Mean Excess Functions For Exponential And Pareto Losses

Why Mean Excess Depends on the Threshold

When a loss is known to exceed a threshold, its severity distribution changes: the relevant quantity is no longer the average loss across all claims, but the average amount by which an above-threshold loss exceeds that threshold. The mean excess function summarizes this conditional amount as the threshold varies. It helps compare tail behavior without confusing the loss itself with the portion above the threshold.

For example, an insurer considering losses above a large reporting threshold may want to know the expected additional amount beyond that threshold. The answer depends on both the severity model and the threshold. Exponential losses have a constant mean excess, whereas the mean excess of Pareto losses increases with the threshold.

Definition: For a nonnegative loss $X$, the mean excess function at threshold $d$ is $$ m_X(d)=E[X-d\mid X>d], $$ provided $P(X>d)>0$ and the conditional expectation is finite. It is the expected amount of loss above $d$, conditional on the loss exceeding $d$; it is not the expected loss conditional on exceeding $d$.

Throughout, $S_X(x)=P(X>x)$ denotes the survival function. The threshold $d$ is nonnegative and expressed in the same units as $X$. If the conditional expectation is infinite, the mean excess is infinite rather than a finite value. We will use continuous severity models, so boundary points have probability zero.

A Tail-Integral Formula

The mean excess can be calculated from the conditional survival function of the amount above the threshold. This representation is particularly useful for exponential and Pareto models.

Result: If $X$ is nonnegative, $S_X(d)>0$, and the integral is finite, then $$ m_X(d)=\int_0^\infty \frac{S_X(d+y)}{S_X(d)}\,dy. $$

To justify the result, condition on $X>d$ and define the excess $W=X-d$. For $y\geq 0$, $$ P(W>y\mid X>d)=P(X>d+y\mid X>d)=\frac{S_X(d+y)}{S_X(d)}. $$ The second equality follows from the definition of conditional probability, since the event $X>d+y$ is contained in $X>d$. For any nonnegative random variable, its expectation equals the integral of its survival function over nonnegative values. Applying that identity to $W$ conditional on $X>d$ gives the stated formula. If the integral diverges, the conditional mean excess is infinite.

Exponential Losses: Constant Mean Excess

Let $X$ have an exponential distribution with rate $\lambda>0$, so $S_X(x)=e^{-\lambda x}$ for $x\geq 0$. Its mean is $1/\lambda$. Substitution into the tail-integral formula gives

m_X(d)=\int_0^\infty \frac{e^{-\lambda(d+y)}}{e^{-\lambda d}}\,dy =\int_0^\infty e^{-\lambda y}\,dy =\frac{1}{\lambda}.

The threshold $d$ cancels. Thus, for an exponential loss, the expected amount above any threshold is the same as the unconditional mean. This is the mean-excess form of the exponential distribution's memoryless property. It does not mean that the conditional expected loss $E[X\mid X>d]$ is constant: that expectation is $d+1/\lambda$.

Worked Example: Exponential Loss Above a Reporting Threshold

A hypothetical portfolio has loss $X$, measured in thousands of dollars, with an exponential distribution of rate $\lambda=0.25$ per thousand dollars. For a loss exceeding $d=6$ thousand dollars, calculate the mean excess and the probability that the loss exceeds 10 thousand dollars, conditional on exceeding 6 thousand dollars.

Method. Use the exponential survival function for the conditional probability and integrate the conditional excess survival function for the mean excess. This keeps the threshold amount separate from the expected amount above it.

The survival function is $S_X(x)=e^{-0.25x}$. First, the conditional probability is

P(X>10\mid X>6)=\frac{S_X(10)}{S_X(6)} =\frac{e^{-0.25(10)}}{e^{-0.25(6)}} =e^{-0.25(4)} =e^{-1} \approx 0.3679.

The mean excess is

m_X(6)=\int_0^\infty \frac{e^{-0.25(6+y)}}{e^{-0.25(6)}}\,dy =\int_0^\infty e^{-0.25y}\,dy =\frac{1}{0.25}=4

in thousands of dollars, or $4,000. Therefore, conditional on exceeding $6,000, the loss is expected to be $4,000 above the threshold; equivalently, $E[X\mid X>6]=6+4=10$ thousand dollars. The conditional probability is 0.3679 to four decimal places. The result checks against the exponential mean $1/\lambda=4$ thousand dollars.

