Expected Loss for Continuous Severity

Introduction

For a continuous claim severity, the mean is not found by averaging a list of possible losses: a continuous model assigns probability across intervals rather than to individual loss amounts. Instead, each possible loss is weighted by how much probability density lies near that amount. The resulting integral gives the expected loss, provided the integral is finite.

The calculation has two checks that should come before the arithmetic: confirm that the density is a valid probability model, and identify its support. A correct-looking integral over the wrong range, or an integral based on a density that does not integrate to one, does not give the model’s expected severity.

Definition: Let $X$ be a nonnegative continuous claim severity with probability density function $f_X(x)$ on support $S$. Its expected severity is $$ E[X]=\int_S x f_X(x)\,dx, $$ provided this integral is finite. If the integral diverges, the expected severity is infinite, not a finite number. A density must satisfy $f_X(x)\geq 0$ and $\int_S f_X(x)\,dx=1$.

Setting Up the Expectation Integral

The factor $x$ in the integrand is the loss amount; $f_X(x)\,dx$ represents the probability assigned to a very small interval near $x$. Thus, $x f_X(x)\,dx$ is that interval’s approximate contribution to the mean. Integrating over the support adds these contributions across all possible losses.

The support determines the bounds. If $X$ can take values only from $a$ to $b$, integrate from $a$ to $b$. If the support is unbounded, such as $x\geq 0$, the expectation is an improper integral: evaluate it to a finite upper bound and then take the limit. The limit must be checked rather than assumed to exist.

Model check: Before calculating $E[X]$, identify the support $S$, confirm that the density is nonnegative, and verify $\int_S f_X(x)\,dx=1$. Then evaluate $\int_S x f_X(x)\,dx$. For a nonnegative severity, this expectation can be infinite even when the density is valid.

A useful justification for the formula comes from approximating the range of possible losses by narrow intervals. In an interval of width $\Delta x$ near $x_j$, the probability is approximately $f_X(x_j)\Delta x$. The contribution of that interval to the average loss is therefore approximately $x_j f_X(x_j)\Delta x$. Adding the contributions over a partition gives a sum, and as the interval widths tend to zero, that sum becomes the integral:

E[X]\ \approx\ \sum_j x_j f_X(x_j)\Delta x \quad\longrightarrow\quad \int_S x f_X(x)\,dx.

This argument explains why the density itself is not averaged as though it were a list of probabilities: it must be multiplied by the loss amount, and the contributions must be integrated over the support.

Worked Calculations

Worked Example: A Uniform Loss on a Bounded Interval

A hypothetical policy has a claim severity $X$, measured in thousands of dollars, with density $f_X(x)=1/4$ for $1\leq x\leq 5$, and zero elsewhere. Verify the density and calculate $E[X]$.

Method and support. The support is $[1,5]$. The density is nonnegative there. Its total probability is

\int_1^5 \frac14\,dx =\frac14(5-1)=1.

So the proposed function is a valid density. Apply the continuous expectation formula on $[1,5]$:

E[X]=\int_1^5 x\left(\frac14\right)\,dx =\frac14\left[\frac{x^2}{2}\right]_1^5 =\frac18(25-1) =3.

The mean severity is $3$ thousand dollars, or $\$3{,}000$. This lies within the model’s support, as expected. Common mistake avoided: integrating $x/4$ over a range other than the stated support would calculate a different quantity.

Worked Example: A Density That Increases Across Its Support

For a hypothetical warranty portfolio, suppose severity $X$ is measured in hundreds of dollars and has density $f_X(x)=x/8$ for $0\leq x\leq 4$, and zero elsewhere. Verify the density and find the mean severity.

Method and support. The density is nonnegative on $[0,4]$. Its integral is

\int_0^4 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^4 =\frac{16}{16}-0=1.

The density is normalized. Weight each loss amount by its density:

E[X]=\int_0^4 x\left(\frac{x}{8}\right)\,dx =\frac18\int_0^4 x^2\,dx =\frac18\left[\frac{x^3}{3}\right]_0^4 =\frac18\left(\frac{64}{3}\right) =\frac83.

Thus the mean is $8/3$ hundreds of dollars, or $\$266.67$ to the nearest cent. The result is between 0 and the maximum possible severity of 4 hundreds of dollars. Common mistake avoided: integrating the density alone gives total probability, not the mean; the loss factor $x$ is essential.

Worked Example: Checking Convergence on an Unbounded Support

A hypothetical claim severity $X$, measured in thousands of dollars, has density $f_X(x)=2/(x+1)^3$ for $x\geq 0$, and zero elsewhere. Verify the density and determine whether the mean is finite.

Method and support. The support is unbounded, so first check normalization using an improper integral:

\int_0^\infty \frac{2}{(x+1)^3}\,dx =\lim_{b\to\infty}\left[-\frac{1}{(x+1)^2}\right]_0^b =\lim_{b\to\infty}\left(1-\frac{1}{(b+1)^2}\right) =1.

