Mean Severity Conditional on Exceeding a Threshold

Why Condition on Exceeding a Threshold?

An insurer may want to know the average size of claims that exceed a reporting level, attachment point, or other specified amount. This is a different question from asking for the average size of all claims: observations at or below the threshold are excluded, and the remaining claim amounts have a conditional distribution.

The distinction between exceeding and meeting or exceeding matters whenever severity has a point mass. With the condition $X>d$, a claim exactly equal to $d$ is excluded. The formulas below use that strict inequality throughout.

Definition: Let $X$ be a nonnegative claim severity, let $d\geq 0$ be a threshold, and suppose $P(X>d)>0$. The conditional severity given that the threshold is exceeded is the distribution of $X$ under the condition $X>d$. Its conditional CDF is $P(X\leq x\mid X>d)$, and its conditional mean is $E[X\mid X>d]$, provided that this expectation is finite.

The Conditional Distribution

Write $F_X(x)=P(X\leq x)$ and $S_X(x)=P(X>x)=1-F_X(x)$. Since the event $X>d$ excludes all values at or below $d$, its probability is $S_X(d)$. For $x>d$, the event $d<X\leq x$ has probability $F_X(x)-F_X(d)$. Dividing by the probability of the conditioning event gives the conditional CDF.

F_{X\mid X>d}(x)= \begin{cases} 0, & x\leq d,\\[4pt] \dfrac{F_X(x)-F_X(d)}{S_X(d)}, & x>d, \end{cases} \qquad S_X(d)>0.

The numerator subtracts $F_X(d)$, rather than $F_X(d-)$, because $F_X(d)$ includes any probability mass at $d$. That mass is not part of the event $X>d$. If $X$ is continuous at $d$, then $P(X=d)=0$, so the distinction between strict and non-strict exceedance does not affect the probability. For a discrete or mixed model, it can change the answer.

If $X$ has a probability mass function, each original mass above the threshold is rescaled by the same conditioning probability. For a continuous model with density $f_X$, the conditional density on $x>d$ is $f_X(x)/S_X(d)$. A mixed distribution retains both kinds of probability: its masses above $d$ are rescaled, and its continuous density above $d$ is rescaled.

P(X=x\mid X>d)=\frac{P(X=x)}{S_X(d)} \quad (x>d), \qquad f_{X\mid X>d}(x)=\frac{f_X(x)}{S_X(d)} \quad (x>d)

The density expression applies to the continuous part of the distribution; it does not represent any point masses. In either case, the conditioning probability must be positive. If $S_X(d)=0$, the event being conditioned on has probability zero, and these ordinary conditional-distribution formulas do not apply.

Conditional Mean: Two Equivalent Calculations

The most direct calculation is to retain only the part of the first moment contributed by values above $d$, then divide by the probability of being above $d$. This works for discrete, continuous, and mixed severities, with the expectation evaluated according to the model.

E[X\mid X>d] = \frac{E[X\,\mathbf{1}_{\{X>d\}}]}{P(X>d)} = \frac{E[X\,\mathbf{1}_{\{X>d\}}]}{S_X(d)}.

For a nonnegative severity with finite mean, there is also a useful survival-function expression. Conditional on exceeding $d$, the excess amount is $X-d$. For any nonnegative random variable $W$, its expectation equals the integral of its survival probabilities: $E[W]=\int_0^\infty P(W>t)\,dt$. Apply this identity to $W=X-d$ under the condition $X>d$.

Result: If $X$ is nonnegative with finite mean and $S_X(d)>0$, then $$ E[X\mid X>d] = d+\frac{\displaystyle\int_d^\infty S_X(x)\,dx}{S_X(d)}. $$ The integral is the unconditional expected excess above $d$, $E[(X-d)\mathbf{1}_{\{X>d\}}]$.

To see why the result holds, for $t\geq 0$ the event $X-d>t$, together with $X>d$, is simply $X>d+t$. Thus

P(X-d>t\mid X>d) = \frac{P(X>d+t)}{P(X>d)} = \frac{S_X(d+t)}{S_X(d)}.

