The Mathematics of Excess Losses

1. Introduction

The concept of excess loss is widely used and appreciated within the casualty-actuarial world, particularly in reinsurance, but also in retrospective and increased-limits rating. This paper will reveal some of its hidden mysteries. In the next section we will define the excess-loss function, and state its sometime advantages over standard probability concepts. This will prepare us, in the third section, to find new applications of the function, applications extending as far as moment generation. Excess losses naturally imply loss layers, whose probability distributions we will show in the fourth section to be more amenable to an excess-loss treatment than to standard probability theory. Here also we will introduce an example involving the first two moments of a mixed exponential distribution. Then we will, in the fifth section, round out the second moment of the example by considering the covariances among loss layers, and will conclude in the sixth section.

2. The excess-loss function

Let X be a non-negative random variable, i.e., a random variable suitable for representing an amount of loss. Its cumulative distribution function F_X(a) = Prob[X≤a] has the following four properties:

1. If a b, then F_X(a) ≤ F_X(b) non-decreasing

2. $\lim _{a \rightarrow \infty} F_{X}(a)=1$ total probability

3. $\lim _{b \rightarrow a^{+}} F_{X}(b)=F_{X}(a)$ continuity from the right

4. $\lim _{a \rightarrow 0^{-}} F_{X}(a)=0$ non-negative

These properties allow for points of probability mass, since

$$\begin{aligned} {Prob}[X=a] & =\lim _{b \rightarrow a^{-}} {Prob}[{b}<{X} \leq {a}] \\ & ={F}_{X}(a)-\lim _{b \rightarrow a^{-}} {F}_{X}(b) \\ & \geq {0} . \end{aligned}$$

Of particular note, the probability for X to equal 0 may be positive. Moreover, let G_X be the survival function, the complement of F_X: G_X(a) = 1 − F_X(a) = Prob[X > a]. Hence, dG_X(a) = −dF_X(a). G_X is also continuous from the right, and Prob[X = a] = $\lim _{b \rightarrow a^{-}}$ G_X(b) − G_X(a). However, it is non-increasing, its limit at infinity is zero, and G_X(a) = 1 for a 0.

For r ≥ 0, the expected portion of loss X in excess of “retention” r is defined as:

$$\begin{aligned} \ Excess_{X}(r) & \equiv \int_{x=-\infty}^{\infty} \max (0, x-r) d F_{X}(x) \\ & =\int_{x=r}^{\infty}(x-r) d F_{X}(x) \end{aligned}$$

In particular, Excess_X(0) = $\int_{x=0}^{\infty}$xdF_X(x) = E[X]. From

integration by parts, we reformulate:

$$\begin{aligned} \ Excess_{X}(r) & =\int_{x=r}^{\infty}(x-r) d F_{X}(x) \\ & =-\int_{x=r}^{\infty}(x-r) d G_{X}(x) \\ & =-\left.(x-r) G_{X}(x)\right|_{r} ^{\infty}+\int_{x=r}^{\infty} G_{X}(x) d(x-r) \\ & =\left.(x-r) G_{X}(x)\right|_{\infty} ^{r}+\int_{x=r}^{\infty} G_{X}(x) d x \\ & =0-0+\int_{x=r}^{\infty} G_{X}(x) d x . \end{aligned}$$

From their studies, casualty actuaries are familiar with this expression.^[1]

The excess-loss function is especially useful in reinsurance: if 0 ≤ a ≤ b, the pure premium for the portion of loss X in the layer [a, b] equals Excess_X(a) − Excess_X(b). Note that since G is dimensionless, the unit of the excess-loss function is the unit of dx, which is the unit of the loss amount and the retention. Two appealing properties of the excess-loss function are (1) that it is everywhere continuous, even where the probability density is discrete, and (2) that if it is positive, it strictly decreases. Moreover, its derivative at r, if it exists, equals −G_X(r). Even if it does not exist, at least the left and right derivatives exist, and the difference of the left derivative from the right is the probability mass at r. It is helpful to extend the definition of the excess-loss function to negative retentions, at which G_X(r) = 1. For r 0:

$$\begin{aligned} \ Excess_{X}(r) & =\int_{x=r}^{\infty} G_{X}(x) d x\\ &=\int_{x=r}^{0} G_{X}(x) d x+\int_{x=0}^{\infty} G_{X}(x) d x \\ & =\int_{x=r}^{0} 1 \cdot d x+\ Excess_{X}(0)\\ &=-r+E[X] . \end{aligned}$$

Often useful is the form, valid for all r, Excess_X(r) = −min(0, r) + $x=\max (0, r)$G_X(x)dx.

It is not difficult to prove two basic theorems. If c > 0, then for all r:

$$\begin{array}{c} \ Excess_{X+c}(r)=\ Excess_{X}(r-c) \\ \ Excess_{c X}(r)=\operatorname{cxcess}_{x}(r / c) . \end{array}$$

The first holds true for c = 0; the second holds true as well in the limit as c → 0⁺, from which it follows that Excess₀(r) = −min(0, r). Furthermore, Excess_c(r) = −min(0, r − c) = max(0, c − r).

