Practice probability calculations using explicit assumptions and worked explanations across three subject areas. This package has no recorded current official syllabus verification, and its topic allocation is editorial rather than an official examination weighting or count. Its format and coverage do not establish current examination logistics or official alignment. Insurance scenarios state their own payment terms.
What this guide covers
Original probability practice questions covering discrete random variables, linear combinations, and insurance payment variance.
Probability foundations and conditional reasoning
1. Translate exactly, at least, and only into events
Event language determines which outcomes belong in a probability calculation. At least one of A and B means A ∪ B; both means A ∩ B. A only means A ∩ Bᶜ, while exactly one means the disjoint union of A ∩ Bᶜ and Aᶜ ∩ B. Neither means (A ∪ B)ᶜ. The word or is ordinarily inclusive in mathematical probability unless the problem explicitly excludes overlap. Translate the statement before substituting numerical probabilities, particularly when eligibility conditions or overlapping insurance coverages are involved.
Worked example: Suppose 40% of accounts use service A, 30% use service B, and 10% use both. Accounts using exactly one service have probability 0.40 + 0.30 − 2(0.10) = 0.50, because the overlap must be excluded from both marginal totals.
Common mistake: Treating exactly one as an ordinary union retains outcomes in both events and therefore answers a different question.
2. Probability axioms and two-event inclusion–exclusion
Probabilities are nonnegative, the whole sample space has probability one, and probabilities add over disjoint events. For overlapping events, P(A ∪ B) = P(A) + P(B) − P(A ∩ B), because the intersection otherwise appears twice. This identity also constrains feasible inputs: the intersection cannot exceed either marginal probability, and the union cannot exceed one. Use these bounds to detect an impossible model before calculating conditional probabilities. Additivity applies directly to mutually exclusive events, but independence is a separate property and does not remove overlap.
Worked example: If P(A) = 0.65 and P(B) = 0.55, their intersection must be at least 0.20, since their union cannot exceed one. An assumed intersection of 0.10 would produce a union of 1.10 and invalidate the model.
Common mistake: Adding overlapping event probabilities without subtracting their intersection double-counts shared outcomes, even when the events are independent.
3. Three-event inclusion–exclusion restores triple overlap
For three events, add the three marginal probabilities, subtract the three pairwise intersections, and add the triple intersection: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(AB) − P(AC) − P(BC) + P(ABC). An outcome in all three events is counted three times initially and removed three times by the pairwise correction, so it must be restored once. Pairwise intersections include the triple intersection unless stated otherwise. This distinction matters when converting survey overlap data into probabilities of at least one or exactly one attribute.
Worked example: Three alerts each occur with probability 0.30. Each pair occurs with probability 0.10, and all three occur with probability 0.04. The probability of at least one alert is 0.90 − 0.30 + 0.04 = 0.64.
Common mistake: Subtracting pairwise overlaps without restoring the triple overlap removes outcomes that should remain in the union once.
4. Choose combinations or permutations consistently
A permutation counts ordered selections; a combination counts subsets whose internal order is irrelevant. Selecting r distinct objects from n without replacement gives n!/(n−r)! ordered outcomes or C(n,r) unordered outcomes. Either framework can solve a probability problem if numerator and denominator use the same convention and equally likely elementary outcomes. With replacement, ordered selections instead have nʳ possibilities. Repeated categories require additional care because category patterns may represent unequal numbers of underlying selections. Decide what a single outcome means before performing the count.
Worked example: A committee of three is selected uniformly from eight people, including two actuaries. Exactly one actuary can be chosen in C(2,1)C(6,2) = 30 ways, out of C(8,3) = 56 committees, giving probability 15/28.
Common mistake: Using an ordered numerator with an unordered denominator changes the counting scale and generally produces an incorrect probability.
5. Conditional probability and the multiplication chain
Conditional probability restricts attention to an event B with positive probability and rescales: P(A | B) = P(A ∩ B)/P(B). Rearrangement gives the multiplication rule P(A ∩ B) = P(B)P(A | B). For three events, P(ABC) = P(A)P(B | A)P(C | AB). Each factor describes the next requirement after earlier requirements have occurred. This rule does not assume independence; independence merely simplifies selected conditional factors. In sequential sampling without replacement, update both the composition and size of the remaining population at every step.
Worked example: A box has three red and two blue tokens. The probability that two draws without replacement are both red is (3/5)(2/4) = 3/10. The second factor changes because the first red token is no longer available.
Common mistake: Multiplying marginal probabilities instead of appropriate conditional probabilities silently assumes independence that the situation may not provide.
6. Total probability across a complete partition
A partition consists of mutually exclusive events that exhaust the sample space. If H₁ through Hₙ form a partition, then P(A) = ΣP(A | Hᵢ)P(Hᵢ). Each term is a joint probability, and the sum combines all possible routes to A. Group-specific event rates must be weighted by the proportions of their groups, rather than averaged equally unless those proportions are equal. The partition must leave no outcomes unclassified and must not double-count outcomes. This method is central to mixture models, claim probabilities, and Bayesian denominators.
Worked example: A company obtains 80% of parts from supplier U and 20% from supplier V. Their defect probabilities are 0.01 and 0.06. The overall defect probability is 0.80(0.01) + 0.20(0.06) = 0.02, not their unweighted average.
Common mistake: Averaging conditional probabilities equally ignores unequal group sizes and generally gives the wrong overall event probability.
7. Bayes’ theorem and likelihood-ratio updating
Bayes’ theorem reverses conditioning: P(H | E) = P(E | H)P(H)/P(E), where the denominator sums evidence probabilities across all possible hypotheses. With two hypotheses, posterior odds equal prior odds multiplied by the likelihood ratio P(E | H)/P(E | Hᶜ). Evidence favors H when that ratio exceeds one, but the posterior still depends on the starting prevalence. A high evidence probability within H does not by itself imply a high probability of H after observing evidence. Use probability or odds form according to which makes the supplied information easier to organize.
Worked example: A condition has prior probability 0.10. An indicator appears with probability 0.80 when the condition is present and 0.20 otherwise. Posterior odds are (1/9)(4) = 4/9, so the posterior probability is 4/13.
Common mistake: Interchanging P(E | H) and P(H | E) discards the base rate and contributions from competing hypotheses.