Integrating over $t\geq 0$ gives $E[X-d\mid X>d]=\int_0^\infty S_X(d+t)/S_X(d)\,dt$. Substituting $x=d+t$ makes the numerator $\int_d^\infty S_X(x)\,dx$. Finally, $X=d+(X-d)$ under the conditioning event, so adding $d$ gives the stated conditional mean. The formula requires a finite integral for a finite conditional mean; if that integral diverges, the conditional mean is infinite.

Worked Example: A Discrete Severity Above a Threshold

A hypothetical equipment insurer models a claim amount $X$, in dollars, with the following distribution: $P(X=100)=0.20$, $P(X=300)=0.30$, $P(X=700)=0.30$, and $P(X=1{,}200)=0.20$. Find the conditional distribution and average claim amount given $X>500$.

Method. Use the discrete conditional probabilities and divide each included probability by $P(X>500)$. The threshold is strict, so only the amounts greater than $500 are retained.

First calculate the conditioning probability:

$$ P(X>500)=P(X=700)+P(X=1{,}200)=0.30+0.20=0.50. $$

The conditional probabilities are therefore

P(X=700\mid X>500)=\frac{0.30}{0.50}=0.60, \qquad P(X=1{,}200\mid X>500)=\frac{0.20}{0.50}=0.40.

They sum to $0.60+0.40=1.00$, as a conditional distribution must. The conditional mean is

E[X\mid X>500] =(700)(0.60)+(1{,}200)(0.40) =420+480 =\$900.

As a check using the truncated-expectation formula, the numerator is $(700)(0.30)+(1{,}200)(0.20)=210+240=450$, so $450/0.50=\$900$. A common mistake is to include the $300 claim because it is close to the threshold. It is excluded: $300$ is not greater than $500$.

Worked Example: A Continuous Severity Model

For a hypothetical loss model, severity $X$, in thousands of dollars, has density $f_X(x)=x/18$ for $0<x<6$, and zero elsewhere. Find the conditional CDF and conditional mean given $X>3$.

Method. Integrate the density to obtain the original CDF, then apply the conditional CDF formula. The stated density is valid because $\int_0^6 x/18\,dx=36/36=1$.

For $0\leq x\leq 6$, integration gives $F_X(x)=x^2/36$. In particular, $F_X(3)=9/36=1/4$, so

S_X(3)=1-F_X(3)=1-\frac14=\frac34.

For $3<x<6$, the conditional CDF is

F_{X\mid X>3}(x) =\frac{x^2/36-9/36}{3/4} =\frac{x^2-9}{27}.

It is zero for $x\leq3$, and equals one for $x\geq6$. Differentiating on $3<x<6$ gives the conditional density $2x/27$. Its integral over the interval is $(36-9)/27=1$, confirming the normalization.

For the mean, first calculate the truncated first moment:

E[X\mathbf{1}_{\{X>3\}}] =\int_3^6 x\frac{x}{18}\,dx =\left[\frac{x^3}{54}\right]_3^6 =\frac{216-27}{54} =\frac{7}{2}.

Divide by $S_X(3)=3/4$:

E[X\mid X>3] =\frac{7/2}{3/4} =\frac{14}{3} \approx 4.667

Thus the average conditional severity is approximately $4,666.67 when expressed in dollars, or $4.667$ thousand dollars. The answer exceeds the threshold of $3$ thousand dollars, as it must. The common mistake of dividing the truncated moment by one would give $3.5$, which is the contribution to the unconditional mean from losses above the threshold, not the conditional mean.

Worked Example: A Mixed Model With an Atom at the Threshold

A hypothetical claim amount $X$, in thousands of dollars, has point masses $P(X=2)=0.20$, $P(X=4)=0.20$, and $P(X=8)=0.20$. The remaining probability $0.40$ is continuous, with density $0.40(2x/100)$ for $0<x<10$. The density integrates to $0.40$, so the total probability is $0.60+0.40=1$. Find $E[X\mid X>4]$.

Method. Add the probability and first-moment contributions from the atom and continuous part strictly above $4$, then divide the latter total by the exceedance probability. The atom at $4$ is excluded.

For the continuous component, the probability above $4$ is

\int_4^{10}0.40\frac{2x}{100}\,dx =0.40\frac{100-16}{100} =0.40(0.84) =0.336.

