Checking A Multi-Topic Probability Solution

Why Audit a Probability Solution?

In an integrated probability problem, a calculation can be arithmetically correct and still answer the wrong question. A density may omit a point mass, a gamma parameter may be read as a scale instead of a rate, or two events may be multiplied as if independent when the problem does not justify that assumption. Each error can produce a neat-looking result.

A useful audit does not begin by redoing every line. It first checks whether the model and its assumptions are coherent, then checks the event being calculated, and only then checks the arithmetic. This tutorial develops that approach for mixed distributions, parameterized models, dependence, and normal approximations. The methods are aligned with the SOA Exam P (Probability) syllabus.

Definition: An audit of a probability solution is a sequence of checks on its model, support, total probability, parameterization, dependence assumptions, event boundaries, approximation conditions, and numerical work. A result is acceptable only if these checks agree with the problem statement.

A Practical Audit Order

Work from the model outward. Confirm what values the random variable can take and whether its distribution assigns probability at isolated points as well as across intervals. Check that all probability mass totals one. Confirm the meaning of every parameter before using a formula. Then verify the event and any assumptions, such as independence. For an approximation, check its conditions before calculating a standardized value.

Check the model and support.
State the possible values of each random variable and identify whether its distribution is discrete, continuous, or mixed.

Check total probability.
Add point masses and integrate density over its support. The result must be one.

Check parameters and assumptions.
For example, distinguish a gamma rate from a scale, and verify independence rather than assuming it.

Check boundaries and numerical method.
Determine whether an endpoint has an atom. For a normal approximation to a count, check adequacy and use a continuity correction.

Mixed Distributions: Mass Plus Density

A mixed random variable can have positive probability at particular points and also have a continuous component. Its probability on an event must include both contributions. A density describes the continuous component; it does not by itself describe any point masses.

Theorem: If a random variable has point masses $p_j=P(X=a_j)$ and a density $f_X$ for its continuous component, then for an event $C$, $$ P(X\in C)=\sum_{j:a_j\in C}p_j+\int_{C} f_X(x)\,dx. $$ The continuous density is integrated only over the continuous part of the distribution's support. The point masses and the integral together must total one.

This formula follows by splitting the event into its disjoint point outcomes and its continuous outcomes. Probability is additive over disjoint events, so the probability of $C$ is the sum of the probabilities at its atoms plus the probability contributed by the continuous component. Integrating a density gives the latter probability; it cannot recover an atom, because an integral over a single point is zero.

Worked Example: Auditing a Mixed Distribution at an Endpoint

A hypothetical claim severity $X$, measured in thousands of dollars, has probability $0.25$ of being zero. Conditional on being positive, it is not given a conditional density; instead, the unconditional density on $0<x<3$ is $f_X(x)=0.25$. There is no probability outside the stated atom and interval. Find $P(X\leq 1)$, $P(X=1)$, and check that the model is normalized. No interest rate or mortality assumption is involved.

Method and support check. The model has an atom at $0$, continuous density on $(0,3)$, and no other mass. The total probability is $$ P(X=0)+\int_0^3 f_X(x)\,dx =0.25+\int_0^3 0.25\,dx =0.25+0.75=1. $$ Thus it is normalized.

For $X\leq1$, include the atom at zero and integrate the density from just above zero through 1: $$ P(X\leq1)=0.25+\int_0^1 0.25\,dx =0.25+0.25=0.50. $$ There is no atom at 1, so $P(X=1)=0$. The integral over a single endpoint is zero. The result checks: $P(X\leq1)$ contains the entire atom and one-third of the continuous mass, giving $0.25+0.25=0.50$.

Common mistake avoided: Treating $f_X$ as the entire distribution gives $0.25$ for $P(X\leq1)$ and misses the atom. Conversely, adding an atom at 1 would invent probability not included in the model.

Parameterization: Read the Formula, Not Just the Name

A distribution name alone may not specify how its parameters are defined. For a gamma model, one standard convention uses shape $\alpha$ and rate $\lambda$; another uses shape and scale $\theta$. The rate and scale are reciprocals. Confusing them changes the mean and variance, even though the shape parameter is unchanged.