8. Independence differs from mutual exclusivity
Events A and B are independent when P(A ∩ B) = P(A)P(B). When conditioning is defined, this means learning one event does not change the probability of the other. Mutually exclusive events instead have an empty intersection. If both have positive probability, mutual exclusivity implies dependence because occurrence of one rules out the other. Independence is preserved when either or both events are replaced by their complements. For collections of three or more events, pairwise independence alone does not establish mutual independence, which requires factorization for every relevant subcollection.
Worked example: For one fair die, A = {1,2} and B = {3,4} are mutually exclusive. Their intersection probability is zero, whereas P(A)P(B) = 1/9. They therefore are not independent, despite having no overlapping outcomes.
Common mistake: Calling events independent because they cannot happen together confuses absence of overlap with absence of probabilistic influence.
9. Conditional independence can disappear after mixing
Conditional independence within a group H means P(A ∩ B | H) = P(A | H)P(B | H). It does not automatically establish independence when groups are combined. Shared group membership can make A and B move together: observing A changes beliefs about the group, which changes the probability of B. Calculate the marginal joint probability by weighting group-specific products, then compare it with the product of marginal probabilities. Conversely, conditioning can also create dependence in some models, so independence must always be checked under the particular information set being used.
Worked example: Two indicators independently occur with probability 0.9 in one group and 0.1 in another equally common group. Each marginal probability is 0.5, but their joint probability is 0.5(0.81) + 0.5(0.01) = 0.41, exceeding 0.25.
Common mistake: Dropping the conditioning label from an independence statement ignores the common group variable and can substantially understate joint occurrence.
10. Use complements for at-least-one calculations
An at-least-one event is often easier to calculate through its complement, none: P(N ≥ 1) = 1 − P(N = 0). For independent trials with success probability p, the result is 1 − (1−p)ⁿ. Without replacement, calculate the probability that every selected item comes from the non-target group using combinations or sequential conditional probabilities. The complement principle itself does not require independence; only the chosen expression for the no-event probability may require it. This method avoids summing several overlapping or separately enumerated positive-count events.
Worked example: A shipment contains five damaged and fifteen undamaged packages. Four are sampled without replacement. The probability of detecting at least one damaged package is 1 − C(15,4)/C(20,4), because the complementary sample contains only undamaged packages.
Common mistake: Using (1−p)ⁿ for sampling without replacement assumes unchanged independent draw probabilities, even though the remaining population changes.
Discrete random variables
11. Validate a probability mass function
A probability mass function assigns a nonnegative probability to every possible value of a discrete random variable, with total mass one. If p(x) contains an unknown constant, normalize over the stated support and then verify nonnegativity. A formula valid on one support may fail on another, so support is part of the distribution rather than an optional annotation. Combine probabilities when different underlying outcomes produce the same variable value. Once the mass function is valid, event probabilities are sums over the values satisfying the event.
Worked example: Suppose p(x) = c(x+1) for x = 0, 1, 2. Normalization gives c(1+2+3) = 1, so c = 1/6. Consequently, P(X ≥ 1) = 2/6 + 3/6 = 5/6.
Common mistake: Solving only for a total of one is insufficient if any resulting probability is negative on the stated support.
12. Read CDF jumps and endpoint inequalities
The cumulative distribution function is F(x) = P(X ≤ x). For a discrete distribution it is nondecreasing, right-continuous, and constant between mass points. The jump at a value a equals P(X = a) = F(a) − F(a−), where F(a−) is the left limit. Consequently, P(a < X ≤ b) = F(b) − F(a), while including the lower endpoint requires its left limit. These distinctions are essential whenever an interval boundary has positive probability. Unlike a continuous distribution, changing a strict inequality to a non-strict one can change the answer.
Worked example: A variable has probabilities 0.2, 0.5, and 0.3 at values zero, one, and two. Then F(1) = 0.7, but P(X < 1) = 0.2. The difference, 0.5, is the jump representing P(X = 1).
Common mistake: Replacing P(X < a) with F(a) includes the mass at a and is generally wrong for discrete variables.
13. Expectation of a function uses original probabilities
For a discrete variable X, E[g(X)] = Σg(x)p(x), provided the expectation exists. This rule evaluates the function at each possible value and uses the original probability weights; deriving the full distribution of g(X) is unnecessary. In general, E[g(X)] does not equal g(E[X]) because nonlinear functions do not commute with averaging. Linearity does hold: E[aX+b] = aE[X]+b, and expectations of sums add without independence. Use direct weighted sums for nonlinear benefits, penalties, squared counts, and other specified transformations.
Worked example: Let X equal zero or two with equal probabilities, and let a fee be g(X) = X² + 3 dollars. The expected fee is 0.5(3) + 0.5(7) = $5, whereas applying the fee formula to E[X] gives only $4.
Common mistake: Evaluating a nonlinear payment or cost formula at the mean generally differs from averaging the actual possible payments.
14. Variance and factorial moments of counts
Variance measures expected squared deviation from the mean and can be calculated as Var(X) = E[X²] − E[X]² when the second moment is finite. For count variables, the factorial moment E[X(X−1)] is often easier to obtain than E[X²], since X² = X(X−1)+X. Thus Var(X) = E[X(X−1)] + E[X] − E[X]². Binomial counts have factorial moment n(n−1)p²; Poisson counts have factorial moment λ². A second raw moment and a variance have the same squared units but represent different quantities.
Worked example: For X binomial with n = 5 and p = 0.2, E[X] = 1 and E[X(X−1)] = 5(4)(0.2²) = 0.8. Therefore E[X²] = 1.8 and Var(X) = 1.8 − 1² = 0.8.
Common mistake: Reporting E[X²] as the variance forgets to remove the squared mean and overstates dispersion whenever the mean is nonzero.
15. Bernoulli indicators build binomial counts
A Bernoulli variable records one success as one and failure as zero, with mean p and variance p(1−p). The sum of n independent Bernoulli variables sharing success probability p is binomial: P(X=k) = C(n,k)pᵏ(1−p)ⁿ⁻ᵏ. Its mean is np and variance np(1−p). Fixed trial count, independence, and a common success probability are the defining requirements. If success probabilities differ, the sum generally is not binomial. Sampling without replacement from a finite population also generally violates the binomial independence assumption.
Worked example: Eight independent applications each have probability 0.25 of requiring review. The probability that exactly two require review is C(8,2)(0.25)²(0.75)⁶. The expected number requiring review is 8(0.25) = 2, although the realized count must be an integer.
Common mistake: A fixed number of yes-or-no observations does not establish a binomial model unless independence and a common success probability also hold.
