Dependence Traps In Joint Risk Calculations

Why Dependence Matters

In a portfolio or a multi-step probability problem, knowing each risk separately does not always tell you how the risks behave together. For events $A$ and $B$, their marginal probabilities $P(A)$ and $P(B)$ do not, by themselves, determine $P(A\cap B)$. Similarly, the means and variances of random variables do not, by themselves, determine the variance of their sum. Those calculations depend on how the variables relate.

This distinction is important in actuarial calculations. If two hypothetical claims can respond to the same event, their occurrences may be dependent. The direction and size of that dependence must come from a model or additional information; it cannot be inferred just from the fact that the claims share a possible cause. Multiplying marginals or adding variances without checking the conditions can produce a precise-looking but unjustified answer.

Definition: Events $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$. For events with positive probabilities, this is equivalent to $P(A\mid B)=P(A)$. Random variables $X$ and $Y$ are independent if their joint distribution factors into their marginal distributions. Independence implies $\operatorname{Cov}(X,Y)=0$ when the variables have finite variances; the converse need not hold.

Joint Probabilities: What the Marginals Do and Do Not Tell You

If independence is established, multiplying marginal probabilities gives the joint probability. Without independence, the product $P(A)P(B)$ is only a comparison value, not necessarily the answer. The possible values of the intersection are constrained by the marginal probabilities:

\max\{0,\ P(A)+P(B)-1\} \leq P(A\cap B) \leq \min\{P(A),P(B)\}.

The upper bound follows because $A\cap B$ is contained in both $A$ and $B$. For the lower bound, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ cannot exceed 1, and an intersection probability cannot be negative. Rearranging gives the stated bounds. If the marginals are known but no dependence information is given, any answer for the joint event must at least be consistent with these limits.

Worked Example: A Shared Risk Regime

A hypothetical pair of claims is exposed to a severe-weather regime $W$, where $P(W)=0.10$. Conditional on severe weather, each claim occurs with probability 0.20; conditional on no severe weather, each occurs with probability 0.02. Assume the two claims are conditionally independent within each regime. Find the probability both occur, compare it with the product of their marginal probabilities, and calculate the probability that at least one occurs.

Method. First find each marginal by conditioning on the regime. Then use the stated conditional independence to calculate the joint probability within each regime and combine the results.

For either claim, the marginal probability is $0.10(0.20)+0.90(0.02)=0.020+0.018=0.038$. Thus the product of the marginals is $0.038^2=0.001444$.

Conditional independence gives a probability $0.20(0.20)=0.04$ that both claims occur during severe weather, and $0.02(0.02)=0.0004$ that both occur without severe weather. Therefore, $P(\text{both})=0.10(0.04)+0.90(0.0004)=0.004+0.00036=0.00436$.

The joint probability $0.00436$ is greater than the product $0.001444$. Finally, by inclusion-exclusion, $P(\text{at least one})=0.038+0.038-0.00436=0.07164$.

Check. The joint probability $0.00436$ is below either marginal $0.038$, as required. The probability of at least one is above $0.038$ and below $0.076$, also consistent with the event bounds. The increase in joint probability here follows from the specified conditional independence and the higher occurrence probabilities in the shared regime; a shared regime alone does not guarantee this direction.

That last qualification matters. Even if both individual risks are more likely during severe weather, their joint probability need not exceed the product of their unconditional marginals. For instance, keep the same regime probabilities and conditional marginal probabilities as in the example, but make the two claims mutually exclusive within each regime. Their unconditional marginals remain 0.038 each, while their joint probability is zero. The product of marginals is still 0.001444. The dependence structure within the regimes changes the answer, so the direction of dependence cannot be inferred from the regime description alone.

Aggregate Variance Includes Covariance

For two random variables with finite variances, expanding the square in the definition of variance gives $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)$. The covariance term records whether high values of one variable tend to occur with high or low values of the other. For a finite sum, every pair contributes:

\operatorname{Var}\left(\sum_{i=1}^{n}X_i\right) = \sum_{i=1}^{n}\operatorname{Var}(X_i) + 2\sum_{1\leq i<j\leq n}\operatorname{Cov}(X_i,X_j).

To see why, write the variance as the expectation of the squared deviation: $\operatorname{Var}(\sum_i X_i)=E[(\sum_i(X_i-E[X_i]))^2]$. Expanding the square produces one squared-deviation term for each variable and two copies of each cross-product. The expected squared-deviation terms are the individual variances; each expected cross-product is $\operatorname{Cov}(X_i,X_j)$. This yields the formula. If the variables are independent in pairs, all these covariances are zero, and the variances add. Without that condition—or another reason each covariance is zero—one cannot simply add the variances.

Theorem: For $T=\sum_{i=1}^{n}X_i$, if all variances are finite, then $E[T]=\sum_i E[X_i]$, regardless of dependence. Its variance includes the pairwise covariance terms shown above. Knowing the marginal means and variances is enough for the expectation of the sum, but not generally enough for its variance.

Worked Example: Aggregate Payment From Two Claims

Two hypothetical claim indicators $I_1$ and $I_2$ are Bernoulli random variables: $P(I_1=1)=0.20$, $P(I_2=1)=0.30$, and $P(I_1=1,I_2=1)=0.10$. The first claim pays $100 if it occurs; the second pays $200. Find the expected aggregate payment and its variance.

Method. Represent each payment as its amount multiplied by its claim indicator. Use the given joint probability to calculate covariance; do not assume the indicators are independent.

The aggregate payment is $T=100I_1+200I_2$. Its expectation is $E[T]=100(0.20)+200(0.30)=20+60=\$80$.

