Introduction
Risk aggregation often starts with a simple question: given two discrete losses, what can their joint probability law tell us about their separate distributions and their dependence? A joint table answers this directly, but only if its probabilities are read and summed carefully. In particular, the marginal distributions alone do not determine joint-event probabilities unless independence is also established.
This tutorial develops a systematic method for discrete paired-loss calculations. The examples use hypothetical one-period losses. They focus on translating a joint table into marginals and conditional laws, calculating covariance and correlation, and finding probabilities involving both losses or their sum.
A Reliable Calculation Method
Treat each cell as the probability of a pair of outcomes, not as a probability belonging to just one loss. A row total gives a marginal probability for the row’s \(X\) value; a column total gives a marginal probability for the column’s \(Y\) value. A conditional probability divides a joint cell by the relevant row or column total. The full table must have nonnegative entries that sum to 1.
Once the joint law is available, expectations can be calculated directly. For example, \(E[XY]\) is the sum of \(xyP(X=x,Y=y)\) across the cells. The marginal expectations \(E[X]\) and \(E[Y]\) can be calculated from the row and column totals. Then covariance is \(E[XY]-E[X]E[Y]\). This subtraction is essential: \(E[XY]\) alone is not the covariance.
The correlation formula requires both variances to be positive. Covariance has units equal to the product of the units of \(X\) and \(Y\); correlation has no units. A table can also be used to calculate an event such as \(P(X+Y\geq c)\): add the probabilities of exactly those cells whose outcomes satisfy the stated condition.
Worked Example: Marginals, a Conditional Probability, and Dependence
A hypothetical insurer models two one-period losses, \(X\) and \(Y\), in thousands of dollars. There is no discounting or mortality assumption; the joint discrete probability table below is the complete model. Find the marginal distributions, \(P(Y=3\mid X=5)\), \(P(X\geq2,Y=3)\), and the covariance and correlation.
| \(X\) (thousands) | \(Y=0\) | \(Y=3\) | Row total |
|---|---|---|---|
| \(X=0\) | 0.20 | 0.10 | 0.30 |
| \(X=2\) | 0.15 | 0.25 | 0.40 |
| \(X=5\) | 0.05 | 0.25 | 0.30 |
| Column total | 0.40 | 0.60 | 1.00 |
Method and table check. The row totals give \(P(X=0)=0.30\), \(P(X=2)=0.40\), and \(P(X=5)=0.30\). The column totals give \(P(Y=0)=0.40\) and \(P(Y=3)=0.60\). The entries are nonnegative and \(0.30+0.40+0.30=1.00\), so the table is a valid joint law.
Conditional and joint events. The conditional probability uses the \(X=5\) row total as its denominator: \(P(Y=3\mid X=5)=0.25/0.30=0.8333\) to four decimals. For \(P(X\geq2,Y=3)\), add the two cells with \(X=2\) or \(5\) and \(Y=3\): \(0.25+0.25=0.5000\).
Moments. From the marginals, \(E[X]=0(0.30)+2(0.40)+5(0.30)=2.30\) and \(E[Y]=0(0.40)+3(0.60)=1.80\). The nonzero contributions to \(E[XY]\) are \(2(3)(0.25)=1.50\) and \(5(3)(0.25)=3.75\), so \(E[XY]=5.25\). Therefore, \(\operatorname{Cov}(X,Y)=5.25-(2.30)(1.80)=1.11\), in squared thousands of dollars.
For the variances, \(E[X^2]=0^2(0.30)+2^2(0.40)+5^2(0.30)=9.10\), so \(\operatorname{Var}(X)=9.10-2.30^2=3.81\). Also \(E[Y^2]=3^2(0.60)=5.40\), so \(\operatorname{Var}(Y)=5.40-1.80^2=2.16\). Thus \(\operatorname{Corr}(X,Y)=1.11/\sqrt{(3.81)(2.16)}=1.11/\sqrt{8.2296}=0.3869\) to four decimals.
Check and common mistake. The positive covariance is consistent with the higher \(X\) values having substantial probability in the \(Y=3\) column. Do not multiply \(P(X\geq2)=0.70\) by \(P(Y=3)=0.60\) to get the joint probability: that would give 0.4200 and would require independence. The table gives the correct joint probability, 0.5000.
Why the Covariance Formula Works
The covariance formula is useful because it can be evaluated using the joint product moment and the two marginal means. Its justification follows by expanding the definition. For a discrete joint law with finite second moments, covariance is the expected product of the deviations from the means:
The third line uses linearity of expectation and the fact that \(E[X]\) and \(E[Y]\) are constants. The same reasoning applies whether the joint law is displayed in a table or written as a list of possible pairs and probabilities.