16. Hypergeometric sampling and finite-population correction
A hypergeometric variable counts successes in n draws without replacement from N objects containing K successes. Its mass function is C(K,k)C(N−K,n−k)/C(N,n), over feasible counts. The mean is nK/N, matching a binomial mean with p = K/N. Its variance is np(1−p)(N−n)/(N−1), which includes the finite-population correction. Removing an object creates dependence between draws and reduces count variability relative to independent sampling. Check both success and failure availability when determining support; some apparently possible counts cannot fit the remaining population.
Worked example: From ten files containing four errors, an auditor samples three without replacement. The probability of two errors is C(4,2)C(6,1)/C(10,3) = 36/120 = 0.30. The expected number of errors remains 3(4/10) = 1.2.
Common mistake: Using binomial probabilities for a substantial sample without replacement ignores dependence and the reduced variability of the hypergeometric count.
17. Geometric waiting times need a counting convention
If independent trials succeed with probability p, the trial number T of the first success has P(T=t) = (1−p)ᵗ⁻¹p for t ≥ 1. Its mean is 1/p and variance is (1−p)/p². If instead F counts failures before the first success, then F = T−1 and its support starts at zero. Memorylessness states P(T>m+n | T>m) = P(T>n) for nonnegative integer m and n. Translate verbal waiting conditions into the chosen variable before applying the formula, especially around strict inequalities.
Worked example: A trial succeeds with probability 0.2. Given five initial failures, the probability of waiting more than three additional trials for success is 0.8³ = 0.512. The earlier failures do not change the success probability on later independent trials.
Common mistake: Switching between trials-until-success and failures-before-success conventions without shifting the support creates systematic off-by-one errors in probabilities.
18. Negative binomial waits for a specified success
Let T be the trial number of the rth success in independent Bernoulli trials with success probability p. Then P(T=t) = C(t−1,r−1)pʳ(1−p)ᵗ⁻ʳ for t ≥ r. The final trial must succeed, while exactly r−1 successes occur among the earlier t−1 trials. The mean is r/p and variance is r(1−p)/p². A failures-count version subtracts r from T. For r greater than one, the waiting time is not geometric and does not generally have the geometric memoryless property.
Worked example: If p = 0.3, the third success occurs on trial seven only when the first six trials contain two successes and the seventh succeeds. Its probability is C(6,2)(0.3)³(0.7)⁴, which differs from merely counting three successes among seven trials.
Common mistake: Using C(t,r) counts arrangements whose final trial fails, even though the rth success must occur on the specified final trial.
19. Poisson counts and exposure scaling
A Poisson variable with parameter λ has mass P(N=k) = e^(−λ)λᵏ/k!, mean λ, and variance λ. In a homogeneous Poisson process, λ equals event rate multiplied by exposure length, and counts over disjoint intervals are independent. Scale the mean to the interval named in the question before calculating probabilities. A Poisson count model specifies counts; process assumptions are additionally needed to justify independence across time intervals or exponential interarrival times. The equality of mean and variance is a model property, not a universal fact about observed event counts.
Worked example: Calls follow a homogeneous Poisson process at six per hour. Over twenty minutes, the count mean is 6(1/3) = 2. The probability of exactly one call during that interval is e^(−2)2¹/1! = 2e^(−2).
Common mistake: Using an hourly Poisson mean for a shorter observation period fails to match the parameter to the stated exposure.
20. Truncated discrete distributions require renormalization
Conditioning a discrete variable to lie in a set A produces conditional masses p(x)/P(X∈A) for x in A and zero elsewhere. The same denominator rescales probabilities and moments. For example, a zero-truncated Poisson variable has positive-count probabilities e^(−λ)λᵏ/[k!(1−e^(−λ))]. Its mean is λ/(1−e^(−λ)), because zero contributes nothing to the original first moment. Truncation differs from censoring: truncation conditions on selection, while censoring changes what is recorded. Always identify whether the requested population includes the excluded outcomes before choosing the denominator.
Worked example: Suppose X has probabilities 0.5, 0.3, and 0.2 at zero, one, and two. Among observations with X > 0, the probabilities become 0.6 and 0.4. The conditional mean is therefore 1(0.6) + 2(0.4) = 1.4.
Common mistake: Deleting excluded outcomes without rescaling leaves probabilities that do not sum to one and produces unnormalized conditional expectations.
Continuous random variables
21. A density gives probability through area
A continuous density is nonnegative and integrates to one over its support. Its value at a point is not a point probability and may exceed one without invalidating the model. The CDF is F(x) = ∫₋∞ˣ f(t)dt, and interval probabilities are differences of CDF values. An ordinary continuous variable has zero probability at every individual point, so strict and non-strict interval endpoints agree. Normalize a proposed density before deriving its CDF, and write the CDF separately below, inside, and above the support.
Worked example: Let f(x) = c on 0 < x < 0.25. Normalization requires c = 4, although this exceeds one. The probability of 0.05 < X < 0.15 is the area 4(0.10) = 0.40, not the density value itself.
Common mistake: Rejecting a density because its height exceeds one confuses probability per unit with probability over an interval.
22. Uniform probabilities are relative lengths
For X uniform on (a,b), the density is 1/(b−a), and probabilities within the support equal interval length divided by total length. The mean is (a+b)/2 and variance is (b−a)²/12. Its qth quantile is a+q(b−a). Intersect a requested interval with the support before measuring length; extending outside the support adds no probability. Uniformity describes equal probability for equal-length intervals, not equal probabilities at individual points. Conditioning on a subinterval preserves uniformity over that subinterval when the conditioning probability is positive.
Worked example: A departure time is uniform between 2:00 and 2:40. The probability of departure between 2:10 and 2:25 is 15/40 = 0.375. Given departure after 2:10, the corresponding probability becomes 15/30 = 0.50.
Common mistake: Dividing by the original support length after conditioning ignores the smaller sample space created by the observed interval restriction.
23. Exponential rates and residual waiting time
An exponential variable with rate λ has density λe^(−λx), survival e^(−λx), mean 1/λ, and variance 1/λ² for x ≥ 0. Memorylessness gives P(X>s+t | X>s) = e^(−λt), so the residual waiting time after survival to s has the original distribution. The rate has reciprocal-time units; a mean of ten hours corresponds to rate 0.1 per hour. This property is specific and should not be transferred to arbitrary lifetime distributions. In particular, elapsed survival does not usually leave other distributions unchanged.
Worked example: A waiting time is exponential with mean twelve minutes. After eight minutes without an arrival, the probability of waiting more than another six minutes is e^(−6/12). The expected additional waiting time is still twelve minutes under this model.
