Deriving Exponential Moments And Memorylessness

Why Moments and Memorylessness Matter

The exponential severity model is useful not only because its probabilities are easy to calculate. Its moments can be derived directly from its density, and its tail has a distinctive property: once a loss has exceeded a threshold, the distribution of the additional amount does not depend on how high that threshold was. This is called memorylessness.

For actuarial calculations, keep the rate parameter and the units of loss explicit. If $X$ is measured in dollars, its exponential rate is measured per dollar. A larger rate means a smaller average loss. The examples below use hypothetical losses and clearly state the model assumptions.

Definition: A nonnegative loss $X$ has an exponential distribution with rate $\lambda>0$ if its density is $f_X(x)=\lambda e^{-\lambda x}$ for $x\geq 0$, and $f_X(x)=0$ for $x<0$. Its distribution and survival functions are $F_X(x)=1-e^{-\lambda x}$ and $S_X(x)=e^{-\lambda x}$ for $x\geq 0$. The rate $\lambda$ has units reciprocal to those of $X$.

The survival function gives $P(X>x)=S_X(x)$. Because this model is continuous, whether a threshold is written as $X>x$ or $X\geq x$ does not change the probability. The formulas in this tutorial assume an exponential model with a fixed, positive rate; they do not automatically apply to other severity distributions.

Deriving the Exponential Moments

For a nonnegative continuous loss, the $n$th raw moment is obtained by integrating $x^n$ against the density. In the exponential model, integration by parts gives a simple recurrence. This is a direct derivation from the density, rather than a formula inferred from a moment generating function.

E[X^n]=\int_0^\infty x^n\lambda e^{-\lambda x}\,dx

For $n\geq 1$, take $u=x^n$ and $dv=\lambda e^{-\lambda x}\,dx$. Then $du=nx^{n-1}\,dx$ and $v=-e^{-\lambda x}$. Integration by parts yields

E[X^n]=\left[-x^n e^{-\lambda x}\right]_0^\infty +n\int_0^\infty x^{n-1}e^{-\lambda x}\,dx =\frac{n}{\lambda}E[X^{n-1}].

The boundary term is zero: at $x=0$, $x^n=0$, and as $x$ tends to infinity, exponential decay dominates the polynomial factor $x^n$. Since $E[X^0]=1$, applying the recurrence repeatedly gives every nonnegative integer raw moment.

E[X^n]=\frac{n!}{\lambda^n},\qquad n=0,1,2,\ldots

In particular, the first raw moment is the mean and the second raw moment is used to find the variance.

E[X]=\frac{1}{\lambda},\qquad E[X^2]=\frac{2}{\lambda^2},\qquad \operatorname{Var}(X)=E[X^2]-\big(E[X]\big)^2=\frac{1}{\lambda^2}.

Worked Example: Deriving Moments for a Hypothetical Loss

A hypothetical portfolio models an individual claim amount $X$, in dollars, as exponential with rate $\lambda=1/400$ per dollar. Use integration by parts to find $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$. State the units and check the variance.

Method and assumptions. Use the exponential density $f_X(x)=(1/400)e^{-x/400}$ for $x\geq 0$. The rate is per dollar, so moments of order $n$ have units of dollars to the $n$th power.

For the mean, integration by parts, with $u=x$ and $dv=(1/400)e^{-x/400}\,dx$, gives

E[X]=\int_0^\infty x\frac{1}{400}e^{-x/400}\,dx =\left[-xe^{-x/400}\right]_0^\infty +\int_0^\infty e^{-x/400}\,dx =0+400=400.

Thus $E[X]=\$400$. Applying integration by parts to the second moment gives $E[X^2]=(2/\lambda)E[X]$. Substituting $\lambda=1/400$ and $E[X]=400$,

E[X^2]=\frac{2}{1/400}(400)=800(400)=320{,}000\text{ dollars}^2.

The variance is the second raw moment minus the square of the mean:

\operatorname{Var}(X)=320{,}000-400^2 =320{,}000-160{,}000=160{,}000\text{ dollars}^2.

Check. The general formulas give $1/\lambda=400$ dollars and $1/\lambda^2=160{,}000$ dollars squared, matching the integrations. The standard deviation is $\$400$, since $\sqrt{160{,}000}=400$. A common mistake is to report $E[X^2]=320{,}000$ as the variance; subtracting $(E[X])^2$ is essential.

Memorylessness: A Conditional Tail Calculation

Suppose a loss has already been observed to exceed $s$, where $s\geq 0$. The probability that it exceeds an additional amount $t\geq 0$ is calculated by dividing the probability of exceeding both thresholds by the probability of exceeding $s$. Since $s+t\geq s$, the event $X>s+t$ is contained in $X>s$.

P(X>s+t\mid X>s) =\frac{P(X>s+t)}{P(X>s)} =\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} =e^{-\lambda t} =P(X>t).

The conditional probability depends on the additional amount $t$, but not on the amount $s$ already exceeded. Equivalently, conditional on $X>s$, the excess $X-s$ has the same exponential distribution with rate $\lambda$ as the original loss. This result follows from the exponential survival function; it is not an assumption that holds for every severity model.

Theorem: If $X$ is exponential with rate $\lambda>0$, then for $s,t\geq 0$, $P(X>s+t\mid X>s)=e^{-\lambda t}$. The conditional probability of exceeding a further amount is the unconditional survival probability for that further amount.

Worked Example: Conditional Exceedance After a Deductible Threshold

A hypothetical insurer models a claim amount $X$, in dollars, as exponential with rate $\lambda=1/800$ per dollar. Given that a claim exceeds $\$400$, calculate the probability it exceeds $\$1{,}000$. Name the method and identify the common mistake to avoid.