Worked Example: Conditional Law and a Joint Exceedance Event
A hypothetical risk team models two one-period losses \(X\) and \(Y\), in thousands of dollars. There is no discounting or mortality assumption. The following joint discrete distribution is the model. Find the conditional distribution of \(X\) given \(Y=5\), its conditional mean, \(P(X+Y\geq5)\), and the covariance and correlation.
| \(X\) (thousands) | \(Y=0\) | \(Y=2\) | \(Y=5\) | Row total |
|---|---|---|---|---|
| \(X=0\) | 0.10 | 0.10 | 0.05 | 0.25 |
| \(X=1\) | 0.10 | 0.20 | 0.10 | 0.40 |
| \(X=4\) | 0.05 | 0.15 | 0.15 | 0.35 |
| Column total | 0.25 | 0.45 | 0.30 | 1.00 |
Conditional law. The conditioning event \(Y=5\) has probability 0.30. Divide each entry in that column by 0.30: \(P(X=0\mid Y=5)=0.05/0.30=0.1667\), \(P(X=1\mid Y=5)=0.10/0.30=0.3333\), and \(P(X=4\mid Y=5)=0.15/0.30=0.5000\). The conditional probabilities sum to 1.0000. The conditional mean is \(E[X\mid Y=5]=0(0.1667)+1(0.3333)+4(0.5000)=2.3333\) thousand dollars, using the exact fractions \(1/6,1/3,1/2\).
Joint event. The condition \(X+Y\geq5\) is satisfied by \((0,5)\), \((1,5)\), \((4,2)\), and \((4,5)\). Adding their joint probabilities gives \(0.05+0.10+0.15+0.15=0.4500\). In particular, \((4,0)\) does not qualify; it is not enough for \(X\) alone to be large.
Covariance and correlation. The marginals give \(E[X]=1(0.40)+4(0.35)=1.80\) and \(E[Y]=2(0.45)+5(0.30)=2.40\). Summing \(xy\) times each cell probability gives \(E[XY]=0.90+4.20=5.10\): the \(X=1\) row contributes \(1(2)(0.20)+1(5)(0.10)=0.90\), and the \(X=4\) row contributes \(4(2)(0.15)+4(5)(0.15)=4.20\). Thus covariance is \(5.10-(1.80)(2.40)=0.78\).
Next, \(E[X^2]=1^2(0.40)+4^2(0.35)=6.00\), so \(\operatorname{Var}(X)=6.00-1.80^2=2.76\). Also \(E[Y^2]=2^2(0.45)+5^2(0.30)=9.30\), so \(\operatorname{Var}(Y)=9.30-2.40^2=3.54\). Therefore \(\operatorname{Corr}(X,Y)=0.78/\sqrt{(2.76)(3.54)}=0.78/\sqrt{9.7704}=0.2495\) to four decimals.
Check and common mistake. Conditional probabilities must be divided by \(P(Y=5)=0.30\), not by 1.00; using the joint column entries directly would not produce a conditional distribution. The positive covariance and correlation indicate positive linear co-movement, but do not imply that every high value of one loss pairs with a high value of the other.
Using the Joint Law to Analyze the Aggregate
For the aggregate loss \(Z=X+Y\), the joint table remains the safest source for probabilities. For example, to find \(P(Z\geq c)\), identify every pair whose sum reaches the threshold and add those cells. For a variance calculation, covariance provides a compact result:
This identity follows by expanding \(\operatorname{Var}(X+Y)=E[((X-E[X])+(Y-E[Y]))^2]\). The square has two squared-deviation terms and a cross-product term; taking expectations gives the two variances and twice the covariance. It applies when the relevant second moments are finite. Omitting the covariance term is justified only when covariance is zero, and assuming independence is stronger than simply having zero covariance.
Worked Example: Aggregate Loss and a Tail Probability
A hypothetical portfolio has two one-period losses \(X\) and \(Y\), measured in thousands of dollars, with the joint discrete law below. There is no discounting or mortality assumption. Calculate \(P(X+Y\geq8)\), the variance of \(X+Y\), and verify its mean and variance from the aggregate distribution.
| \(X\) (thousands) | \(Y=0\) | \(Y=2\) | \(Y=5\) | Row total |
|---|---|---|---|---|
| \(X=0\) | 0.10 | 0.15 | 0.15 | 0.40 |
| \(X=3\) | 0.15 | 0.10 | 0.05 | 0.30 |
| \(X=6\) | 0.05 | 0.10 | 0.15 | 0.30 |
| Column total | 0.30 | 0.35 | 0.35 | 1.00 |
Tail probability. The pairs meeting \(X+Y\geq8\) are \((3,5)\), \((6,2)\), and \((6,5)\), with aggregate values 8, 8, and 11. Hence \(P(X+Y\geq8)=0.05+0.10+0.15=0.3000\). The pair \((6,0)\) is excluded because its sum is only 6.