Common mistake: Using the mean as the exponential rate reverses the parameter’s units and changes both survival probabilities and expected waiting times.
24. Gamma shape, rate, and accumulated waits
A gamma variable with shape α and rate λ has density λᵅxᵅ⁻¹e^(−λx)/Γ(α) for x > 0, mean α/λ, and variance α/λ². Under a scale parameter θ = 1/λ, these become αθ and αθ². For integer shape r, the distribution models the sum of r independent exponential waits sharing rate λ, or the waiting time to the rth event in a homogeneous Poisson process. Only shape one is exponential. Identify whether a problem uses rate or scale before substituting parameters into a formula.
Worked example: Events arrive at rate two per hour in a homogeneous Poisson process. Waiting for the third event has gamma shape three and rate two. Its mean is 3/2 hours and variance is 3/4 hours squared, rather than those of a single arrival wait.
Common mistake: Treating every gamma waiting time as memoryless overlooks that accumulated progress toward multiple events changes the residual waiting distribution.
25. Standardize normal variables using standard deviation
If X is normal with mean μ and variance σ², then Z = (X−μ)/σ is standard normal. Consequently, interval probabilities become differences of standard normal CDF values after transforming both endpoints. The denominator is the standard deviation, not the variance. Symmetry gives Φ(−z) = 1−Φ(z), which is useful when tables report only positive arguments. Standardization preserves the order of inequalities because σ is positive. A normal model assigns probability across the entire real line, so its suitability for naturally bounded quantities remains a modeling assumption.
Worked example: Suppose X is normal with mean 60 and standard deviation eight. The event 52 < X < 76 becomes −1 < Z < 2. Using Φ(2) = 0.9772 and Φ(−1) = 0.1587 gives probability 0.8185.
Common mistake: Dividing by σ² rather than σ produces a quantity with incorrect units and the wrong standard normal cutoffs.
26. Invert normal probabilities with the correct tail
An inverse normal problem starts with a probability and solves for a cutoff. If P(X≤c) = q, then c = μ+σΦ⁻¹(q). If the supplied probability is an upper tail p, first convert it to cumulative probability 1−p. A central interval or two-sided condition may require splitting probability between two tails before using quantiles. Check the answer’s direction relative to the mean: a small upper-tail probability should produce a cutoff above the mean. Keep the original units when transforming the standardized quantile back.
Worked example: A normal measurement has mean 200 and standard deviation ten. To leave 2.5% above a cutoff, use the supplied standard normal quantile 1.96. The cutoff is 200 + 10(1.96) = 219.6 measurement units.
Common mistake: Using an upper-tail probability directly as the CDF argument selects the opposite side of the normal distribution.
27. Integrate transformed values to obtain moments
For a continuous variable with density f, E[g(X)] = ∫g(x)f(x)dx over the support, provided the expectation exists. The function g may represent a square, logarithm, cost, or benefit; finding its separate distribution is unnecessary. Raw moments use g(x)=xʳ, and variance is E[X²]−E[X]² when the second moment is finite. Split the integral wherever the function or density changes form. As with discrete expectations, a nonlinear function of the mean generally differs from the mean of the nonlinear function.
Worked example: For X uniform on (0,2), E[X²] = ∫₀² x²(1/2)dx = 4/3. Since E[X] = 1, Var(X) = 4/3 − 1 = 1/3. Squaring the mean alone would miss the variation in X.
Common mistake: Integrating g(x) without multiplying by the density weights locations equally even when the probability model does not.
28. Quantiles solve cumulative probability equations
A q-quantile is commonly defined as inf{x:F(x)≥q}. For a continuous strictly increasing CDF on its support, it is the unique solution of F(x)=q. The survival function S(x)=1−F(x) instead describes upper-tail probability, so a cutoff exceeded with probability p solves F(x)=1−p. Quantiles are expressed in the variable’s original units and are not generally determined by its mean alone. For piecewise distributions, first determine which support segment contains the desired cumulative probability before inverting the appropriate expression.
Worked example: Suppose F(x)=x³/8 for 0≤x≤2. The seventy-fifth percentile solves x³/8=0.75, giving x=∛6. This cutoff is exceeded with probability 0.25, and its value remains within the stated support from zero to two.
Common mistake: Solving F(x)=p when p is an exceedance probability returns the complementary percentile rather than the requested cutoff.
29. Conditional densities under truncation
Conditioning on a continuous variable lying in a positive-probability region A restricts its density to A and divides by P(X∈A). For lower truncation at d, the density becomes f(x)/S(d) for x>d. Conditional moments similarly divide truncated integrals by S(d): E[X | X>d] = ∫d∞ xf(x)dx/S(d). The excess X−d is a different variable from the conditional underlying value X. Memorylessness simplifies this excess only for exponential distributions, not for every continuous model. Truncation also differs from recording all small observations as a boundary value.
Worked example: Let X be uniform on (0,10). Given X>6, its conditional density is 1/4 on (6,10), and its mean is eight. The mean excess above six is therefore two, while the conditional underlying value remains eight.
Common mistake: A truncated first-moment integral must be divided by the selection probability; otherwise it is not a conditional mean.
30. Hazard links survival to instantaneous failure
For an absolutely continuous lifetime with positive survival, the hazard is h(x)=f(x)/S(x). It describes the limiting failure rate near x among items still surviving at x, rather than an unconditional probability at a point. Since S′(x)=−f(x), survival satisfies S(x)=exp(−∫₀ˣ h(t)dt) for a nonnegative lifetime starting with survival one. The exponential model has constant hazard λ, explaining its memorylessness. Other distributions can have increasing or decreasing hazard, so survival to an age may change the remaining risk even when the original mean is unchanged.
Worked example: Suppose a hypothetical lifetime has hazard h(t)=2t for t≥0, with time measured in years. Its survival is exp(−t²). Conditional survival from year one to year two is S(2)/S(1)=e^(−3), showing that elapsed age affects the remaining risk.
Common mistake: Treating hazard as an ordinary interval probability ignores its rate interpretation and the conditioning on survival to the specified age.
Joint distributions and dependence
31. Marginalize a joint probability table
A joint mass function assigns probabilities to pairs of values. All cells must be nonnegative and sum to one. To obtain the marginal distribution of X, sum over every possible Y value; to obtain the marginal of Y, sum over X. These marginal probabilities retain no condition on the other variable. Event probabilities follow by summing cells satisfying all stated restrictions. Label rows and columns carefully because a conditional denominator may be a row total or column total depending on which variable is observed. Marginalization does not require independence.