Method and assumptions. Use the exponential survival function and condition on $X>400$. The claim must exceed an additional $1{,}000-400=\$600$ beyond the known threshold.

First calculate the two unconditional tail probabilities needed for the conditional ratio:

P(X>1{,}000)=e^{-1{,}000/800}=e^{-1.25}\approx 0.2865, \qquad P(X>400)=e^{-400/800}=e^{-0.5}\approx 0.6065.

Therefore,

P(X>1{,}000\mid X>400) =\frac{e^{-1.25}}{e^{-0.5}} =e^{-0.75} \approx 0.4724.

Equivalently, memorylessness gives $e^{-(1{,}000-400)/800}=e^{-0.75}\approx 0.4724$. The probability is reported to four decimal places. The common mistake is to use $e^{-1.25}\approx0.2865$, which is the unconditional probability of exceeding $\$1{,}000$, not the probability conditional on already exceeding $\$400$. Check: the conditional probability exceeds the unconditional probability, as expected when the condition $X>400$ is already known.

Conditional Probabilities for an Interval

Memorylessness also simplifies probabilities of an excess falling within an interval. For $0\leq a<b$, given $X>s$, the event $s+a<X\leq s+b$ is equivalent to $a<X-s\leq b$. Its probability is the difference between two conditional tail probabilities.

P(s+a<X\leq s+b\mid X>s) =P(X-s>a\mid X>s)-P(X-s>b\mid X>s) =e^{-\lambda a}-e^{-\lambda b}.

The subtraction is necessary: the first tail includes every excess above $a$, including excesses above $b$. Subtracting the second tail leaves only the interval between $a$ and $b$. The same result can be obtained directly by dividing the unconditional interval probability by $P(X>s)$.

Worked Example: An Excess Within a Specified Range

A hypothetical loss $X$, in dollars, follows an exponential distribution with rate $\lambda=0.001$ per dollar. Given that the loss exceeds $\$700$, find the probability that it is greater than $\$1{,}200$ and no more than $\$2{,}500$. Give the answer to four decimal places.

Method and assumptions. The conditional event corresponds to an excess above $\$700$ that is greater than $1{,}200-700=\$500$ and at most $2{,}500-700=\$1{,}800$. Use the difference of conditional tails.

The conditional probability of an excess above $\$500$ is

P(X>1{,}200\mid X>700)=e^{-0.001(500)} =e^{-0.5}\approx0.6065.

The conditional probability of an excess above $\$1{,}800$ is

P(X>2{,}500\mid X>700)=e^{-0.001(1{,}800)} =e^{-1.8}\approx0.1653.

Subtract the second tail from the first:

P(1{,}200<X\leq2{,}500\mid X>700) =e^{-0.5}-e^{-1.8} \approx0.6065307-0.1652989 =0.4412318 \approx0.4412.

Check. The result is positive and less than $0.6065$, the probability of exceeding the lower endpoint given the condition. A common mistake is to subtract unconditional tails without dividing by $P(X>700)$. The excess thresholds in this calculation are measured from the known $\$700$ threshold.

Common Mistakes and Practical Checks

The exponential model’s short formulas can make a calculation look simpler than it is. Most errors come from confusing a rate with a mean, mixing up conditional and unconditional tails, or failing to translate the thresholds into the same units.

Using the rate as the mean. The mean is $1/\lambda$, not $\lambda$. If loss is measured in dollars, check that $\lambda$ is per dollar and $1/\lambda$ is in dollars.
Confusing the second raw moment with variance. The variance is $E[X^2]-(E[X])^2$, not $E[X^2]$ itself.
Forgetting the condition. The unconditional tail $P(X>s+t)=e^{-\lambda(s+t)}$ is not the conditional probability given $X>s$. Divide by $P(X>s)$, or use $e^{-\lambda t}$.
Using the wrong additional amount. If a loss already exceeds $s$ and the target is $c>s$, the additional amount is $c-s$, not $c$.
Subtracting interval tails in the wrong order. For an interval, subtract the tail above the upper endpoint from the tail above the lower endpoint.

Memorylessness is mathematically distinctive and useful for conditional calculations, but it is specific to the exponential distribution among continuous severity models. Do not apply the cancellation $e^{-\lambda(s+t)}/e^{-\lambda s}=e^{-\lambda t}$ to a different model without establishing that model’s own survival function and conditional probabilities. These derivations and calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical loss model, state the relevant exponential calculation and check that the rate and thresholds use consistent units.

A loss $X$, in dollars, is exponential with rate $0.002$ per dollar. Find $E[X^2]$, $\operatorname{Var}(X)$, and $P(X>900\mid X>300)$. (Answer: $E[X^2]=500{,}000$ dollars squared, $\operatorname{Var}(X)=250{,}000$ dollars squared, and probability $0.3010$.)
A hypothetical equipment claim is exponential with mean $\$1{,}000$. Find the probability that it exceeds $\$2{,}500$, given that it already exceeds $\$1{,}000$. (Answer: $e^{-1.5}\approx0.2231$.)
A loss is exponential with mean $\$250$. Find the probability it exceeds $\$500$, given that it exceeds $\$200$. (Answer: $e^{-1.2}\approx0.3010$.)
A hypothetical loss $X$, in dollars, is exponential with rate $0.004$ per dollar. Given $X>\$500$, find $P(\$700<X\leq\$1{,}500\mid X>\$500)$. (Answer: $e^{-0.8}-e^{-4}\approx0.4310$.)
A loss is exponential with rate $0.0015$ per dollar. Find the additional amount $t$ such that, given $X>\$900$, the probability of exceeding $\$900+t$ is $0.25$. (Answer: $t=-\ln(0.25)/0.0015\approx\$924.20$.)

What You'll Learn