The atom at $8$ contributes another $0.20$ to the exceedance probability. Therefore

$$ P(X>4)=0.336+0.20=0.536. $$

The continuous part's contribution to the truncated first moment is

\int_4^{10}x\left(0.40\frac{2x}{100}\right)\,dx =\frac{0.008}{3}(10^3-4^3) =\frac{0.008}{3}(936) =2.496.

The atom at $8$ contributes $(8)(0.20)=1.600$. Hence the total truncated first moment is $2.496+1.600=4.096$, and

E[X\mid X>4]=\frac{4.096}{0.536} \approx 7.6418

The conditional mean is $7.64$ thousand dollars, or $7,641.79, rounded to the nearest cent. As a check, the conditioning probability is $0.536$, not $0.736$: adding the atom at $4$ would incorrectly change the condition from $X>4$ to $X\geq4$. The computed mean is also above $4$, consistent with the event being conditioned on.

Exam Traps and Practical Checks

A reliable calculation begins by stating the event precisely. Then identify its probability and the values or portions of the model that belong to it. This prevents an error in the conditioning event from carrying through both the conditional distribution and its mean.

Including the threshold atom. For $X>d$, an atom at $d$ is excluded. Use $P(X>d)=1-F_X(d)$, not $1-F_X(d-)$. The latter is $P(X\geq d)$.
Forgetting to renormalize. The truncated first moment $E[X\mathbf{1}_{\{X>d\}}]$ is not the conditional mean. Divide it by $S_X(d)$.
Using a continuous density formula for a mixed model. A density describes only the continuous component. Add any point-mass contributions above the threshold separately.
Ignoring feasibility checks. Confirm $S_X(d)>0$, that conditional probabilities sum to one, and that a finite conditional mean is at least $d$. In fact, because $X>d$ on the conditioning event, the conditional mean is strictly greater than $d$ when finite.

These conditional calculations are useful whenever an average is reported for claims selected by a threshold. They also clarify why an average above a threshold need not equal the overall average: it is calculated from a different population of losses.

Key takeaway: For strict exceedance, condition on $S_X(d)=P(X>d)$. The conditional CDF is $[F_X(x)-F_X(d)]/S_X(d)$ for $x>d$, and the conditional mean is the truncated first moment divided by $S_X(d)$. Treat atoms at the threshold separately.

Check Your Understanding

For each problem, pay attention to the strict exceedance condition and show enough intermediate work to verify the result.

A hypothetical claim severity takes values $0, $200, and $600 with probabilities $0.25$, $0.45$, and $0.30$, respectively. Find $P(X>200)$ and $E[X\mid X>200]$. (Answer: $P(X>200)=0.30$; $E[X\mid X>200]=\$600$.)
A severity in thousands of dollars has density $f_X(x)=x/32$ for $0<x<8$, and zero elsewhere. Find $P(X>4)$ and $E[X\mid X>4]$. (Answer: $P(X>4)=0.75$; $E[X\mid X>4]=56/9\approx 6.222$ thousand dollars.)
A mixed severity has $P(X=3)=0.25$, $P(X=6)=0.25$, and continuous density $0.50/4$ for $6<x<10$, with no other probability. Find $P(X>6)$ and $E[X\mid X>6]$. (Answer: $P(X>6)=0.50$; $E[X\mid X>6]=8$.)
Suppose $S_X(5)=0.40$ and $\int_5^\infty S_X(x)\,dx=2.8$, with $X$ nonnegative and finite-mean. Calculate the mean severity conditional on $X>5$. (Answer: $12$.)
A discrete severity has probabilities $P(X=100)=0.10$, $P(X=400)=0.20$, $P(X=900)=0.50$, and $P(X=1{,}400)=0.20$. Find the conditional probabilities at each possible value given $X>400$, and find the conditional mean. (Answer: $P(X=900\mid X>400)=0.50/0.70=5/7$, $P(X=1{,}400\mid X>400)=0.20/0.70=2/7$; $E[X\mid X>400]=\$1{,}042.86$, rounded to the nearest cent.)

What You'll Learn