeds $\$400$ given $X>\$800$, the conditional expected excess, and the unconditional expected amount above $\$800$. No interest or mortality assumption is involved.

Method. Apply the hazard and conditional excess formulas. Since $\alpha=2.5>1$, the two expected excess quantities are finite.

First, the hazard is

h_X(1{,}600) = \frac{2.5}{1{,}600} = 0.0015625

per dollar. The conditional tail probability is

P(Z_{800}>400) = \left(\frac{800}{800+400}\right)^{2.5} = \left(\frac23\right)^{2.5} \approx 0.3629.

The conditional mean excess is

E[X-800\mid X>800] = \frac{800}{2.5-1} = \frac{800}{1.5} \approx \$533.33.

The probability of exceeding $\$800$ is

P(X>800) = \left(\frac{200}{800}\right)^{2.5} = \left(\frac14\right)^{2.5} = \frac{1}{32}.

Thus the unconditional expected amount above $\$800$ is

E[(X-800)_+] = \frac{1}{32}\cdot\frac{800}{1.5} = \frac{800}{48} \approx \$16.67.

Check. The conditional mean is larger than the unconditional expectation because the latter includes all losses that do not exceed $\$800$, for which the excess payment is zero. The probability and dollar results are rounded to four decimals and the nearest cent, respectively.

Common mistake avoided. The expected excess is $u/(\alpha-1)$, not $\theta/(\alpha-1)$. The relevant scale for the conditional excess is the threshold $u$.

Common Traps and Practical Interpretation

The hazard and the conditional excess answer different questions. The hazard $h_X(x)$ describes the local rate at a particular loss level. The conditional excess distribution describes the full distribution of the amount above a fixed threshold, given that the threshold has been crossed. Neither quantity is itself the probability of exceeding a level; for that, use the survival function or the conditional survival ratio.

Using the original scale in the conditional mean. The conditional mean excess is $u/(\alpha-1)$, not $\theta/(\alpha-1)$, because the excess distribution has scale $u$.
Confusing conditional and unconditional expectations. Multiply the conditional mean excess by $P(X>u)$ to obtain $E[(X-u)_+]$.
Ignoring the shape condition. If $\alpha\leq1$, the conditional expected excess is infinite, even though the conditional survival probability is well-defined.
Applying the Pareto formula below the support. The stated conditional formulas assume $u\geq\theta$. A threshold below $\theta$ does not satisfy that setup.
Reading a decreasing hazard as a decreasing loss. The hazard decreases with the level $x$; this does not assert that any individual loss decreases.

Key takeaway: For Pareto Type I, the hazard is $\alpha/x$, and the excess above $u$, given $X>u$, has survival $(u/(u+y))^\alpha$. Its mean is $u/(\alpha-1)$ only when $\alpha>1$; the unconditional expected excess also includes the probability of exceeding $u$.

Check Your Understanding

Use the Pareto Type I formulas and state whether each requested expected excess is conditional or unconditional.

A hypothetical loss has $\theta=\$250$ and $\alpha=2$. Find its hazard at $x=\$1{,}000$. (Answer: $0.002$ per dollar.)
A loss has $\theta=\$600$, $\alpha=3$, and $u=\$1{,}200$. Find the probability that the conditional excess exceeds $\$600$. (Answer: $8/27\approx0.2963$.)
A loss has $\theta=\$1{,}000$, $\alpha=4$, and $u=\$2{,}000$. Find the conditional expected excess. (Answer: $\$666.67$.)
A loss has $\theta=\$500$, $\alpha=2$, and $u=\$1{,}500$. Find the unconditional expected amount above $u$. (Answer: $\$166.67$.)
A loss has $\theta=\$100$, $\alpha=0.8$, and $u=\$500$. Is its conditional expected excess finite? (Answer: No; it is infinite because $\alpha\leq1$.)
A loss has $\theta=\$400$, $\alpha=2.5$, and $u=\$1{,}600$. Find both the probability that the conditional excess exceeds $\$800$ and the unconditional expected amount above $u$. (Answer: $0.3629$ and $\$33.33$, respectively.)

What You'll Learn