As an example, let X be exponentially distributed with mean θ. Hence, $G_{x}(x)=\left\{\begin{array}{ll} e^{-\frac{x}{\theta}} & 0 \leq x \\ 1 & x \leq 0 \end{array}\right.$

So for r ≥ 0, $Excess_{X}(r)$ $=$ $\int_{x=r}^{\infty} G_{X}(x) d x$ $=$ $\int_{x=r}^{\infty} e^{-\frac{x}{\theta}} d x=\left.\theta e^{-\frac{x}{\theta}}\right|_{\infty} ^{r}=\theta e^{-\frac{r}{\theta}}$ and with negative retentions:

$$\ Excess_{X}(x)=\left\{\begin{array}{ll} \theta e^{-\frac{x}{\theta}} & 0 \leq x \\ -x+\theta & x<0 \end{array}\right. \text {. }$$

Figure 1 graphs this function over the domain [−θ, 4θ]. In preparation for the next section, we’ve also extended as a dotted line the negative-retention line, i.e., f(x) = −x + θ.

Figure 1.Excess-loss function of an Exponential(θ) loss

The straight line itself is the graph of Excess_θ(x). It is indicative of positive variance that the excess-loss function “pulls up and comes in for a landing” to the right of the dotted line.^[2] This is the clue for extracting more information from the excess-loss function.

3. Higher moments and the excess-loss function

The excess-loss function $\ Excess_{X}(r)=\int_{x=r}^{\infty} G_{X}(x) d x$ is not only elegant and useful, it is “full-informational” in the sense that one can derive from it all the moments of X. Above, we saw that the mean of X equals Excess_X(0). That the dimension of the area under the excess-loss function is the square of the loss unit suggests that the area has something to do with the second moment. The following derivation relies on the validity of inverting the order of integration over the region A, which is the part of the first quadrant of the Cartesian plane above the line y = x:

$$\begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d x & =\int_{x=0}^{\infty} \int_{y=x}^{\infty} G_{X}(y) d y \ d x \\ & =\iint_{A} G_{x}(y) d A \\ & =\int_{y=0}^{\infty} \int_{x=0}^{y} G_{x}(y) d x \ d y \\ & =\int_{y=0}^{\infty} G_{x}(y)\left(\int_{x=0}^{y} d x\right) d y \\ & =\int_{y=0}^{\infty} G_{x}(y) y \ d y \\ & =\frac{1}{2} \int_{y=0}^{\infty} G_{X}(y) d\left(y^{2}\right) \\ & =\left.\frac{y^{2} G_{X}(y)}{2}\right|_{0} ^{\infty}-\frac{1}{2} \int_{y=0}^{\infty} y^{2} \ d G_{X}(y) \\ & =0-0+\frac{1}{2} \int_{y=0}^{\infty} y^{2} \ d F_{X}(y) \\ & =\frac{1}{2} E\left[X^{2}\right] . \end{aligned}$$

So the area under the excess-loss function in the first quadrant equals half the second moment. Therefore, the area under the excess-loss function but above the dotted line and the x-axis is

$$\begin{aligned} \frac{1}{2} E\left[X^{2}\right]-\frac{1}{2} E[X] E[X] & =\frac{1}{2}\left(E\left[X^{2}\right]-E[X]^{2}\right) \\ & =\frac{1}{2} \operatorname{Var}[X] . \end{aligned}$$

Hence, for the excess-loss function to “land” to the right of E[X] indicates a non-trivial variance.

In general, for h(x) continuous on [0, ∞),

$$\small{ \begin{aligned} \int_{x=0}^{\infty} \ Excess(x) d h(x) & =\int_{x=0}^{\infty} \int_{y=x}^{\infty} G_{X}(y) d y \ d h(x) \\ & =\int_{y=0}^{\infty} \int_{x=0}^{y} G_{x}(y) d h(x) d y \\ & =\int_{y=0}^{\infty} G_{x}(y)\{h(y)-h(0)\} d y \\ & =\int_{y=0}^{\infty} G_{x}(y) h(y) d y-\int_{y=0}^{\infty} G_{x}(y) h(0) d y \\ & =\int_{y=0}^{\infty} G_{x}(y) h(y) d y-E[X] h(0) . \end{aligned} }$$

Instead of the double integration, we may use integration by parts:

$$\small{ \begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d h(x)= & \ Excess_{X}(x) h(x)_{0}^{\infty} \\ & -\int_{x=0}^{\infty} h(x) d \ Excess_{X}(x) \\ = & 0-\ Excess_{X}(0) h(0) \\ & -\int_{x=0}^{\infty} h(x)\left(-G_{X}(x) d x\right) \\ = & \int_{x=0}^{\infty} G_{X}(x) h(x) d x-E[X] h(0) . \end{aligned} }$$

Now let H′(x) = h(x), or dH(x) = h(x)dx. The following derivation relies on a formula from Appendix A, viz., $E[h(X)]=h(0)+\int_{x=0}^{\infty} G_{X}(x) d h(x)$

$$\begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d h(x)= & \int_{x=0}^{\infty} G_{X}(x) h(x) d x-E[X] h(0) \\ = & \int_{x=0}^{\infty} G_{X}(x) d H(x)-E[X] H^{\prime}(0) \\ = & H(0)+\int_{x=0}^{\infty} G_{X}(x) d H(x) \\ & -H(0)-E[X] H^{\prime}(0) \\ = & E[H(X)]-H(0)-E[X] H^{\prime}(0) . \end{aligned}$$