Common mistake avoided: reporting $4,000 as $E[X\mid X>6]$. That is the mean excess, not the conditional loss itself; the latter is $10,000.

Pareto Type I: Mean Excess Proportional to the Threshold

For Pareto Type I, use scale $\theta>0$ and shape $\alpha>0$, with support $x\geq\theta$ and survival function $S_X(x)=(\theta/x)^\alpha$ on that support. At thresholds $d\geq\theta$, the conditional excess survival function is

P(X-d>y\mid X>d) =\frac{S_X(d+y)}{S_X(d)} =\left(\frac{d}{d+y}\right)^\alpha,\qquad y\geq 0.

Integrating this expression gives $m_X(d)=d/(\alpha-1)$, provided $\alpha>1$. The condition matters: if $0<\alpha\leq1$, the integral diverges, so the mean excess is infinite. When $\alpha>1$, Pareto Type I mean excess grows linearly with the threshold. For thresholds $0\leq d<\theta$, every loss exceeds $d$; if $\alpha>1$, $m_X(d)=E[X]-d=\alpha\theta/(\alpha-1)-d$. This joins the formula at $d=\theta$.

Worked Example: Pareto Type I Conditional Excess

A hypothetical loss $X$, in thousands of dollars, follows Pareto Type I with $\theta=2$ and $\alpha=3$. At threshold $d=6$ thousand dollars, find the mean excess and the conditional probability that the loss exceeds 12 thousand dollars.

Method. Since $d=6\geq\theta=2$, use the Pareto Type I tail ratio. The shape satisfies $\alpha=3>1$, so the mean excess is finite.

The mean excess is

m_X(6)=\frac{6}{3-1}=3

thousand dollars. Thus the expected excess is $3,000, and the conditional expected loss is $6+3=9$ thousand dollars.

For the probability, divide the survival probability at 12 by that at 6:

P(X>12\mid X>6) =\frac{(2/12)^3}{(2/6)^3} =\left(\frac{6}{12}\right)^3 =\frac18 =0.1250.

The calculation checks: an above-threshold loss has probability $1/8$ of exceeding twice the threshold, under this Pareto Type I model. The mean excess is $3,000, not the conditional mean loss of $9,000.

Common mistake avoided: using $\alpha d/(\alpha-1)$ as the mean excess. That expression is $E[X\mid X>d]$; subtracting $d$ gives the correct excess $d/(\alpha-1)$.

Pareto Type II: A Shifted Linear Mean Excess

Pareto Type II, also called the Lomax model, has scale $\theta>0$, shape $\alpha>0$, support $x\geq0$, and survival function $S_X(x)=(1+x/\theta)^{-\alpha}$. Its threshold can be any $d\geq0$. The conditional excess survival function is

\frac{S_X(d+y)}{S_X(d)} =\left(\frac{\theta+d}{\theta+d+y}\right)^\alpha.

This has the same form as the Pareto Type I conditional tail with $d$ replaced by $\theta+d$. Consequently, when $\alpha>1$,

m_X(d)=\frac{\theta+d}{\alpha-1}.

Unlike the Type I expression for thresholds $d\geq\theta$, the Type II mean excess includes the scale $\theta$ as well as the threshold. It increases linearly with $d$, and its value at $d=0$ is the unconditional mean $\theta/(\alpha-1)$. For $0<\alpha\leq1$, the mean is infinite and so is the mean excess at every threshold with positive survival probability.

Worked Example: Comparing Two Pareto Type II Thresholds

A hypothetical loss $X$, measured in thousands of dollars, has a Pareto Type II distribution with $\theta=2$ and $\alpha=3$. Calculate the mean excess at thresholds $d=4$ and $d=8$, and find the probability that a loss above 4 thousand dollars also exceeds 10 thousand dollars.

Method. Apply the Type II formula at each threshold and use a ratio of survival probabilities for the conditional probability. Since $\alpha=3>1$, both mean excesses are finite.