The function is nonnegative and integrates to one. For the mean, substitute $u=x+1$, so $x=u-1$, and the bounds become $1$ to $\infty$:

E[X] =\int_0^\infty \frac{2x}{(x+1)^3}\,dx =2\int_1^\infty \left(\frac{1}{u^2}-\frac{1}{u^3}\right)\,du =2\left(1-\frac12\right) =1.

The integral converges, and the mean severity is $1$ thousand dollars, or $\$1{,}000$. Common mistake avoided: a density on an unbounded support does not automatically have an infinite mean; the expectation integral must be evaluated to determine convergence.

Worked Example: A Valid Density With an Infinite Mean

Suppose a hypothetical severity $X$, measured in thousands of dollars, has density $f_X(x)=1/(x+1)^2$ for $x\geq 0$, and zero elsewhere. Check the density and decide whether $E[X]$ is finite.

Method and support. The function is nonnegative. Its total probability is

\int_0^\infty \frac{1}{(x+1)^2}\,dx =\lim_{b\to\infty}\left[-\frac{1}{x+1}\right]_0^b =\lim_{b\to\infty}\left(1-\frac{1}{b+1}\right) =1.

It is therefore a valid density. For the mean, integrate to $b$ and use $u=x+1$:

\int_0^b \frac{x}{(x+1)^2}\,dx =\int_1^{b+1}\left(\frac1u-\frac1{u^2}\right)\,du =\ln(b+1)+\frac{1}{b+1}-1.

As $b\to\infty$, the logarithm tends to infinity, so the expectation integral diverges. The density is valid, but $E[X]=\infty$; there is no finite mean severity under this model. Common mistake avoided: a normalized density does not guarantee a finite expected loss.

Practical Checks and Exam Traps

The calculation is short once the model is set up correctly, but several errors can invalidate an otherwise correct integration. In particular, check the density before using it. Consider the claim that $f_X(x)=2/21$ on $2\leq x\leq 5$ is a density. Its integral is

\int_2^5 \frac{2}{21}\,dx =\frac{2}{21}(5-2) =\frac{2}{7}\ne 1.

This function does not define a probability density on that support, so integrating $x f_X(x)$ cannot produce the expected severity of the claimed model. If the intended model is uniform on $[2,5]$, its density is $1/3$, since $(1/3)(5-2)=1$. Its mean is then

E[X]=\int_2^5 \frac{x}{3}\,dx =\frac13\left[\frac{x^2}{2}\right]_2^5 =\frac{25-4}{6} =3.5.

That is $3.5$ in the stated loss units. This illustrates why normalization must be verified before calculating a mean.

Other practical checks are equally important:

Use the support, not just convenient bounds. A density is zero outside its support, and the expectation integral must cover the possible loss values.
Keep the loss factor. $\int_S f_X(x)\,dx=1$ checks probability; it is not the expectation. The expectation integrand is $x f_X(x)$.
Check improper integrals by taking a limit. Finding an antiderivative is not enough; the limit at an infinite endpoint determines whether the mean is finite.
Carry the stated units through the answer. If $x$ is measured in thousands of dollars, the computed expectation is also in thousands of dollars.
Check plausibility without treating it as proof. For bounded support $[a,b]$, a finite mean must lie between $a$ and $b$. For unbounded support, this check alone does not establish convergence.

These steps are useful in severity problems aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus: they separate a valid model from an invalid one and a finite expected loss from a divergent integral.

Key takeaway: For a continuous severity, calculate $E[X]$ by integrating $x f_X(x)$ over the full support. Verify that $f_X$ is a probability density first, and, for unbounded support, prove that the expectation integral converges before reporting a finite mean.

Check Your Understanding

For each problem, check the density and its support before evaluating the expectation.

A hypothetical severity $X$, in thousands of dollars, is uniform with density $1/6$ on $[3,9]$ and zero elsewhere. Find $E[X]$. (Answer: $6$ thousand dollars.)
A severity $X$, in hundreds of dollars, has density $c(2x+1)$ on $0\leq x\leq 1$, and zero elsewhere. Find $c$, verify the density, and calculate $E[X]$. (Answer: $c=1/2$, $E[X]=7/12$ hundred dollars.)
A hypothetical severity $X$, in thousands of dollars, has density $3x^2/8$ on $0\leq x\leq 2$, and zero elsewhere. Verify it and find the mean. (Answer: $E[X]=1.5$ thousand dollars.)
A hypothetical severity has density $3/(x+1)^4$ for $x\geq 0$, and zero elsewhere. Verify it and determine whether the expected severity is finite. (Answer: The density integrates to 1 and $E[X]=1/2$ in the stated loss units.)
A severity $X$ has density $8/x^3$ for $x\geq 2$, and zero elsewhere. Check normalization and calculate its mean, accounting for the unbounded support. (Answer: The density integrates to 1 and $E[X]=4$ in the stated loss units.)

What You'll Learn