The invariance of this formula to the addition of a linear function to H confirms its correctness. For if H(x) ← H(x) + cx + d, then h(x) ← h(x) + c and

$$\begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d(h(x)+c)= & E[H(X)+c X+d] \\ & -(H(0)+c \cdot 0+d) \\ & -E[X]\left(H^{\prime}(0)+c\right) \\ = & E[H(X)]+c E[X] \\ & +d-H(0)-d \\ & -E[X] H^{\prime}(0)-c E[X] \\ = & \int_{x=0}^{\infty} E x c e s s_{X}(x) d h(x) . \end{aligned}$$

So, for example, if h(x) = x^k, where k must be positive for continuity at zero, then $H(x)=\frac{x^{k+1}}{k+1}$ and

$$\begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d x^{k}= & E\left[\frac{X^{k+1}}{k+1}\right]-\frac{0^{k+1}}{k+1} \\ & -E[X] \cdot 0^{k}=\frac{1}{k+1} E\left[X^{k+1}\right] . \end{aligned}$$

So, the second and higher moments result from integrals of the excess-loss function. Of course, the first moment is just Excess_X(0).

Especially interesting is the integral $\int_{x=0}^{\infty} \ Excess_{X}(x)$dte^tx, for which h(x) = te^tx and H(x) = e^tx:

$$\begin{aligned} \int_{x=0}^{\infty} \ Excess_{X}(x) d t e^{x x} & =E\left[e^{x}\right]-e^{t \cdot 0}-E[X] \cdot t e^{t \cdot 0} \\ & =E\left[e^{t x}\right]-1-E[X] \cdot t \\ & =M_{X}(t)-1-M_{X}^{\prime}(0) \frac{t^{1}}{1 !} \\ & =\sum_{i=2}^{\infty} M_{X}^{[i]}(0) \frac{t^{i}}{i !} \end{aligned}$$

This integral of the excess-loss function reproduces the moment-generating function of X, except for the removal of the first two terms of its Maclaurin series. The whole moment-generating function of X is recovered as $M_{x}(t)=1+\ Excess_{x}(0) \cdot t+\int_{x=0}^{\infty} \ Excess_{x}(x) d t e^{t x} .$ This implies that for t ≠ 0, $\int_{x=0}^{\infty} \ Excess_{x}(x) d e^{x}=\sum_{i=2}^{\infty} M_{x}^{[i]}(0) \frac{t^{i-1}}{i !}=\sum_{i=1}^{\infty} \frac{M_{x}^{[i+]}(0)}{i+1} \frac{t^{i}}{i !} \text {, }$ which holds true even for t = 0. Therefore, $1+\int_{x=0}^{\infty} \ Excess_{X}(x) d e^{t x}$is the moment-generating function of the random variable Y whose moments relate to those of X as $E\left[Y^{i}\right]=\frac{E\left[X^{i+1}\right]}{i+1}$ To use a musical analogy, $\int_{x=0}^{\infty} \ Excess_{X}(x) d e^{t x}$ is a transformation that “plays” X one octave lower.^[3]

4. Excess-loss functions of layered losses

If X is a non-negative random variable and 0 ≤ a ≤ b, then define the portion of loss X in the layer [a, b] as Layer(X; a, b) ≡ min(b − a, max(0, X − a)). The graph in Figure 2 shows that the layer function is flat except in the interval [a, b], in which Layer(x; a, b) = x − a.

Figure 2.Layer function

Our purpose in this section is to express Excess_Y in terms of Excess_X, which requires expressing G_Y in terms of G_X.

Of course, if r 0, then G_Y(r) = 1. For b − a ≤ r, G_Y(r) = Prob[Y > r] = Prob[Y > b − a] = 0. And in the middle, for 0 r b − a:

$$\begin{aligned} G_{Y}(r) & =\ Prob[Y>r]=\ Prob[X-a>r] \\ & = \ Prob[X>r+a]=G_{X}(r+a) . \end{aligned}$$

By continuity from the right, G_Y(0) = G_Y(0⁺) = G_X(0⁺ + a) = G_X(a). So the final conditional formula is

$${G}_{Y}(y)=\left\{\begin{array}{rc} {1} & {y}<{0} \\ {G}_{x}(y+a) & {0} \leq {y}<{b}-{a}. \\ {0} & {b-a} \leq {y} \end{array}\right.$$

Hence, for 0 ≤ r b − a:

$$\begin{aligned} \ Excess_{Y}(r) & =\int_{y=r}^{\infty} G_{Y}(y) d y \\ & =\int_{y=r}^{b-a} G_{Y}(y) d y \\ & =\int_{y=r}^{b-a} G_{X}(y+a) d y \\ & =\int_{y+a=r+a}^{b-a+a} G_{X}(y+a) d(y+a) \\ & =\int_{x=r+a}^{b} G_{X}(x) d x \\ & =\ Excess_{X}(r+a)- \ Excess_{X}(b) . \end{aligned}$$

The form Excess_Y(r) = Excess_X(min(b,r + a)) − Excess_X(b) is valid for all r ≥ 0; it also accommodates the three limiting cases a = 0, b = a, and b = ∞.^[4] Furthermore, E[Y] = Excess_Y(0) = Excess_X(a) − Excess_X(b), as mentioned in Section 2.