At $d=4$,

m_X(4)=\frac{2+4}{3-1}=3

thousand dollars. At $d=8$,

m_X(8)=\frac{2+8}{3-1}=5

thousand dollars. Raising the threshold by 4 thousand dollars raises the mean excess by 2 thousand dollars. The conditional expected losses are respectively $4+3=7$ and $8+5=13$ thousand dollars.

For the conditional probability,

P(X>10\mid X>4) =\frac{(1+10/2)^{-3}}{(1+4/2)^{-3}} =\left(\frac{3}{6}\right)^3 =\frac18 =0.1250.

The probability is 0.1250 to four decimal places. The calculations are consistent with the model: both the conditional excess and the conditional expected loss increase with the threshold, while the probability calculation uses the ratio of tails rather than an unconditional survival probability.

Common mistake avoided: applying the Type I expression $d/(\alpha-1)$ to Type II. At $d=4$, that would give 2 thousand dollars, but the correct Type II value includes the scale: $(\theta+d)/(\alpha-1)=3$ thousand dollars.

Comparing the Threshold Behavior

The models differ in what the mean excess tells us as the threshold rises. For exponential severity it remains fixed. For either Pareto form with finite mean, it increases linearly, but the intercept and the correct threshold formula depend on the parameterization.

Model	Mean excess	Threshold condition
Exponential with rate $\lambda$	$1/\lambda$	$\lambda>0,\ d\geq0$
Pareto Type I with scale $\theta$ and shape $\alpha$	$d/(\alpha-1)$ for $d\geq\theta$	$\alpha>1$; if $d<\theta$, use $E[X]-d$
Pareto Type II with scale $\theta$ and shape $\alpha$	$(\theta+d)/(\alpha-1)$	$\alpha>1,\ d\geq0$

For both Pareto models, the increase in mean excess as the threshold rises reflects a heavier tail than the exponential model. This is a statement about these model formulas, not a guarantee about every pair of loss distributions. Always verify the Pareto type, support, shape parameter, and finiteness condition before interpreting the result.

Key takeaway: Mean excess is expected loss above the threshold, conditional on exceeding it. It equals $1/\lambda$ for an exponential loss, $d/(\alpha-1)$ for Pareto Type I when $d\geq\theta$ and $\alpha>1$, and $(\theta+d)/(\alpha-1)$ for Pareto Type II when $\alpha>1$. Keep mean excess distinct from conditional expected loss: $E[X\mid X>d]=d+m_X(d)$.

Check Your Understanding

For each problem, state the distribution and use the specified threshold and parameterization.

A loss in thousands of dollars is exponential with rate $\lambda=0.5$. Find its mean excess at $d=3$ and $P(X>7\mid X>3)$. (Answer: mean excess $=2$ thousand dollars; probability $=0.1353$.)
A loss follows Pareto Type I with $\theta=3$, $\alpha=4$, and threshold $d=9$, all amounts in thousands of dollars. Find the mean excess and $P(X>18\mid X>9)$. (Answer: mean excess $=3$ thousand dollars; probability $=0.0625$.)
A loss follows Pareto Type II with $\theta=4$, $\alpha=3$, and threshold $d=8$, in thousands of dollars. Find the mean excess at $d=8$, the mean excess at zero, and $P(X>20\mid X>8)$. (Answer: mean excesses $=6$ and $2$ thousand dollars, respectively; probability $=0.1250$.)
Two hypothetical losses have Pareto parameters $\theta=2$, $\alpha=3$, and threshold $d=4$, all in thousands of dollars. One is Pareto Type I and the other Pareto Type II. Find each mean excess and the difference between them. (Answer: Type I $=2$ thousand dollars; Type II $=3$ thousand dollars; Type II exceeds Type I by $1$ thousand dollars.)

Model	Mean excess	Threshold condition
Exponential with rate \(\lambda\)	\(1/\lambda\)	\(\lambda>0,\ d\geq0\)
Pareto Type I with scale \(\theta\) and shape \(\alpha\)	\(d/(\alpha-1)\) for \(d\geq\theta\)	\(\alpha>1\); if \(d<\theta\), use \(E[X]-d\)
Pareto Type II with scale \(\theta\) and shape \(\alpha\)	\((\theta+d)/(\alpha-1)\)	\(\alpha>1,\ d\geq0\)

What You'll Learn