Inversion Method

Study guide: Exam MAS-I

The Inversion Method (inverse transform method) simulates a draw from a distribution with CDF $F$ by generating $U$ from a Uniform $(0, 1)$ distribution and returning the value $x$ at which the CDF reaches $U$ .

$X = F^{-1}(U), \quad U \sim \text{Unif}(0, 1)$

$X = x_j \quad \text{if} \quad F(x_{j-1}) \leq U < F(x_j)$

Why it works: $F$ is nondecreasing, so $F^{-1}(U) \leq x$ exactly when $U \leq F(x)$ , and a uniform falls below $F(x)$ with probability $F(x)$ . Hence $P(X \leq x) = F(x)$ .
Continuous case: solve $F(x) = u$ for $x$ . For an exponential with mean $\theta$ , $x = -\theta \ln(1-u)$ ; Pareto and Weibull also invert in closed form, others need numerical root-finding.
Discrete case (second block): lay the probabilities end to end on $(0, 1)$ and return the value whose interval contains $u$ . A jump in a mixed CDF — a point mass at a policy limit, say — catches every $u$ in the jump. Conventions differ on which value a $u$ landing exactly on a cumulative probability maps to, so follow the problem's rule.
Because $1 - U$ is also uniform, some texts invert the survival function instead, $X = S^{-1}(U)$ . The two give different draws from the same $u$ — read which one a question wants.
$F^{-1}(0.99)$ is the 99th percentile, so inversion is also how simulated quantiles are read. Applied to exponential interarrival times and cumulated, it simulates a Poisson Process.

Simulating Severities (a) Losses are exponential with mean $\$1{,}000$ . Simulate a loss from $u = 0.70$ .

(b) Losses are Pareto with $F(x) = 1 - \left(\frac{2{,}000}{x + 2{,}000}\right)^3$ . Simulate a loss from $u = 0.875$ .
Answer (a) $x = -1{,}000 \ln(1 - 0.70) = -1{,}000 \ln(0.30) = \$1{,}203.97$ .

(b) Set $F(x) = 0.875$ :
 $\begin{align*} \left(\frac{2{,}000}{x + 2{,}000}\right)^3 &= 1 - 0.875 \\ \frac{2{,}000}{x + 2{,}000} &= 0.125^{1/3} \\ &= 0.5 \\ x &= 2{,}000 \end{align*}$ 
The simulated loss is $\$2{,}000$ , the $87.5$ th percentile of the Pareto.

Simulating Annual Claim Counts A policy's annual claim count has $P(0) = 0.50$ , $P(1) = 0.30$ , $P(2) = 0.15$ , $P(3) = 0.05$ . Using the rule $X = x_j$ if $F(x_{j-1}) \leq u < F(x_j)$ , simulate four years from $u = 0.62,\ 0.97,\ 0.18,\ 0.83$ .

Answer The cumulative probabilities are $0.50,\ 0.80,\ 0.95,\ 1.00$ , giving intervals $[0, 0.50) \to 0$ , $[0.50, 0.80) \to 1$ , $[0.80, 0.95) \to 2$ , $[0.95, 1) \to 3$ .

$0.62 \to 1$

$0.97 \to 3$

$0.18 \to 0$

$0.83 \to 2$

The simulated counts are $1, 3, 0, 2$ : six claims in four years, against an expected $4 \times 0.75 = 3$ .

Simulating Claim Arrival Times Claims arrive as a Poisson process at $\lambda = 2$ per month. Using $T = -\ln(1 - u)/\lambda$ with $u_1 = 0.40$ and $u_2 = 0.80$ , simulate the first two arrival times. How many claims fall in the first month?
Answer
 $\begin{align*} T_1 &= -\ln(0.60)/2 \\ &= 0.2554 \\ T_2 &= -\ln(0.20)/2 \\ &= 0.8047 \\ S_2 &= T_1 + T_2 \\ &= 1.0601 \end{align*}$ 
The first claim arrives at $0.26$ months and the second after the month ends, so this path has exactly one claim in month 1.