The tables provide an example. First, we mixed four exponential distributions (with weights w and means θ at the top of Table 1).^[5] The excess-loss function of the mixed exponential distribution (the “Mxd-Exp Excess” column) is $\ Excess_{M X}(r)=\sum_{i=1}^{4} w_{i} \theta_{i} e^{-\frac{r}{\theta_{i}}}$ Its values are shown for retentions from zero to 50 million (consider the unit of loss as USD) in steps of one million; the values are also equal to the values of the previous four columns (gray-shaded), weighted according 0.500, 0.250, 0.125, and 0.125. The mean loss is 1,375,000. The final column of Table 1 shows the area under the Excess_MX curve from r to infinity. Its formula is

$$\small{ \begin{aligned} \int_{x=r}^{\infty} \ Excess_{M X}(x) d x & =\int_{x=r}^{\infty} \sum_{i=1}^{4} w_{i} \theta_{i} e^{-\frac{r}{\theta_{i}}} d x \\ & =\sum_{i=1}^{4} w_{i} \theta_{i} \int_{x=r}^{\infty} e^{-\frac{r}{\theta_{i}}} d x=\sum_{i=1}^{4} w_{i} \theta_{i}^{2} e^{-\frac{r}{\theta_{i}}} . \end{aligned} }$$

The total area, which according to Section 3 is E[MX²]/2, is 4.000 × 10¹² (USD squared). Therefore the variance of the mixed exponential loss is 2×4.000×10¹² − 1,375,000², for a standard deviation of 2,471,715.

Table 2 partitions the support of MX into four non-overlapping layers: [0, 5M], [5M, 10M], [10M, 20M], and [20M, ∞).^[6] Let Y_i denote the portion of loss MX in the i^th layer. Because of the non-overlapping and complete partition, $M X=\sum_{i} Y_{i}$ The excess losses are calculated according to the formula Excess_Y(r) = Excess_X(min(b,r+a)) − Excess_X(b). The formula for the infinite top layer is simpler: Excess_Y(r) = Excess_X(min(∞, r + 20M)) − Excess_X(∞) = Excess_X(r + 20M). As expected, the sum of the means of the layered losses, E$\left[Y_i\right]=$ Excess $_{Y_i}(0)$, equals 1,375,000; the partitioning conserves the first moment of the loss.

Table 2, like Table 1, derives the second moment of each layered loss from the area under its excess-loss curve. But, as one of the advantages of the excess-loss function, it is not necessary to do this anew; it is implicit in Table 1. For algebraically:

$$\begin{array}{l} \frac{E\left(Y_{i}^{2}\right)}{2}=\int_{y=0}^{\infty} \ Excess_{Y_{i}}(y) d y \\ \quad = \int_{y=0}^{\infty}\left\{\ Excess_{M X}\left(\min \left(b_{i}, y+a_{i}\right)\right)-\ Excess_{M X}\left(b_{i}\right)\right\} d y \\ \quad= \int_{y=0}^{b_{i}-a_{i}}\left\{\ Excess_{M X}\left(y+a_{i}\right)-\ Excess_{M X}\left(b_{i}\right)\right\} d y \\ \quad = \int_{y=0}^{b_{i}-a_{i}} \ Excess_{M X}\left(y+a_{i}\right) d y-\left(b_{i}-a_{i}\right) \ Excess_{M X}\left(b_{i}\right) \\ \quad = \int_{x=a_{i}}^{b_{i}} \ Excess_{M X}(x) d x-\left(b_{i}-a_{i}\right) \ Excess_{M X}\left(b_{i}\right) \\ \quad = \int_{x=a_{i}}^{\infty} \ Excess_{M X}(x) d x-\int_{x=b_{i}}^{\infty} \ Excess_{M X}(x) d x \\ \qquad-\left(b_{i}-a_{i}\right) \ Excess_{M X}\left(b_{i}\right) . \end{array}$$

The values of the last two integrals are those of the last column of Table 1 at retentions a_i and b_i.^[7] The “Area” row at the bottom of Table 2 contains these E[Y²_i]/2 values, from which follow the variances and standard deviations. It is well known among reinsurance actuaries that the coefficients of variation, CV = Std/E, increase as the layers ascend.^[8] The sum of the four areas, 2.193 × 10¹², does not equal the 4.000 × 10¹² of the MX area; nor is the sum of the four variances, 3.093 × 10¹², equal to the variance of MX, 6.109 × 10¹². What is lacking in the conservation of the second moment is the covariance among the layered losses, to which we now turn.