Worked example: Suppose joint masses for (X,Y)=(0,0),(0,1),(1,0),(1,1) are 0.1, 0.2, 0.3, and 0.4. Then P(X=1)=0.7 and P(Y=1)=0.6. The joint probability P(X=1,Y=1) remains 0.4, distinct from either marginal.
Common mistake: A cell probability describes both variables simultaneously; interpreting it as a marginal omits other compatible outcomes in its row or column.
32. Joint-density support determines integration limits
A joint density integrates to one over its two-dimensional support. Marginal densities are found by integrating out the other variable, but the integration limits may depend on the retained variable. For triangular support 0<x<y<1, fX(x)=∫x¹ f(x,y)dy and fY(y)=∫₀ʸ f(x,y)dx. Extending a positive formula outside its support creates probabilities for impossible pairs. Sketching or algebraically describing the support helps establish valid bounds. Density values themselves are not joint point probabilities; probabilities arise from integrating over regions of positive area.
Worked example: If f(x,y)=2 on 0<x<y<1, then fX(x)=2(1−x) for 0<x<1 and fY(y)=2y for 0<y<1. Each marginal integrates to one, although their shapes differ because the available slice lengths differ.
Common mistake: Using fixed unit-square limits for a triangular support includes impossible pairs and produces incorrectly normalized marginal densities.
33. Condition a discrete joint distribution by a slice
For a joint mass function, P(X=x | Y=y)=p(x,y)/pY(y), provided pY(y)>0. Conditioning selects the row or column compatible with the observed value and rescales it to total one. Conditional expectations then weight X values using those new probabilities. The denominator is the probability of the observed condition, not the probability of the event being predicted. Comparing conditional and marginal distributions can reveal dependence: if the conditional distribution changes with y, X and Y are not independent. A zero-probability conditioning value cannot use this discrete ratio.
Worked example: Suppose P(X=0,Y=2)=0.15 and P(X=1,Y=2)=0.35, with X taking only zero and one. Then P(Y=2)=0.50, so P(X=1 | Y=2)=0.70 and E[X | Y=2]=0.70 because X is binary.
Common mistake: Dividing by the marginal probability of the predicted variable reverses the conditioning and generally answers a different question.
34. Conditional densities use marginal density ratios
For continuous variables, a conditional density is fX|Y(x|y)=fXY(x,y)/fY(y) where the marginal density is positive. Although P(Y=y)=0, this density ratio defines a conditional distribution through local limiting behavior. Integrate it over x to obtain conditional probabilities or moments, respecting the support available at that y. The denominator is a density, not a point probability. A conditional density must integrate to one over its conditional support. It may depend on y through both its formula and its integration limits.
Worked example: For joint density two on 0<x<y<1, fY(y)=2y. Hence fX|Y(x|y)=1/y for 0<x<y: X given Y=y is uniform on (0,y). At y=0.8, its conditional mean is 0.4.
Common mistake: Declaring continuous conditioning impossible because P(Y=y)=0 overlooks the conditional density ratio available when the marginal density is positive.
35. Check independence using both density and support
Random variables are independent when their joint distribution factors into their marginal distributions. For mass functions this means p(x,y)=pX(x)pY(y) at every pair; for densities, equality is required almost everywhere. A separable-looking density formula is insufficient if support restrictions couple x and y. Independent continuous variables generally have support described by the product of their separate supports, apart from zero-density exceptions. One valid counterexample to factorization establishes dependence. For discrete tables, a single mismatched cell can therefore settle the question, although all cells are needed to establish independence.
Worked example: The formula f(x,y)=2 on 0<x<y<1 is constant, but X and Y are dependent. Knowing Y=0.2 forces X<0.2, whereas the marginal X distribution allows larger values. The triangular support prevents independence despite the simple formula.
Common mistake: Factoring only the algebraic expression while ignoring a coupled support can falsely classify dependent variables as independent.
36. Covariance measures centered co-movement
Covariance is Cov(X,Y)=E[(X−E[X])(Y−E[Y])]=E[XY]−E[X]E[Y] when the needed moments exist. Positive covariance indicates that above-mean values tend to accompany above-mean values, while negative covariance indicates opposing centered movements. It carries the product of the variables’ units and therefore depends on scale. Constants do not affect covariance, and Cov(aX+b,cY+d)=acCov(X,Y). Independence implies zero covariance when the relevant moments exist, but zero covariance alone does not prove independence. Calculate E[XY] from the joint distribution rather than multiplying means without justification.
Worked example: Suppose E[X]=2, E[Y]=5, and E[XY]=13. Cov(X,Y)=13−2(5)=3. If X is converted to a unit ten times smaller, the numerical variable becomes 10X and its covariance with Y becomes 30.
Common mistake: Replacing E[XY] with E[X]E[Y] assumes the very zero-covariance relationship that a calculation may be intended to test.
37. Correlation standardizes linear association
Correlation is ρ=Cov(X,Y)/(σXσY) when both standard deviations are finite and positive. It is dimensionless and lies between −1 and one. Positive rescaling leaves it unchanged, while multiplying one variable by a negative constant reverses its sign. Values near either endpoint indicate strong linear association, not necessarily strong association of every possible form. Correlation is undefined if either variable has zero variance. A magnitude of one means an exact affine relationship almost surely under these moment conditions; zero correlation can coexist with substantial nonlinear dependence.
Worked example: If covariance is six and standard deviations are two and five, correlation is 6/(2×5)=0.6. Replacing Y by −3Y changes covariance to −18 and its standard deviation to fifteen, so correlation becomes −0.6 rather than −1.8.
Common mistake: Interpreting correlation as a probability or as a complete measure of dependence overlooks its specific focus on standardized linear association.
38. Symmetric nonlinear dependence can have zero covariance
Zero covariance means the centered product averages to zero; it does not require the joint distribution to factor. A symmetric variable X and its square illustrate the distinction. When the relevant moments exist and X is symmetric about zero, E[X]=E[X³]=0, so Cov(X,X²)=0. Nevertheless, X² is determined by X and ordinarily is dependent on it. A separate factorization or conditional-distribution argument is required to establish independence. Special results linking zero covariance and independence, such as for jointly normal variables, require their full additional assumptions.
Worked example: Let X equal −2, zero, or two with equal probabilities and let Y=X². Then E[XY]=E[X³]=0 and E[X]=0, giving covariance zero. Yet observing Y=0 identifies X=0 exactly, demonstrating dependence through information rather than linear co-movement.