Definition: Under the shape-rate convention, $X\sim\operatorname{Gamma}(\alpha,\lambda)$ has density $$ f_X(x)=\frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)},\qquad x>0, $$ where $\alpha>0$ is the shape and $\lambda>0$ is the rate. Its mean is $\alpha/\lambda$ and its variance is $\alpha/\lambda^2$. Under the shape-scale convention, the scale is $\theta=1/\lambda$, so the mean is $\alpha\theta$.

Worked Example: Auditing a Gamma Rate Calculation

A hypothetical processing time $X$, in hours, has a gamma distribution with shape $\alpha=2$ and rate $\lambda=3$ per hour. Find $E[X]$, $\operatorname{Var}(X)$, and $P(X>1)$. No interest rate or mortality assumption is involved.

Method and parameter check. The problem says rate $3$, so the mean is shape divided by rate: $$ E[X]=\frac{2}{3}\text{ hours},\qquad \operatorname{Var}(X)=\frac{2}{3^2}=\frac{2}{9}\text{ hours}^2. $$ For shape 2, the density is $f_X(x)=9xe^{-3x}$ for $x>0$. Integrating the density from 1 to infinity gives the survival probability: $$ P(X>1)=\int_1^\infty 9xe^{-3x}\,dx. $$ An antiderivative is $-(3x+1)e^{-3x}$, since its derivative is $9xe^{-3x}$. Therefore, $$ P(X>1)=\left[-(3x+1)e^{-3x}\right]_1^\infty =0-(-4e^{-3})=4e^{-3}\approx0.1991. $$ The probability is between zero and one, and it is plausible that the probability of exceeding one hour is below one-half when the mean is $2/3$ hour.

Common mistake avoided: If 3 were a scale, the mean would be $2(3)=6$, not $2/3$. That would be a different distribution. Always identify the density or explicitly stated rate/scale convention before using formulas.

Independence Must Be Checked

The multiplication rule $P(A\cap B)=P(A)P(B)$ applies when $A$ and $B$ are independent. Knowing the two marginal probabilities is not enough to establish independence. If the joint probability is provided, compare it with the product of the marginals. If the full joint distribution is available, its cells should also be nonnegative and sum to one.

Worked Example: Detecting an Unsupported Independence Assumption

In a hypothetical portfolio, let $A$ be the event that a policy has a water claim and $B$ the event that it has a theft claim during a specified month. Suppose $P(A)=0.40$, $P(B)=0.50$, and $P(A\cap B)=0.30$. Find $P(B\mid A)$ and $P(A\cup B)$, and determine whether the events are independent. No interest rate or mortality assumption is involved.

Method and dependence check. The joint probability $0.30$ is supplied; do not replace it with a product without checking independence. The conditional probability is $$ P(B\mid A)=\frac{P(A\cap B)}{P(A)} =\frac{0.30}{0.40}=0.75. $$ The union probability follows from inclusion-exclusion: $$ P(A\cup B)=P(A)+P(B)-P(A\cap B) =0.40+0.50-0.30=0.60. $$ For independence, compare the joint probability with the product: $$ P(A)P(B)=0.40(0.50)=0.20\ne0.30=P(A\cap B). $$ Thus $A$ and $B$ are dependent. The joint model is coherent: the four cell probabilities are $0.30$ for both, $0.10$ for $A$ only, $0.20$ for $B$ only, and $0.40$ for neither; they are nonnegative and sum to 1.

Common mistake avoided: Multiplying the marginal probabilities would produce $0.20$ for the intersection, contradicting the supplied joint probability. In a solution audit, an explicit joint value takes precedence over an unsupported independence assumption.

Normal Approximation: Check Conditions and Boundaries

A binomial random variable $X\sim\operatorname{Binomial}(n,p)$ counts successes in $n$ independent trials, each with the same success probability $p$. Its mean is $np$ and variance is $np(1-p)$. A normal approximation uses a normal distribution with this same mean and variance. It is more credible when both expected counts, $np$ and $n(1-p)$, are sufficiently large; a common exam-level screening rule is that each is at least 5. This check is a guideline, not a guarantee of exact accuracy.

Because a binomial variable takes integer values and a normal variable is continuous, use a continuity correction. For example, $P(a\leq X\leq b)$ is approximated by the normal probability from $a-0.5$ to $b+0.5$. The half-unit adjustment aligns the continuous interval with the integer outcomes being counted.