5. Covariances among non-overlapping layers

Since Cov[Y_i, Y_j] = E[Y_iY_j] − E[Y_i]E[Y_i], and the means are known, we need a formula for the product moments E[Y_iY_j], where i ≠ j. The actual formula, based on the loss variable of which they are layers, is $E\left[Y_{i} Y_{j}\right]=\int_{x=-\infty}^{\infty}$y_i(x)y_j(x)dF(x). A formal derivation is not necessary; the following argument will suffice. Since the layers are different but non-overlapping, one is above the other. The range of the integration may be restricted to the range over which the integrand y_i(x)y_j(x) is non-zero, which range is the intersection of the non-zero ranges of y_i(x) and y_j(x) separately. However, due to the non-overlapping layering, the range of the higher layer must be a subset of that of the lower. Therefore, the range of integration may be restricted to the range over which the higher layer is non-zero. But over this range the lower layer is exhausted, or equal to its width. Hence, the product moment of two different layers equals the product of the width of the lower layer and the mean of the higher.^[9]

The 4×4 blue-shaded block at the bottom of Table 3 contains the product moments. Down its diagonal are E[Y²_i], or twice the values of the “Area” row of Table 2. Off the diagonal are the lower-width-and-higher-mean products. The augmenting margin (unshaded) pertains to the loss from ground up, or MX; it contains row and column sums of the blue-shaded block, since

$$\small{ \begin{aligned} E\left[Y_{i} M X\right] & =E\left[Y_{i}\left(\sum_{j=1}^{4} Y_{j}\right)\right]=E\left[\sum_{j=1}^{4} Y_{i} Y_{j}\right]=\sum_{j=1}^{4} E\left[Y_{i} Y_{j}\right] \\ E\left[M X^{2}\right] & =E\left[\left(\sum_{i=1}^{4} Y_{i}\right) M X\right]=E\left[\left(\sum_{i=1}^{4} Y_{i} M X\right)\right] \\ & =\sum_{i=1}^{4} E\left[Y_{i} M X\right]=\sum_{i=1}^{4} \sum_{j=1}^{4} E\left[Y_{i} Y_{j}\right] . \end{aligned} }$$

The soundness of our logic is confirmed inasmuch as E⎣MX²⎦ = 8.000 × 10¹², which is twice the area under the Excess_MX curve, according to Table 1.

Table 3 gives the label ‘Z’ to the 5 × 1 vector whose first element is MX and remaining four are the $Y_i$s. So the augmented product-moment matrix is E[ZZ′], as labeled. The box above it is the outer product of the means, E[Z]E[Z]′. The vector variance, Var[Z] = E[ZZ′] − E[Z]E[Z]′, is shown under the heading “Variance.” The “Std” column contains the square roots of the diagonal elements of the variance matrix, which values agree with those of Tables 1 and 2. When covariance is taken into account, the second moment of the loss is conserved. Finally, removing the standard-deviation scale from the variance matrix results in the “Correlation” matrix. It bears out something else well known to reinsurance actuaries, namely, that layered losses are positively correlated, although the correlation diminishes as the distance between the layers increases.^[10]

6. Conclusion

The mathematics of excess losses is not only beautiful; it is powerful. The excess-loss function impounds all the information of the probability distribution of its loss. Therefore, although from the beginning actu-aries and underwriters have found it convenient for the calculation of the pure premiums of layered losses, it is just as serviceable for the calculation of higher moments, whether the integrals involved be calculated analytically (as done in our example) or approximated numerically. The versatile mixed exponential distribution lessens the difficulty of such calculations.

Appendix

Appendix A

Stieltjes integrals: Watch your step!

The expectation of h(X) is $\sum_{i=1}^{\infty}$h(x_i) ⋅ Prob[X = x_i], if X is discreet, and $\prod_{x=-\infty}^{\infty}$h(x)f_X(x)dx, if X is continuous. But random variables may be mixed, i.e., continuous with discrete steps.^[11] For the sake of generality we can employ the Stieltjes integral [1, p. 12]: E[h(X)] = $\prod_{x=-\infty}^{\infty}$h(x)dF_X(x), where F_X(x) is the cumulative distribution function of X. Of course, if F is differentiable, dF_X(x) = f_X(x)dx, and the Stieltjes integral reverts back to the familiar (Cauchy-Riemann) integral $\prod_{x=-\infty}^{\infty}$h(x)f_X(x)dx. The Stieltjes integral is defined as

$$\int_{x=a}^{b} h(x) d F(x)=\lim _{\max \left\langle\Delta x_{i}\right\rangle \rightarrow 0} \sum_{i=1}^{n} h\left(\xi_{i}\right) \Delta F_{i}$$

where the interval is partitioned as a = x₀ x₁ x_n = b, Δx_i = x_i − $x_{i-1}$, and ΔF_i = F(x_i) − ($F(x_{i-1})$). Each $\xi_{i}$ is arbitrarily chosen from the subinterval x_i−1 ≤ ξ_i ≤ x_i.