Common mistake: Applying the jointly normal zero-correlation implication to arbitrary distributions ignores the assumption that makes that special implication valid.
39. Iterated expectation averages conditional means
The law of iterated expectation states E[X]=E[E[X|Y]] when X is integrable. First calculate the mean within each conditional distribution, then average those means over Y’s distribution. This can replace a complicated joint calculation with simpler group-level calculations. The conditional mean E[X|Y] is itself a function of Y, not a single universal constant. The same principle applies to indicators: P(A)=E[P(A|Y)]. Independence is unnecessary, since the conditional means explicitly preserve any dependence between X and the grouping variable Y.
Worked example: A portfolio contains 30% group A policies with expected loss $400 and 70% group B policies with expected loss $100. The overall expected loss is 0.30(400)+0.70(100)=$190, without needing either group’s complete loss distribution.
Common mistake: Averaging group means without the group probabilities replaces the actual population distribution with an unintended equally weighted mixture.
40. Intersect comparison events with joint support
Probabilities such as P(X<Y), P(X+Y<c), or P(aX+bY>d) are integrals of the joint density over the event intersected with its support. Solve the inequalities for one variable to construct inner limits, then determine the outer range where those limits are compatible. Boundaries may change order, requiring a split integral. For ordinary joint densities, a boundary curve itself has probability zero, but discrete or singular models may behave differently. Independence permits replacing the joint density with a product, yet it does not eliminate the need to establish correct event limits.
Worked example: For independent uniform variables on (0,1), the event X+Y<0.6 is a triangle. Its probability is ∫₀^0.6∫₀^(0.6−x)1 dy dx=0.18. A rectangular integration region would include pairs whose sum exceeds the stated limit.
Common mistake: Integrating each variable over its marginal event separately cannot generally represent a coupled inequality involving both variables together.
Functions, combinations, and extremes of random variables
41. Monotone transformations require an inverse Jacobian
If Y=g(X) and g is differentiable and strictly monotone on X’s support, its density is fY(y)=fX(g⁻¹(y))|d g⁻¹(y)/dy| on the transformed support. The absolute derivative adjusts for stretching or compressing intervals and handles decreasing transformations. A CDF approach provides an alternative by translating {g(X)≤y} back into an event involving X. Always transform the support as well as the density formula. Simply substituting the inverse into fX omits the change in interval length and generally fails normalization.
Worked example: If X is uniform on (0,2) and Y=3X+1, then Y lies in (1,7). Its inverse is X=(Y−1)/3 with derivative 1/3, so fY(y)=(1/2)(1/3)=1/6 throughout that transformed interval.
Common mistake: Keeping the original support or omitting the inverse derivative fails to account for how the transformation redistributes probability across intervals.
42. Nonmonotone transformations sum valid inverse branches
When a transformation is not one-to-one, each transformed value may arise from several original values. For differentiable branches, sum fX(xᵢ)|dxᵢ/dy| over all inverse roots xᵢ that lie in X’s support. The number of valid roots can change with y, creating a piecewise density. Squaring is a common example: both positive and negative roots may contribute near zero, while one branch may disappear farther out. The CDF method is equally valid and often handles support restrictions more transparently than memorized transformation formulas.
Worked example: Let X be uniform on (−1,3) and Y=X². For 0<y<1, both roots contribute, giving density 1/(4√y). For 1<y<9, only the positive root remains, giving density 1/(8√y); the negative root lies outside the original support.
Common mistake: Doubling a single-root contribution everywhere assumes both inverse roots remain in the original support for every transformed value.
43. Linear combinations preserve means but require covariance
For L=aX+bY+c, the mean is aE[X]+bE[Y]+c without any independence assumption. Its variance is a²Var(X)+b²Var(Y)+2abCov(X,Y). The constant c shifts the mean without affecting variability. Independence makes the covariance term zero, but uncorrelatedness alone is sufficient for that particular simplification. A negative coefficient can make positive covariance reduce the variance of a difference. In combinations of more variables, include a covariance term for each distinct pair. Keep coefficient signs in cross terms and square coefficients only on individual variance terms.
Worked example: Two costs have variances nine and sixteen and covariance three. For their difference X−Y, variance is 9+16−2(3)=19. For their sum, variance is 9+16+2(3)=31, even though the two individual variances are identical in both calculations.
Common mistake: Subtracting variances when subtracting random variables ignores squared coefficients and usually understates the variability of the resulting difference.
44. Discrete convolution adds every compatible pair
For independent discrete variables, the sum S=X+Y has mass P(S=s)=Σx pX(x)pY(s−x), with terms restricted to valid support values. Each term represents one mutually exclusive way to obtain the target total. If the variables are dependent, replace the product by the corresponding joint probability. Build the support of the sum before calculating its masses, and check that the resulting distribution totals one. Convolution is useful even when the sum has no familiar named distribution. It also clarifies why multiplying just one selected pair misses other routes to the same total.
Worked example: Independent X and Y each equal zero or one, with success probabilities 0.3 and 0.6. Their sum equals one through (1,0) or (0,1), so its probability is 0.3(0.4)+0.7(0.6)=0.54, not either individual product.
Common mistake: Multiplying probabilities for only one decomposition of a total omits other disjoint pairs that produce the same sum.
45. Independent binomial sums need a common success rate
If X and Y are independent binomial variables with trial counts n and m and the same success probability p, then X+Y is binomial with parameters n+m and p. The combined count simply pools independent Bernoulli trials sharing one success rate. If the success probabilities differ, the total generally is not binomial, although its mean and variance still add under independence. If X and Y share trials or otherwise are dependent, even a common p is insufficient. Establish the underlying trial structure before invoking the named closure rule.
Worked example: Two independent teams inspect five and seven items, with each item independently defective with probability 0.1. Their combined defect count is binomial with twelve trials and probability 0.1. Its mean is 1.2 and variance is 12(0.1)(0.9)=1.08.
Common mistake: Averaging different binomial success probabilities and declaring the total binomial generally changes its distribution and variance.
46. Poisson sums add means under independence
Independent Poisson variables with means λ₁ and λ₂ have a Poisson sum with mean λ₁+λ₂. Unlike binomial closure, their individual means need not match. The resulting variance also equals the summed mean. Independence is essential to the usual closure statement; common shocks can make marginal Poisson counts dependent and change the total distribution. A related identity is that X conditional on X+Y=n is binomial with parameters n and λ₁/(λ₁+λ₂). This converts an allocation problem with a known total into a familiar conditional count model.