Worked Example: Auditing a Binomial Normal Approximation

A hypothetical sensor independently flags a defective component with probability $p=0.25$. In a batch of $n=80$ components, let $X$ be the number flagged. Approximate $P(16\leq X\leq26)$ with a normal model, and assess whether the approximation conditions are reasonable. There is no interest rate or mortality assumption.

Method and condition check. The model is $X\sim\operatorname{Binomial}(80,0.25)$, with independent trials and constant probability $0.25$. The expected counts are $$ np=80(0.25)=20,\qquad n(1-p)=80(0.75)=60. $$ Both exceed 5, so the usual screening check is satisfied. The mean and standard deviation are $$ \mu=np=20,\qquad \sigma=\sqrt{np(1-p)}=\sqrt{80(0.25)(0.75)}=\sqrt{15}\approx3.873. $$ Apply the continuity correction: approximate the desired probability by $P(15.5<Y<26.5)$, where $Y\sim N(20,15)$. Standardizing gives $$ z_{\text{lower}}=\frac{15.5-20}{\sqrt{15}}\approx-1.1619,\qquad z_{\text{upper}}=\frac{26.5-20}{\sqrt{15}}\approx1.6783. $$ Using standard normal probabilities, $\Phi(1.6783)\approx0.9534$ and $\Phi(-1.1619)\approx0.1227$. Hence $$ P(16\leq X\leq26)\approx0.9534-0.1227=0.8307. $$ The approximation is $0.8307$ to four decimal places. It is an approximation, not the exact binomial probability.

Common mistake avoided: Using the uncorrected limits 16 and 26 would omit part of the continuous area corresponding to the endpoint counts. Also, a normal calculation should not be accepted merely because a binomial distribution has a mean and variance; the expected-count check matters.

Final Audit Checks

Before accepting a multi-topic answer, compare each line with the model that licenses it. A compact final review is especially useful when the arithmetic looks reasonable but several ideas have been combined.

Support: Are impossible values excluded, and are all stated atoms included?
Normalization: Do all point masses plus integrated density sum to one?
Parameterization: Are rate, scale, and distribution parameters interpreted according to the stated convention?
Dependence: Is independence given, established, or incorrectly assumed?
Boundaries: Does an endpoint contain an atom, or is its probability zero under a continuous component?
Approximation: Are the conditions reasonable, and has a continuity correction been used for integer counts?

These checks are not a substitute for solving the problem. They are a way to make sure the chosen model, event, and method actually fit the data in the question. In exam work, a brief statement of the model and the relevant assumption can prevent a correct-looking calculation from resting on an invalid step.

Check Your Understanding

For each problem, audit the model as well as the calculation.

A mixed variable $X$ has an atom of $0.30$ at zero and density $0.35$ on $0<x<2$, with no other mass. Check normalization and find $P(X\leq1)$ and $P(X=1)$. (Answer: total probability $0.30+0.35(2)=1$; $P(X\leq1)=0.65$; $P(X=1)=0$.)
A waiting time has a gamma distribution with shape 3 and rate 2 per hour. Find its mean and variance, and name the parameterization trap. (Answer: mean $3/2=1.5$ hours; variance $3/2^2=0.75$ hours squared; treating the rate as a scale would be incorrect.)
Suppose $P(A)=0.60$, $P(B)=0.50$, and $P(A\cap B)=0.40$. Find $P(B\mid A)$, $P(A\cup B)$, and decide whether independence holds. (Answer: $P(B\mid A)=2/3$; $P(A\cup B)=0.70$; dependent, since $0.40\ne(0.60)(0.50)=0.30$.)
For $X\sim\operatorname{Binomial}(50,0.40)$, use a normal approximation with continuity correction for $P(X\geq24)$. (Answer: $\mu=20$, $\sigma=\sqrt{12}$; approximate $P(Y>23.5)=P(Z>1.0104)\approx0.1562$.)
A proposed normal approximation is used for $X\sim\operatorname{Binomial}(30,0.05)$. Does the usual expected-count screening rule support it? (Answer: No; $np=1.5<5$, so the normal approximation is not supported by that screening rule.)

What You'll Learn