The individual variances are $\operatorname{Var}(100I_1)=100^2(0.20)(0.80)=10{,}000(0.16)=1{,}600$ and $\operatorname{Var}(200I_2)=200^2(0.30)(0.70)=40{,}000(0.21)=8{,}400$. For Bernoulli indicators, $\operatorname{Cov}(I_1,I_2)=E[I_1I_2]-E[I_1]E[I_2]=0.10-(0.20)(0.30)=0.04$. Thus the payment covariance is $\operatorname{Cov}(100I_1,200I_2)=100(200)(0.04)=800$.

Therefore, $\operatorname{Var}(T)=1{,}600+8{,}400+2(800)=11{,}600$ dollars squared.

Check. The implied joint probability is feasible: the four indicator outcomes have probabilities 0.10 for both claims, 0.10 for only the first, 0.20 for only the second, and 0.60 for neither. They are nonnegative and sum to 1. The positive covariance increases aggregate variance above the sum $1{,}600+8{,}400=10{,}000$. The common mistake is to omit the covariance term and report $10{,}000$.

Same Marginals, Different Portfolio Risk

A useful way to test whether information is sufficient is to ask whether different joint models could have the same stated marginals. If they can, but produce different answers, the desired quantity is not determined by the marginals alone.

Worked Example: Three Dependence Structures, Three Variances

Two hypothetical claim indicators $I_1$ and $I_2$ each have marginal probability 0.50. Let the aggregate claim count be $N=I_1+I_2$. Calculate $E[N]$ and $\operatorname{Var}(N)$ under (a) independence, (b) perfect positive dependence with $I_1=I_2$, and (c) mutual exclusivity.

Method. In each model, specify the joint outcomes. The marginals determine the mean $E[N]=E[I_1]+E[I_2]=0.50+0.50=1$, but the distribution of the sum—and therefore its variance—depends on the joint outcomes.

(a) Independent indicators. Each of the four pairs $(0,0),(0,1),(1,0),(1,1)$ has probability 0.25. Thus $N$ is 0, 1, or 2 with probabilities 0.25, 0.50, and 0.25. Its second moment is $E[N^2]=0^2(0.25)+1^2(0.50)+2^2(0.25)=0+0.50+1=1.50$. Therefore $\operatorname{Var}(N)=1.50-1^2=0.50$.

(b) Perfect positive dependence. The outcomes $(0,0)$ and $(1,1)$ each have probability 0.50. Hence $N$ is 0 or 2, each with probability 0.50. Then $E[N^2]=0^2(0.50)+2^2(0.50)=2$, so $\operatorname{Var}(N)=2-1^2=1$.

(c) Mutual exclusivity. The outcomes $(1,0)$ and $(0,1)$ each have probability 0.50. Then $N=1$ with probability 1, so $E[N]=1$, $E[N^2]=1$, and $\operatorname{Var}(N)=1-1^2=0$.

Check. All three models have the same individual claim probabilities and the same aggregate mean, but their aggregate variances are 0, 0.50, and 1. A variance calculation cannot be selected from the marginal probabilities alone.

Common Dependence Traps

Multiplying marginals automatically. The calculation $P(A\cap B)=P(A)P(B)$ requires independence. Otherwise, use joint or conditional information, or report that the probability is not determined.
Assuming a shared cause fixes the direction. A common regime can raise the joint probability under a specified model, such as conditional independence with both risks elevated in that regime. But a different conditional dependence structure can produce a smaller joint probability. State the assumptions that justify the conclusion.
Adding variances automatically. The variance of a sum includes twice each pairwise covariance. Variances add when the relevant covariances are zero, including when the variables are independent; they do not add merely because the individual variances are known.
Confusing zero covariance with independence. Zero covariance is enough to remove the covariance term for a two-variable variance calculation, but it does not generally establish independence or justify multiplying event probabilities.
Assuming the expectation also needs independence. For a finite sum with finite expectations, $E[\sum_i X_i]=\sum_i E[X_i]$ regardless of dependence. It is the variance and joint-event calculations that require dependence information.

Key takeaway: Before calculating a joint probability, ask what establishes independence or supplies the joint distribution. Before calculating an aggregate variance, identify the covariance terms. If the problem gives only marginals, check whether the requested quantity is actually determined.

Check Your Understanding

For each problem, identify the dependence information being used and check whether the requested result follows from the stated model.

Events $A$ and $B$ have probabilities 0.27 and 0.41. Give the smallest and largest possible values of $P(A\cap B)$ based on these marginals alone. (Answer: 0.0000 to 0.2700)
A regime $R$ has probability 0.20. Two events are conditionally independent given $R$, with conditional probabilities 0.30 and 0.10. Given $R^c$, they are conditionally independent with probabilities 0.05 and 0.04. Find their unconditional joint probability and compare it with the product of their marginals. (Answer: joint probability $0.0076$; product of marginals $0.0052$)
Two claim indicators have probabilities 0.40 and 0.50 of occurring, and their joint occurrence probability is 0.25. The first payment is $50 and the second is $100. Find the expected aggregate payment and its variance. (Answer: expectation $\$70$; variance $3{,}600$ dollars squared)
Random variables $X_1,X_2,X_3$ have means 5, 8, and 2; variances 4, 9, and 16; and pairwise covariances $\operatorname{Cov}(X_1,X_2)=1$, $\operatorname{Cov}(X_1,X_3)=-2$, and $\operatorname{Cov}(X_2,X_3)=3$. Find the mean and variance of their sum. (Answer: mean $15$; variance $33$)
Two Bernoulli claim indicators each have probability 0.50. In one model they are independent; in another they are mutually exclusive. For the aggregate count $N$, calculate the variance under each model and state whether the marginals alone determine it. (Answer: independent model $0.50$; mutually exclusive model $0$; the marginals alone do not determine the variance)

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