If u is continuous over the interval, nothing is problematic about this definition. Now for our purposes X is a non-negative random variable; hence, for x 0, dF_X(x) = 0, and E[h(X)] = $\lim _{\varepsilon \rightarrow 0^{-}} \int_{x=\varepsilon}^{\infty}$h(x)dF_X(x). One is tempted to simplify this to $\int_{x=0}^{\infty}$h(x)dF_X(x) would miss any discrete step at zero, since the cumulative distribution function is continuous from the right. Therefore, the Stieltjes integral $\int_{x=a}^{b}$h(x)dF_x(x) counts probability mass at the upper limit, but not at the lower. This asymmetry ensures that $\int_{x=a}^{b}$h(x)dF_x(x) + $\int_{x=b}^{c}$h(x)dF_x(x) = $\int_{x=a}^{c}$h(x)dF_x(x); otherwise, probability mass at the endpoints might be either ignored or double-counted.

Therefore, the correct formulation for a non-negative random variable X is E[h(X)] = h(0) Prob[X = 0] + $\int_{x=0}^{\infty}$h(x)dF_X(x). From integration by parts, we derive the form for the survival function G_X(x) = 1 − F_X(x), which also is continuous from the right:

$$\begin{aligned} E[h(X)]= & h(0) \ Prob[X=0]+\int_{x=0}^{\infty} h(x) d F_{x}(x) \\ = & h(0) \ Prob[X=0]+\int_{x=\infty}^{\infty} h(x) d G_{x}(x) \\ = & h(0) \ Prob[X=0]+\left.h(x) d G_{x}(x)\right|_{\infty} ^{0} \\ & -\int_{x=\infty}^{\infty} G_{x}(x) d h(x) \\ = & h(0) \ Prob[X=0]+h(0) G_{x}(0)-0 \\ & +\int_{x=0}^{\infty} G_{x}(x) d h(x) . \end{aligned}$$

$$\begin{array}{l} =h(0) \ Prob[X \geq 0]+\int_{x=0}^{\infty} G_{X}(x) d h(x) \\ =h(0)+\int_{x=0}^{\infty} G_{X}(x) d h(x) . \end{array}$$

For this reason we subtitled this appendix “Watch your step!” For the [0, ∞] Stieltjes integrals in this paper do not count probability mass at zero. For two reasons it is easy to overlook this subtlety. First, if Prob[X = 0] = 0, E[h(X)] = $\int_{x=0}^{\infty}$h(x)dF_X(x), although the other form is still E[h(X)] = h(0) + $\int_{x=0}^{\infty}$G_X(x)dh(x). And second, it is common for h(x) to be a positive power of x, in which case h(0) = 0.

Watching one’s step at zero also consistently handles a constant shift in h(x):

$$\begin{aligned} E[h(X)+c]= & E[h(X)]+c \\ = & h(0) \ Prob[X=0]+\int_{x=0}^{\infty} h(x) d F_{X}(x)+c \\ = & h(0) \ Prob[X=0]+\int_{x=0}^{\infty} h(x) d F_{X}(x) \\ & +c\left\{Prob[X=0]+\int_{x=0}^{\infty} d F_{X}(x)\right\} \\ = & (h(0)+c) \ Prob[X=0] \\ & +\int_{x=0}^{\infty}(h(x)+c) d F_{X}(x) . \end{aligned}$$

And

$$\begin{aligned} E[h(X)+c] & =E[h(X)]+c \\ & =h(0)+\int_{x=0}^{\infty} G_{X}(x) d h(x)+c \\ & =(h(0)+c)+\int_{x=0}^{\infty} G_{X}(x) d(h(x)+c) . \end{aligned}$$

Appendix B

Two theorems about reinsurance layers

Here we will give proofs of the two facts which this paper claims to be “well known to reinsurance actuaries”:

1. that the coefficients of variation, CV = Std/E, increase as the layers ascend, and

2. that layered losses are positively correlated, although the correlation diminishes as the distance between the layers increases.

Our proofs will begin with “differential” layers, i.e., to layers whose width is dx. But, as we shall show, one can integrate such layers into layers of any width.

Let X be a non-negative random variable, whose survival function (the complement of the cumulative distribution function) is G_X. The probability that X x is G_X(x); therefore, the probability of a non-zero loss in the interval [x, x + Δx] is G_X(x). In the limit, as Δx → 0⁺, the expected loss in the layer, E[Layer(X; x, x + Δx)], approaches G_X(x)Δx. Defining dY(x) as the portion of X in the differential layer [x, x + dx], we may say that dY(x) ∼ Bernoulli(G_X(x)) ⋅ dx. Accordingly, E[dY(x)] = G_X(x)dx and E⎣(dY(x))²⎦ = G_X(x)(dx)². Arguing as we did in Section 5, we have E[dY(x₁)dY(x₂)] = min(G_X(x₁), G_X(x₂))dx₁dx₂, of which E[(dY(x))²] = G_X(x)(dx)² is a special instance in which x₁ = x₂ = x.¹²

Before we prove the two theorems, it will be instructive to see how a layer can be integrated from differential layers. If Y is the portion of X in layer [a,b], then Y = Layer(X;a,b) = $\int_{x=a}^{b}$dY(x). Hence,

$$E[Y]=E\left[\int_{x=a}^{b} d Y(x)\right]=\int_{x=a}^{b} E[d Y(x)]=\int_{x=a}^{b} G_{X}(x) d x$$

