Sampling From Tabulated Cumulative Distributions

Why a Cumulative Table Can Generate Outcomes

Simulation often begins with a uniform random number $U$ on $[0,1)$. A cumulative distribution function converts that number into an outcome: low values of $U$ select lower outcomes, while high values select outcomes farther into the upper tail. When a distribution is available as a table rather than a formula, the key questions are how to locate the appropriate cumulative probability and whether to select a tabulated outcome or interpolate between tabulated values.

The distinction matters. A table for a discrete distribution lists jumps in the cumulative distribution, so a draw selects one of the distribution's possible values. A table of points used to approximate a continuous distribution may instead be interpolated. Treating a discrete table as continuous, or a coarse continuous table as exact, changes the distribution being simulated.

Definition: For a cumulative distribution function $F_X(x)=P(X\leq x)$, the generalized inverse at $0<u<1$ is $$ F_X^{-1}(u)=\inf\{x:F_X(x)\geq u\}. $$ If $U$ is uniform on $(0,1)$, then $X=F_X^{-1}(U)$ has cumulative distribution function $F_X$. In a table, this rule means finding the first cumulative probability that is at least $u$, then using the corresponding outcome or interpolating within the corresponding interval, as appropriate.

Discrete Tables: Select an Outcome, Do Not Interpolate

Suppose the possible outcomes are $x_1<x_2<\cdots<x_n$, with probabilities $p_j=P(X=x_j)$. Define $F_j=P(X\leq x_j)=\sum_{k=1}^j p_k$, and let $F_0=0$. The outcome $x_j$ is selected when $F_{j-1}<U\leq F_j$. For a continuously generated uniform variate, the chance of landing on any one boundary is zero. Nevertheless, an implementation should use a consistent boundary rule, particularly when the supplied uniform values are fixed or rounded.

The standard rule is to select the first row with cumulative probability greater than or equal to the draw. Thus a draw exactly equal to $F_j$ selects $x_j$, not $x_{j+1}$. The probability of selecting $x_j$ is the width of its uniform interval:

P(F_{j-1}<U\leq F_j)=F_j-F_{j-1}=p_j.

This interval-width argument is the reason inverse-CDF sampling reproduces the table's probabilities. Each possible outcome receives exactly the length of the uniform interval assigned to it. The argument also shows why using the next outcome at a boundary is not the standard generalized-inverse convention, even though changing the result at isolated boundaries does not change the theoretical distribution for an ideal continuous uniform draw.

Worked Example: Mapping Draws to Discrete Claim Amounts

A hypothetical small-loss model has possible claim amounts of $0, $1,000, $3,000, and $7,000, with probabilities 0.20, 0.35, 0.30, and 0.15, respectively. There is no interest or mortality assumption in this simulation. Use the cumulative table to map the uniform draws $0.08, 0.20, 0.41, 0.55, 0.84,$ and $0.96$ to claim amounts.

Claim amount $x_j$	Probability $p_j$	Cumulative probability $F_j$
$0	0.20	0.20
$1,000	0.35	0.55
$3,000	0.30	0.85
$7,000	0.15	1.00

Method. Apply the discrete inverse-CDF rule: for each draw, take the first cumulative probability at least as large as the draw. The intervals are $(0,0.20]$ for $0, $(0.20,0.55]$ for $1,000, $(0.55,0.85]$ for $3,000, and $(0.85,1]$ for $7,000.

For $U=0.08$, the first cumulative probability at least 0.08 is 0.20, so the amount is $0. For $U=0.20$, the first cumulative probability at least 0.20 is again 0.20, so the amount is $0. For $U=0.41$, the first cumulative probability at least 0.41 is 0.55, so the amount is $1,000.

For $U=0.55$, the first cumulative probability at least 0.55 is 0.55, so the amount is $1,000. For $U=0.84$, the first cumulative probability at least 0.84 is 0.85, so the amount is $3,000. For $U=0.96$, the first cumulative probability at least 0.96 is 1.00, so the amount is $7,000.

