Exponential Distribution

Study guides: Exam P-1, Exam MAS-I

The Exponential Distribution X \sim \text{Exp}(\theta) is a continuous distribution on (0, \infty) commonly used to model the time between events or the size of insurance losses.

f(x) = \frac{1}{\theta}\,e^{-x/\theta}, \quad x > 0

\text{where } \theta > 0 = \text{mean (scale parameter)}

• E[X] = \theta, \text{Var}(X) = \theta^2, and F(x) = 1 - e^{-x/\theta}
• Its defining property is memorylessness: P(X > s + t \mid X > s) = P(X > t), making it the unique continuous distribution with no aging effect
• A direct consequence: given X > d, the excess X - d is again \text{Exp}(\theta). So E[X \mid X > d] = d + \theta and E[(X-d)_+] = \theta e^{-d/\theta} — the two identities most deductible questions reduce to
• Capped at a limit u: E[\min(X,u)] = \theta\left(1 - e^{-u/\theta}\right)
• Work from the survival function S(x) = e^{-x/\theta} rather than the CDF; almost every exponential question is a tail probability
• Some texts parameterize by rate \lambda = 1/\theta, so f(x) = \lambda e^{-\lambda x} and E[X] = 1/\lambda. Check which convention a problem uses before substituting

Probability a Loss Exceeds the Deductible Ground-up losses follow X \sim \text{Exp}(\theta = 500). A deductible of d = 300 applies. What proportion of losses result in a claim payment?

Answer A payment is triggered only when X > 300: P(X > 300) = 1 - F(300) = e^{-300/500} = e^{-0.6} \approx 0.5488 About 54.9% of losses exceed the deductible and generate a payment.

Expected Payment per Loss vs. per Payment Losses are X \sim \text{Exp}(\theta = 500) with a deductible d = 300. Find the average payment across all losses, and the average payment across only those losses that produce a payment.

Answer Averaged over all losses (zeros included):

\begin{align*}
E[(X - 300)_+] &= \theta e^{-d/\theta} \\
               &= 500\,e^{-0.6} \\
               &\approx 274.4
\end{align*}

Averaged over payments only, condition on X > 300 and use memorylessness — the excess is again \text{Exp}(500): E[X - 300 \mid X > 300] = \theta = 500 The two differ by the factor P(X > 300) = e^{-0.6} \approx 0.5488, since 274.4 = 0.5488 \times 500. Read carefully whether a question asks per loss or per payment — it is the most common way to lose a mark on this material.

Combining a Deductible and a Limit Losses are X \sim \text{Exp}(\theta = 1000). A policy pays the excess over a deductible of 200, capped at a maximum payment of 800. Find the expected payment per loss.

Answer The payment is Y = \min\bigl((X-200)_+,\ 800\bigr), which pays out only for 200 < X < 1000 and pays a flat 800 above that. Write it as a difference of two excess-loss terms:

\begin{align*}
E[Y] &= E[(X-200)_+] - E[(X-1000)_+] \\
     &= 1000\,e^{-0.2} - 1000\,e^{-1.0} \\
     &= 818.73 - 367.88 \\
     &\approx 450.85
\end{align*}

Layering as a difference of excess-loss expectations avoids splitting the integral into three pieces, and works for any severity distribution — not just the exponential.