Moreover, the second moment is

$$\begin{array}{l} E\left[Y^{2}\right]=E\left[\left(\int_{x=a}^{b} d Y(x)\right)^{2}\right] \\ =E\left[\int_{x_{1}=a}^{b} d Y\left(x_{1}\right) \int_{x_{2}=a}^{b} d Y\left(x_{2}\right)\right] \\ =E\left[\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{b} d Y\left(x_{2}\right) d Y\left(x_{1}\right)\right] \\ =\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{b} E\left[d Y\left(x_{2}\right) d Y\left(x_{1}\right)\right] \\ =\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{b} \min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right) d x_{2} d x_{1} \\ =\int_{x_{1}=a}^{b}\left\{\int_{x_{2}=a}^{x_{1}} \min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right) d x_{2}\right. \\ \left.+\int_{x_{2}=x_{1}}^{b} \min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right) d x_{2}\right\} d x_{1} \\ =\int_{x_{1}=a}^{b}\left\{\int_{x_{2}=a}^{x_{1}} G_{X}\left(x_{1}\right) d x_{2}+\int_{x_{2}=x_{1}}^{b} G_{X}\left(x_{2}\right) d x_{2}\right\} d x_{1} \\ =\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{x_{1}} G_{X}\left(x_{1}\right) d x_{2} d x_{1}+\int_{x_{1}=a}^{b} \int_{x_{2}=x_{1}}^{b} G_{X}\left(x_{2}\right) d x_{2} d x_{1} \\ =\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{x_{1}} G_{X}\left(x_{1}\right) d x_{2} d x_{1}+\int_{x_{2}=a}^{b} \int_{x_{1}=a}^{x_{2}} G_{X}\left(x_{2}\right) d x_{1} d x_{2} \\ =2 \int_{x_{1}=a}^{b} \int_{x_{2}=a}^{x_{1}} G_{X}\left(x_{1}\right) d x_{2} d x_{1} . \\ \end{array}$$

$$\begin{array}{l} =2 \int_{x=a}^{b} G_{X}(x)(x-a) d x \\ =\int_{x=a}^{b} G_{X}(x) d(x-a)^{2} . \end{array}$$

For ease of understanding, the derivation proceeded in many small steps; nevertheless, line seven deserves an explanation. Since G_X is non-increasing, min(G_X(x₁), G_X(x₂)) = G_X(max(x₁, x₂)). So by dividing the inner integral into the two regions, we can identify the minimum.¹³ The reproduction of the moments of Y confirms the legitimacy of the formula $Y=\int_{x=a}^{b} d Y(x)$

One more notion is required for our proofs, a notion which we will call the coefficient of covariance. It is the covariance between two random variables whose means have been normalized to unity, i.e.,

$$\begin{aligned} \operatorname{CoefCov}[X, Y] & =\operatorname{Cov}\left[\frac{X}{E[X]}, \frac{Y}{E[Y]}\right] \\ & =\frac{\operatorname{Cov}[X, Y]}{E[X] E[Y]}=\frac{E[X Y]-E[X] E[Y]}{E[X] E[Y]} \\ & =\frac{E[X Y]}{E[X] E[Y]}-1 . \end{aligned}$$

Of course, CoefCov[X, X] = CV²[X]. The coefficient of covariance between two differential layers is

$$\begin{array}{l} \operatorname{CoefCov}\left[d Y\left(x_{1}\right), d Y\left(x_{2}\right)\right]=\frac{E\left[d Y\left(x_{1}\right) d Y\left(x_{2}\right)\right]}{E\left[d Y\left(x_{1}\right)\right] E\left[d Y\left(x_{2}\right)\right]}-1 \\ \quad =\frac{\min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right) d x_{1} d x_{2}}{G_{X}\left(x_{1}\right) d x_{1} G_{X}\left(x_{2}\right) d x_{2}}-1 \\ \quad =\frac{\min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right)}{G_{X}\left(x_{1}\right) G_{X}\left(x_{2}\right)}-1 \\ \quad =\frac{\min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right)}{\min \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right) \cdot \max \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right)}-1 \\ \quad =\frac{1}{\max \left(G_{X}\left(x_{1}\right), G_{X}\left(x_{2}\right)\right)}-1 \\ \quad =\frac{1}{G_{X}\left(\min \left(x_{1}, x_{2}\right)\right)}-1 . \end{array}$$

This coefficient is well defined when G_X(x₁) and G_X(x₂) are non-zero; loss in the differential layers must be possible. Furthermore, CV²[dY(x)] = CoefCov[dY(x), dY(x)] $=\frac{1}{G_{X}(x)}-1$. Due to the properties of G_X, CV²[dY(x)] is a non-decreasing function in x.