Worked example: Independent arrival streams have hourly Poisson means two and three. Their combined hourly count is Poisson with mean five. Given a total of four arrivals, the number from the first stream is binomial with four trials and success probability 2/5.
Common mistake: Adding Poisson parameters without checking independence can overlook shared events or common causes that alter the total count distribution.
47. Independent normal differences remain normal
A linear combination of independent normal variables is normal. For D=X−Y, its mean is μX−μY and its variance is σX²+σY², so comparison probabilities reduce to standardizing D. More generally, jointly normal variables allow covariance terms in the variance formula. Normal marginal distributions alone do not guarantee that every linear combination is normal when dependence is unspecified. Means carry coefficient signs, but individual variance contributions use squared coefficients. This method avoids a two-dimensional probability integral when the joint normal structure or independence assumptions justify it.
Worked example: Independent normal completion times have means ten and twelve minutes and standard deviations two and three minutes. Their difference is normal with mean −2 and variance thirteen. Thus P(X>Y)=P(Z>2/√13), using the standard deviation √13.
Common mistake: Subtracting standard deviations or variances for a difference ignores that independent sources of random variation still accumulate.
48. Minimum distributions are easiest through survival
For M=min(X₁,…,Xₙ), the event M>t requires every observation to exceed t. Independence therefore gives P(M>t)=∏Sᵢ(t), and FM(t)=1−∏Sᵢ(t). For identically distributed observations this becomes 1−[1−F(t)]ⁿ. Independent exponential minima are exponential with rate equal to the sum of rates. Independence must be stated or otherwise justified before multiplying survivals. This construction models first failure or earliest completion; it differs from requiring only one observation to survive past t, which concerns a maximum.
Worked example: Three independent lifetimes have survival probability 0.8 at five years. Their minimum exceeds five years with probability 0.8³=0.512. Therefore the probability of at least one failure by five years is 1−0.512=0.488 under the stated independence assumption.
Common mistake: Multiplying CDFs for a minimum calculates the probability that all observations are small, which instead describes a maximum event.
49. Maximum distributions multiply cumulative probabilities
For M=max(X₁,…,Xₙ), the event M≤t requires all observations to be at most t. For independent variables, FM(t)=∏Fᵢ(t); with identical continuous distributions, the density is n[F(t)]ⁿ⁻¹f(t). Maximum quantiles can be found by solving [F(t)]ⁿ=q. The maximum becomes stochastically larger as more observations are included, even though each individual distribution stays unchanged. Distinguish a largest observation from a sum or an average, since these statistics have different supports and distribution rules. Dependence invalidates the product formula unless additional structure justifies it.
Worked example: Four independent values are uniform on (0,10). Their maximum has CDF (t/10)⁴ for 0≤t≤10. Its median solves (t/10)⁴=0.5, giving t=10(0.5)^(1/4), which exceeds the individual median of five.
Common mistake: Using the individual CDF as the maximum’s CDF ignores the requirement that every observation satisfy the cutoff simultaneously.
50. Comparison probabilities can integrate one conditional tail
For independent continuous variables, P(X>Y)=∫S X(y)fY(y)dy, where S X denotes X’s survival function. Conditioning on Y=y turns the comparison into a one-variable tail probability, then averaging removes the condition. Equivalently, integrate FX(x)fY(x) for P(X<Y) with the appropriate variable roles. For independent exponentials, P(X<Y)=λX/(λX+λY), linking comparison probabilities to competing event rates. Discrete comparisons require care with ties, because strict and non-strict inequalities can differ. For dependent variables, use the conditional distribution rather than a marginal tail.
Worked example: Independent exponential waits have rates two and five per hour. The probability that the first wait finishes before the second is ∫₀∞2e^(−2t)e^(−5t)dt=2/7. The faster second rate makes the first finish less often.
Common mistake: Comparing means alone cannot generally determine the probability that one random variable exceeds another; their full distributions matter.
Insurance loss and payment applications
51. Translate an ordinary deductible into a payment function
For a nonnegative loss X and ordinary deductible d, an unlimited insurer payment is Y=(X−d)₊=max(X−d,0). Losses at or below d generate zero payment; larger losses generate only the excess above d. This piecewise function is the starting point for payment probabilities and moments. It differs from subtracting d from every loss, which would produce impossible negative payments, and from paying the entire loss once a threshold is crossed. When additional contract terms apply, write their precise order explicitly before combining them with the deductible.
Worked example: With a $400 ordinary deductible, losses of $250, $400, and $1,100 produce payments of zero, zero, and $700. A distribution calculation must map each possible loss through this function before averaging or calculating payment probabilities.
Common mistake: Subtracting the deductible from the expected loss ignores zero-payment losses and can produce an incorrect or even negative expected payment.
52. Positive-payment probability is a loss survival probability
With an ordinary deductible d and a positive insurer participation rate and payment cap, a positive payment occurs when X>d. Thus P(Y>0)=S X(d)=1−FX(d). If the loss distribution has mass exactly at d, that mass produces zero payment and must remain excluded. For a continuous loss, endpoint equality has zero probability, but the strict inequality remains the correct event. This probability measures payment frequency per modeled loss; it does not describe payment severity or the number of losses an exposure produces. Those are separate model components.
Worked example: A loss is uniform from zero to $5,000 and has a $1,200 ordinary deductible. Positive payment probability is (5,000−1,200)/5,000=0.76. Adding 50% coinsurance leaves this probability unchanged, provided the insurer’s payment cap is strictly positive.
Common mistake: Reducing payment probability by the coinsurance percentage confuses the size of positive payments with whether any payment occurs.
53. Payment per loss differs from payment per payment
A payment-per-loss distribution includes zero payments, whereas a payment-per-payment distribution conditions on Y>0. For nonnegative Y, E[Y]=P(Y>0)E[Y|Y>0] when the mean exists and positive payment is possible. Therefore the conditional positive-payment mean is the unconditional payment mean divided by payment frequency. The same relationship applies to positive raw moments, but variances require recomputing both first and second moments. Always identify the denominator population: all modeled losses or only losses generating insurer payments. Neither quantity automatically equals the conditional underlying loss before subtracting a deductible.
Worked example: Suppose 40% of losses generate a payment and expected payment across all losses is $300. The average positive payment is $300/0.40=$750. The $300 figure remains appropriate when costing all losses, including those below the deductible.