The amounts in draw order are therefore $0, $0, $1,000, $1,000, $3,000, and $7,000. Check: the boundary draw 0.55 is assigned to the outcome whose cumulative probability is 0.55, and every draw falls in exactly one interval. The common mistake avoided here is interpolating between claim amounts: this is a discrete model, so only the listed amounts can occur.

Continuous Tables: Interpolate Only Under an Explicit Approximation

A table for a continuous distribution may provide pairs $(x_i,F_i)$, where $F_i$ is the cumulative probability at $x_i$. If the table is being used with linear interpolation, then between adjacent points $(x_i,F_i)$ and $(x_{i+1},F_{i+1})$, the cumulative distribution is approximated by a straight line. For a draw $u$ with $F_i\leq u\leq F_{i+1}$, inversion gives:

x=x_i+\frac{u-F_i}{F_{i+1}-F_i}(x_{i+1}-x_i).

This formula follows by setting the fraction of the vertical distance, $(u-F_i)/(F_{i+1}-F_i)$, equal to the fraction of the horizontal distance, $(x-x_i)/(x_{i+1}-x_i)$, on the straight-line segment, then solving for $x$. Linear interpolation of the CDF is equivalent to assuming a constant density within that interval: the probability added across equal-width subintervals is proportional to their widths. It is an approximation unless the true CDF is actually linear there.

If $F_{i+1}=F_i$, the interpolation formula has a zero denominator and must not be used. That interval has no probability increase in the tabulated approximation. A search for the first cumulative value at least $u$ should handle a draw at a plateau using the generalized-inverse rule, rather than dividing by zero.

Worked Example: Interpolating a Continuous Loss Table

A hypothetical continuous loss model is tabulated at $(x,F_X(x))=(0,0.00),(100,0.20),(300,0.80),(500,1.00)$, where losses are in dollars. Use linear interpolation to generate losses for uniform draws 0.10, 0.50, and 0.90. No interest or mortality assumption applies.

Method. Locate the two cumulative probabilities bracketing each draw, then apply the linear-interpolation formula. For $u=0.10$, the draw lies between 0.00 at $x=0$ and 0.20 at $x=100$:

x=0+\frac{0.10-0.00}{0.20-0.00}(100-0)=0+0.5(100)=50.

For $u=0.50$, the bracketing points are $(100,0.20)$ and $(300,0.80)$:

x=100+\frac{0.50-0.20}{0.80-0.20}(300-100) =100+\frac{0.30}{0.60}(200)=100+100=200.

For $u=0.90$, the bracketing points are $(300,0.80)$ and $(500,1.00)$:

x=300+\frac{0.90-0.80}{1.00-0.80}(500-300) =300+\frac{0.10}{0.20}(200)=300+100=400.

The generated losses are $50, $200, and $400, respectively. Check: substituting each loss into its interpolated CDF segment gives 0.10, 0.50, and 0.90. The common mistake avoided is using a discrete-table rule and returning only the grid values $0, $100, $300, or $500; under the stated interpolation method, losses between grid points are allowed.

What a Coarse Grid Changes

A table is not the distribution itself; it is information about the distribution at selected points. For a discrete model with all outcomes and probabilities listed, selecting a listed outcome exactly represents that model. For a continuous model approximated by straight segments, a coarse grid imposes a piecewise-linear CDF and piecewise-constant density. If the true CDF curves substantially between listed points, that approximation can change generated quantiles, interval probabilities, and simulated averages.

A coarser table does not automatically make every estimate larger or smaller. The direction and size of the difference depend on the shape of the underlying distribution and on where the grid points lie. In particular, a coarse segment can smooth over a region where probability accumulates rapidly or slowly. Adding grid points can represent such changes more closely, but does not guarantee an exact result unless the table and interpolation rule capture the true CDF.

Worked Example: Measuring the Effect of a Coarse Grid

Suppose a hypothetical continuous severity distribution is available at the points $(0,0.00),(1,0.10),(2,0.50),(3,0.90),(4,1.00)$. Compare the generated values for $U=0.25$ and $U=0.75$ when using all five points with linear interpolation, versus using only the coarse endpoints $(0,0.00)$ and $(4,1.00)$. Losses are in thousands of dollars. No interest or mortality assumption applies.