Variance using covariance. The marginal means are \(E[X]=3(0.30)+6(0.30)=2.70\) and \(E[Y]=2(0.35)+5(0.35)=2.45\). The product moment is \(E[XY]=3(2)(0.10)+3(5)(0.05)+6(2)(0.10)+6(5)(0.15)=7.05\), so \(\operatorname{Cov}(X,Y)=7.05-(2.70)(2.45)=0.435\).
The second moments are \(E[X^2]=9(0.30)+36(0.30)=13.50\) and \(E[Y^2]=4(0.35)+25(0.35)=10.15\). Thus \(\operatorname{Var}(X)=13.50-2.70^2=6.21\) and \(\operatorname{Var}(Y)=10.15-2.45^2=4.1475\). Therefore, \(\operatorname{Var}(X+Y)=6.21+4.1475+2(0.435)=11.2275\) squared thousands of dollars.
Direct check. The possible aggregate values and probabilities are \(0:0.10\), \(2:0.15\), \(3:0.15\), \(5:0.25\), \(6:0.05\), \(8:0.15\), and \(11:0.15\). These probabilities sum to 1.00. Their mean is \(2(0.15)+3(0.15)+5(0.25)+6(0.05)+8(0.15)+11(0.15)=5.15\), which equals \(E[X]+E[Y]=2.70+2.45\). Their second moment is \(4(0.15)+9(0.15)+25(0.25)+36(0.05)+64(0.15)+121(0.15)=37.75\). Hence the variance is \(37.75-5.15^2=11.2275\), matching the covariance calculation.
Common mistake avoided. Adding the two marginal variances alone would give \(6.21+4.1475=10.3575\), which is not the aggregate variance here. The omitted \(2\operatorname{Cov}(X,Y)=0.87\) matters because the joint law shows nonzero covariance.
Common Calculation Traps
- Confusing joint and conditional probabilities. A cell is a joint probability. To condition on a row or column event, divide by that row or column total.
- Using marginal probabilities as if they determine the joint law. Multiplying \(P(A)P(B)\) gives \(P(A\cap B)\) only when the events are independent.
- Stopping at the product moment. The covariance is \(E[XY]-E[X]E[Y]\), not \(E[XY]\).
- Including the wrong cells in an event. Translate the event condition into a test for each outcome pair before summing probabilities.
- Dropping the covariance term from aggregate variance. In general, \(\operatorname{Var}(X+Y)\) includes \(2\operatorname{Cov}(X,Y)\).
Check Your Understanding
For each problem, use the stated discrete joint law and keep probability calculations to four decimals where needed.
- A hypothetical pair of losses has \(X\in\{0,4\}\) and \(Y\in\{0,2\}\), with joint probabilities \(P(0,0)=0.30\), \(P(0,2)=0.20\), \(P(4,0)=0.10\), and \(P(4,2)=0.40\). Find \(P(Y=2\mid X=4)\), \(E[X]\), \(E[Y]\), and \(\operatorname{Cov}(X,Y)\). (Answer: \(0.8000,\ 2,\ 1.2,\ 0.8\).)
- For a hypothetical joint law, \(X\in\{0,2\}\), \(Y\in\{1,3\}\), with cell probabilities \(P(0,1)=0.20\), \(P(0,3)=0.30\), \(P(2,1)=0.10\), \(P(2,3)=0.40\). Calculate \(\operatorname{Cov}(X,Y)\) and \(\operatorname{Corr}(X,Y)\). (Answer: covariance \(=0.20\); correlation \(=0.2182\).)
- A joint table gives \(P(X=1,Y=4)=0.12\), \(P(X=3,Y=4)=0.18\), and \(P(Y=4)=0.50\). Find \(P(X\geq3\mid Y=4)\) and \(P(X\geq3,Y=4)\). (Answer: \(0.3600\) and \(0.1800\).)
- For two losses, \(\operatorname{Var}(X)=5.2\), \(\operatorname{Var}(Y)=3.1\), and \(\operatorname{Cov}(X,Y)=-0.6\). Find \(\operatorname{Var}(X+Y)\). (Answer: \(7.1\).)
- A hypothetical joint law has \(X\in\{0,2\}\), \(Y\in\{0,3\}\), with \(P(0,0)=0.20\), \(P(0,3)=0.30\), \(P(2,0)=0.10\), and \(P(2,3)=0.40\). Find \(P(X+Y\geq5)\), \(P(X\geq2)\), and \(P(Y\geq3)\). Is multiplying the last two marginal probabilities a valid way to obtain the first probability? (Answer: \(0.4000,\ 0.5000,\ 0.7000\); no, their product is \(0.3500\) and independence is not given.)