Common mistake: Using a conditional positive-payment mean as expected payment per loss omits the chance of a zero payment and overstates cost.
54. Expected deductible payment equals integrated survival
For nonnegative loss X, expected unlimited payment above deductible d is E[(X−d)₊]=∫d∞S X(x)dx when finite. This follows by expressing a nonnegative expectation as area under its survival function. An equivalent density calculation is ∫d∞(x−d)fX(x)dx for a continuous loss. The survival form automatically handles zero-payment losses and often simplifies exponential or piecewise models. It is an unconditional expected payment per loss, not a conditional positive-payment mean. To obtain the latter, divide by S X(d) when that probability is positive.
Worked example: For an exponential loss with mean $600 and deductible $150, survival is e^(−x/600). Expected payment per loss is ∫₁₅₀∞e^(−x/600)dx=600e^(−0.25). Dividing by e^(−0.25) gives a conditional positive-payment mean of $600.
Common mistake: Integrating survival from zero without shifting for the deductible calculates expected underlying loss rather than expected excess payment.
55. Deductibles and caps create payment point masses
A continuous loss need not produce a continuous payment distribution. For Y=min((X−d)₊,u), losses at or below d collapse to payment zero, creating mass FX(d). Losses at or above d+u collapse to payment u, creating mass P(X≥d+u). Between zero and u, payment follows the shifted loss distribution. For a continuous loss the upper mass is S X(d+u). A complete payment distribution must include both endpoint masses and its interior density. Treating it as an ordinary continuous density alone loses probability and distorts moments.
Worked example: Let X be uniform from zero to $10,000, with deductible $2,000 and maximum payment $3,000. Payment has mass 0.20 at zero and 0.50 at $3,000. Its remaining probability 0.30 is spread continuously between those endpoints.
Common mistake: Assuming continuous losses imply continuous payments overlooks the many loss values mapped to the same deductible or cap boundary.
56. Coinsurance and limits depend on their stated order
Coinsurance scales the insurer’s share of a covered amount, but the meaning of a limit must be specified. A maximum insurer payment u after coinsurance gives Y=min(c(X−d)₊,u). A limit u on the deductible-adjusted covered loss before coinsurance gives Y=c min((X−d)₊,u). These contracts reach their caps at different loss levels and have different maximum payments. With no cap, multiplying payment by c multiplies its mean by c and variance by c². Positive c does not change whether an ordinary-deductible payment is positive.
Worked example: For loss $10,000, deductible $1,000, coinsurance 80%, and limit $5,000, an insurer-payment cap produces $5,000. A covered-excess-loss limit produces 0.8($5,000)=$4,000. Writing the two formulas exposes the distinction before any expectation calculation.
Common mistake: Applying an unspecified limit before or after coinsurance by habit can change both the cap threshold and maximum insurer payment.
57. Payment variance requires squared payment moments
Once the payment function g is specified, compute E[Y]=E[g(X)] and E[Y²]=E[g(X)²], then subtract E[Y]² to obtain variance. For continuous losses, integrate the squared payment, not merely the loss squared over the paying region. An alternative for nonnegative payments is E[Y²]=2∫₀∞yP(Y>y)dy. Caps make the upper integration range finite. Zero-payment losses remain part of unconditional moments even though they contribute zero to sums. Conditional payment variance requires dividing raw moments by positive-payment probability before calculating the conditional squared mean.
Worked example: Suppose payment equals zero with probability 0.6 and $1,000 with probability 0.4. Its mean is $400 and second moment is 400,000 dollars squared. Variance is 400,000−400²=240,000 dollars squared, although every positive payment has the same size.
Common mistake: Squaring expected payment does not give the second moment; variability comes from the difference between those two quantities.
58. Inflation changes a fixed deductible’s effective level
If every loss is multiplied by a positive inflation factor a while deductible d stays fixed, payment becomes (aX−d)₊=a(X−d/a)₊. Positive-payment probability becomes P(X>d/a), so inflation can increase both frequency and severity of payments. Expected payment is aE[(X−d/a)₊], not generally a times the original expected payment at deductible d. If losses, deductibles, and payment limits all scale together, the entire payment scales by a and variance scales by a². State exactly which monetary terms change before applying any scaling shortcut.
Worked example: A loss is uniform from zero to $4,000 with a fixed $1,000 deductible. After losses rise by 25%, positive-payment probability becomes P(X>800)=0.80, compared with 0.75 originally. The deductible remains unchanged in dollars but falls relative to loss size.
Common mistake: Multiplying original expected payment by the inflation factor assumes contract terms scale with losses, which a fixed deductible does not.
59. Deductible sensitivity equals negative survival
Let m(d)=E[(X−d)₊]=∫d∞S X(x)dx. Where differentiable, m′(d)=−S X(d). A small deductible increase therefore reduces expected payment approximately by the increase multiplied by the probability of exceeding the original deductible. For a finite change from d₁ to d₂, the exact reduction is ∫d₁ᵈ₂S X(x)dx. Since survival is nonincreasing, expected payment decreases at a diminishing absolute rate as the deductible rises. This result concerns unlimited ordinary-deductible payment; additional caps or changing participation terms require a payment-specific calculation.
Worked example: At a deductible of $500, suppose loss survival is 0.30 and changes little over the next $10. Raising the deductible to $510 reduces expected payment by approximately $10(0.30)=$3 per loss, rather than by $10 across every loss.
Common mistake: Assuming a deductible increase reduces every expected payment dollar-for-dollar ignores losses that already produce no payment below the deductible.
60. A capped layer is a difference of stop-loss payments
A payment layer attaching at deductible d with width u has Y=min((X−d)₊,u). Pointwise, Y=(X−d)₊−(X−d−u)₊. Consequently, expected layer payment is the difference between two unlimited excess-loss expectations, or ∫dᵈ⁺ᵘS X(x)dx. This representation avoids separate uncapped and capped density integrals and applies to nonnegative discrete, continuous, or mixed losses. The upper attachment point is d+u, not u alone. If coinsurance follows the layer calculation, multiply the resulting layer payment by the stated participation fraction.
Worked example: For an exponential loss with mean $1,000, a layer above $500 with width $2,000 has expected payment 1,000e^(−0.5)−1,000e^(−2.5). The two terms subtract the excess above $2,500 from the excess above $500, leaving only the covered layer.
Common mistake: Subtracting stop-loss expectations at d and u treats the layer width as its upper loss boundary and misplaces the coverage.