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Concept

Exponential Distribution

The exponential distribution models waiting time to the next event in a Poisson process. It is the cleanest entry point into survival functions, hazard rates, and memorylessness.

What exponential answers

Exponential answers a waiting-time question: how long until the next claim, call, failure, or arrival? It pairs naturally with Poisson counts. Poisson counts events in a fixed window; exponential measures the waiting time until the next one.

The parameter lambda is a rate, not a mean. A larger lambda means events arrive faster and the expected waiting time is shorter.

Core formulas

For a rate lambda, the survival function gives the chance that the waiting time exceeds t. The density starts high and decays as time passes.

Density
fT(t)=λe−λt,t≥0f_T(t)=\lambda e^{-\lambda t},\quad t\ge 0
Distribution function
FT(t)=1−e−λtF_T(t)=1-e^{-\lambda t}
Survival function
P(T>t)=e−λtP(T>t)=e^{-\lambda t}
Mean and variance
E[T]=1λ,Var⁡(T)=1λ2E[T]=\frac{1}{\lambda},\qquad \operatorname{Var}(T)=\frac{1}{\lambda^2}

Waiting-time worked example

If claim arrivals average 0.5 per day, the chance the next claim takes more than 3 days is about 0.223. The expected waiting time is 2 days because the mean is 1 divided by the rate.

The same model says the chance the next claim arrives within 3 days is about 0.777. Survival and distribution probabilities are complements, so do not calculate both from scratch.

More than 3 days
P(T>3)=e−0.5(3)≈0.223P(T>3)=e^{-0.5(3)}\approx 0.223
Within 3 days
P(T≤3)=1−e−0.5(3)≈0.777P(T\le 3)=1-e^{-0.5(3)}\approx 0.777

Memoryless property

Memorylessness means the conditional remaining waiting time has the same distribution as a fresh waiting time. If no claim has arrived after 2 days, the chance of waiting more than 3 additional days is still the same as the original chance of waiting more than 3 days.

This property is mathematically clean but often unrealistic for lifetimes and claim reporting delays. It is useful as a baseline, not as a promise that real hazards never change.

Memorylessness
P(T>s+t∣T>s)=P(T>t)P(T>s+t\mid T>s)=P(T>t)

Rate versus scale

Some courses parameterize exponential with a rate lambda. Others use a scale theta, where theta equals 1 divided by lambda. Exam P problems usually make the intended form clear, but FAM and ASTAM distribution tables can use scale conventions for related severity models.

The safest move is to translate the words. If arrivals average 0.5 per day, that is a rate. If the mean waiting time is 2 days, that is a scale or mean.

Rate-scale relationship
θ=1λ\theta=\frac{1}{\lambda}

Exam traps

The common mistakes are using the mean as the rate, reversing within-time and beyond-time probabilities, and treating memorylessness as if it applied to every lifetime model.

  • Use survival for more than t and the distribution function for within t.
  • Check whether lambda is given directly or must be inferred from the mean.
  • Connect arrival counts to Poisson and waiting times to exponential.
  • Do not apply memorylessness to gamma, Weibull, or general survival models unless it is stated.

Where exponential connects next

Gamma generalizes exponential waiting times to the time until multiple events. Survival functions and hazard rates turn the same notation into life, reliability, and claim-reporting models.

Practice

Original exam practice

4 questions built from syllabus outcomes and released-exam patterns. The prompts and answers are original, so they train the skill without copying official exam text.

Exponential Waiting-Time Drill

Original exponential checks for survival, distribution probabilities, rate conversion, and memorylessness.

Exam P - 15 min
Source pattern: SOA Exam P exponential distribution outcomes; original waiting-time prompts.
  1. Question 1/Calculation

    More than three days

    Claim arrivals average 0.5 per day. With an exponential waiting-time model, find the probability that the next claim takes more than 3 days.

    Solution and grading points

    Use the survival function with rate 0.5 and time 3. The probability is e to the negative 1.5, about 0.223.

    • Identifies 0.5 as a rate.
    • Uses the survival probability for more than 3 days.
    • Reports a probability near 0.223.
  2. Question 2/Calculation

    Within two hours

    A help desk receives actuarial-model tickets at an average rate of 0.25 per hour. Under an exponential waiting-time model, find the probability the next ticket arrives within 2 hours.

    Solution and grading points

    Within 2 hours is the complement of waiting more than 2 hours. The probability is 1 minus e to the negative 0.5, about 0.393.

    • Uses the distribution function rather than survival alone.
    • Multiplies the rate by the 2-hour time window.
    • Reports a probability near 0.393.
  3. Question 3/Calculation

    Rate from mean waiting time

    The mean waiting time to the next claim is 5 days. What rate should be used in the exponential survival function?

    Solution and grading points

    For an exponential model, mean waiting time is 1 divided by the rate. A 5-day mean gives rate 0.2 per day.

    • Uses mean equal to 1 divided by the rate.
    • Converts 5 days into rate 0.2 per day.
    • Keeps the rate unit attached.
  4. Question 4/Written Answer

    Remaining wait

    In an exponential model, no claim has arrived after 4 days. How does that affect the distribution of the additional waiting time?

    Solution and grading points

    It does not change the remaining waiting-time distribution. The exponential model is memoryless, so the additional wait has the same distribution as a fresh wait.

    • States the memoryless property.
    • Uses additional waiting time, not total waiting time.
    • Avoids saying the event becomes impossible or guaranteed.

References and official sources