With this preparation, the first fact is easily proven. If Y = Layer(X; a,b) = $\int_{x=a}^{b}$dY(x) and G_X(a) ≥ G_X(b) 0, then

$$\begin{array}{l} C V^{2}[Y]=\operatorname{Var}[Y] / E[Y]^{2} \\ =\frac{\int_{x_{1}=a} \int_{x_{2}=a}^{b} \operatorname{Cov}\left[d Y\left(x_{1}\right), d Y\left(x_{2}\right)\right]}{\left(\int_{x=a}^{b} E[d Y(x)]\right)^{2}} \\ =\frac{\int_{x_{1}=a x_{2}=a}^{b} E\left[d Y\left(x_{1}\right)\right] E\left[d Y\left(x_{2}\right)\right] \operatorname{CoefCov}\left[d Y\left(x_{1}\right), d Y\left(x_{2}\right)\right]}{\int_{x_{1}=a}^{b} \int_{x_{2}=a}^{b} E\left[d Y\left(x_{1}\right)\right] E\left[d Y\left(x_{2}\right)\right]} . \end{array}$$

Hence, CV²[Y] is a weighted average (weighted over two dimensions) of the coefficients of the layer’s covariances. And since the weights are non-negative, the weighted average must be bounded by the minimum and maximum coefficients, which are at the endpoints:

$$\begin{aligned} C V^{2}[d Y(a)]= & \ CoefCov[d Y(a), d Y(a)] \leq C V^{2}[Y] \\ & \leq \ CoefCov[d Y(b), d Y(b)] \\ & =C V^{2}[d Y(b)] . \end{aligned}$$

Therefore, of two layers, [a, b] and [c, d], where a b ≤ c d, the CV² of the lower will be less than or equal to that of the higher. And if probability is consumed anywhere in these layers,¹⁴ the inequality will be strict. Even if the two layers overlap (i.e., a ≤ c b and b ≤ d, but not both a = c and b = d), one can consider three intervals, the middle interval being the overlap. Then, as above, CV²(A) ≤ CV²(B) ≤ CV²(C). Because the unions involve weighted-averaging, CV²(A) ≤ CV² (A ∪ B) ≤ CV² (B) ≤ CV² (B ∪ C) ≤ CV²(C). Therefore, the CV² of a higher layer is greater than or equal to that of a lower layer, even if there is some overlap; the inequality is strict, if probability is consumed. Finally, since CV ≥ 0, the inequalities are as valid for CV as for CV². Note that the widths of the layers do not need to be equal.

Second, as to correlation, let Y₁ = Layer(X;a,b) = $\int_{x=a}^{b}$dY(x) and Y₂ = Layer(X;c,d) = $\int_{x=c}^{d}$dY(x), for a b ≤ c d. Therefore, we know that the minimum G_X will be in the higher interval [c, d].

Under these conditions,

$$\small{ \begin{aligned} & \operatorname{Cov}\left[Y_1, Y_2\right]=E\left[Y_1 Y_2\right]-E\left[Y_1\right] E\left[Y_2\right] \\ & \quad =\int_{x_1=a}^b \int_{x_2=c}^d E\left[d Y\left(x_1\right) d Y\left(x_2\right)\right]-E\left[Y_1\right] E\left[Y_2\right] \\ & \quad =\int_{x_1=a}^b \int_{x_2=c}^d \min \left(G_X\left(x_1\right), G_X\left(x_2\right)\right) d x_2 d x_1-E\left[Y_1\right] E\left[Y_2\right] \\ & \quad =\int_{x_1=a}^b \int_{x_2=c}^d G_X\left(x_2\right) d x_2 d x_1-E\left[Y_1\right] E\left[Y_2\right] . \\ & \quad \quad =(b-a) \int_{x_2=c}^d G_X\left(x_2\right) d x_2-E\left[Y_1\right] E\left[Y_2\right] \\ & \quad \quad =(b-a) E\left[Y_2\right]-E\left[Y_1\right] E\left[Y_2\right] \\ & \quad \quad =\left(b-a-E\left[Y_1\right]\right) E\left[Y_2\right] . \end{aligned} }$$

Because E[Y₁] = $\int_{x=a}^{b}$G_X(x)dx ≤ $\int_{x=a}^{b}$₁⋅ dx = b − a, b − a − E[Y₁] ≥ 0; hence, Cov[Y₁, Y₂] ≥ 0. So the correlation coefficient between the portions of X in the two layers is

$$\begin{aligned} \operatorname{Corr}\left[Y_{1}, Y_{2}\right] & =\frac{\operatorname{Cov}\left[Y_{1}, Y_{2}\right]}{\sigma_{Y_{1}} \sigma_{Y_{2}}} \\ & =\frac{b-a-E\left[Y_{1}\right]}{\sigma_{Y_{1}}} \cdot \frac{E\left[Y_{2}\right]}{\sigma_{Y_{2}}} \\ & =\frac{\left(\frac{b-a-E\left[Y_{1}\right]}{\sigma_{Y_{1}}}\right)}{C V\left[Y_{2}\right]} . \end{aligned}$$

Now consider shifting [c, d] to the right, i.e., to [c + ξ, d + ξ], where ξ ≥ 0. And let Y₂(ξ) = Layer(X; c + ξ, d + ξ) = $\int_{x=c+\xi}^{d+\xi}$dY(x). From the first proof we know that CV[Y₂(ξ)] is non-decreasing. Since the numerator of Corr is constant, Corr[Y₁, Y₂(ξ)] is non-increasing; strictly decreasing if probability is consumed. Therefore, as the retention of the upper layer so moves away from that of the lower as to consume probability, the correlation decreases. This implies that in the absence of compensating risk premiums, a reinsurer should not underwrite neighboring layers.