Method using the full table. For $U=0.25$, the bracketing points are $(1,0.10)$ and $(2,0.50)$:

x=1+\frac{0.25-0.10}{0.50-0.10}(2-1) =1+\frac{0.15}{0.40}=1.375.

For $U=0.75$, the bracketing points are $(2,0.50)$ and $(3,0.90)$:

x=2+\frac{0.75-0.50}{0.90-0.50}(3-2) =2+\frac{0.25}{0.40}=2.625.

Method using only the coarse endpoints. Linear interpolation from cumulative probability 0 at 0 to cumulative probability 1 at 4 gives $x=4u$. Thus $U=0.25$ produces $x=1.000$, and $U=0.75$ produces $x=3.000$.

The coarse-grid values differ from the full-table values: 1.000 rather than 1.375, and 3.000 rather than 2.625, in thousands of dollars. Check: on the full table, the first value lies between 1 and 2, and the second lies between 2 and 3; on the coarse table, the two values equal $4(0.25)=1$ and $4(0.75)=3$. The coarse endpoints alone imply a uniform distribution over 0 to 4, an assumption not supported by the additional table points. The common mistake avoided is treating a coarse-grid result as if it were an exact inversion of the fuller distribution.

Implementation Checks and Exam Traps

A reliable table-based simulation begins by identifying what the rows represent. If the rows are atoms of a discrete distribution, use cumulative intervals and return the corresponding outcome. If they are CDF values for a continuous distribution, use the specified interpolation rule. Before drawing, verify that cumulative probabilities are nondecreasing and that the final cumulative probability is 1 for a complete distribution.

A table that stops at $F_X(x_m)<1$ does not specify what to do for draws above $F_X(x_m)$. Those draws represent omitted tail probability. Do not silently assign them to $x_m$, discard them, or rescale the listed probabilities unless that treatment is explicitly part of the model. A tail rule or a table covering the full distribution is needed. Similarly, rounding cumulative probabilities can create small inconsistencies, including apparent decreases or a final value slightly different from 1; use the prescribed precision or an explicitly stated adjustment rather than making an undocumented change.

Key takeaway: For a discrete CDF table, select the first outcome whose cumulative probability is at least the uniform draw. For a continuous table, interpolate only when that approximation is intended. The grid and interpolation rule define the simulated approximation, so check both before interpreting results.

These methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus because they test how to transform uniform draws into values from a specified distribution and how to recognize assumptions introduced by a numerical table.

Check Your Understanding

Apply the appropriate table rule in each case. Unless a question says otherwise, use the first cumulative probability at least as large as the draw, and use linear interpolation for a continuous table.

A discrete hypothetical payment has outcomes $0, $500, $1,500, and $4,000 with probabilities 0.15, 0.25, 0.40, and 0.20. Find the outcomes for draws 0.12, 0.40, 0.63, and 0.99. (Answer: $0, $500, $1,500, and $4,000, respectively.)
A continuous hypothetical loss table gives $(0,0.00),(50,0.25),(150,0.75),(250,1.00)$. Use linear interpolation to find the generated loss for $U=0.60$. (Answer: $120.)
A continuous CDF table gives $(0,0.00),(20,0.10),(80,0.70),(100,1.00)$. Under linear interpolation, find $P(20<X\leq80)$, and state the generated value when $U=0.10$. (Answer: Probability 0.60; generated value 20.)
A coarse continuous table has endpoints $(0,0.00)$ and $(8,1.00)$. A refined table adds $(2,0.10)$ and $(6,0.90)$. Find the generated value for $U=0.30$ using the coarse table and then the refined table. (Answer: Coarse value 2.4; refined value 3.)
A discrete hypothetical severity has outcomes $100, $200, and $500 with probabilities 0.30, 0.50, and 0.20. Find the outcome for $U=0.80$, and explain whether a draw of exactly 1 occurs when the generator produces values on $[0,1)$. (Answer: $200; no, 1 is excluded from $[0,1)$.)

What You'll Learn