Why Measure Co-Movement?
When an insurer considers two sources of loss together, knowing each loss separately does not show how their outcomes tend to occur together. Two portfolios can have the same expected losses and the same marginal loss distributions, yet differ in whether large outcomes tend to coincide or offset one another. Covariance summarizes this co-movement by comparing each paired outcome with the two losses’ respective averages.
Covariance is useful for describing the direction and the amount of joint variation. It is not a measure with a universal scale: its numerical size depends on the units used for each loss. This tutorial develops the definition, shows how to calculate it from a joint distribution, and explains how to interpret the result without treating it as a probability or a dollar amount.
Calculating Covariance From a Joint Distribution
For a discrete pair of losses, list each possible paired outcome \((x_i,y_j)\) and its joint probability \(p_{ij}=P(X=x_i,Y=y_j)\). First calculate each mean using the probabilities in the joint distribution. Then multiply the two deviations from their means in each outcome and take the probability-weighted sum.
The same definition applies to continuous losses: replace the probability-weighted sum with an integral over the joint density, when the required moments exist. In either setting, the calculation requires the joint distribution. Separate marginal distributions give the means, but do not in general tell us the average product of the paired deviations.
Why the Product-of-Means Formula Works
A useful calculation form is \(E[XY]-E[X]E[Y]\). It follows directly from the definition and the linearity of expectation. Write \(\mu_X=E[X]\) and \(\mu_Y=E[Y]\), then expand the product inside the expectation:
This form can save work, but it does not remove the need for the joint distribution: calculating \(E[XY]\) requires the probabilities of paired outcomes. The product \(E[X]E[Y]\) alone is not the covariance.
Worked Example: Positive Co-Movement in Two Annual Losses
A hypothetical insurer models one-year losses \(X\) and \(Y\), in dollars, from two coverages. No discounting is applied. The joint distribution is:
| \(X\) | \(Y\) | Joint probability |
|---|---|---|
| 0 | 0 | 0.40 |
| 0 | 80 | 0.10 |
| 100 | 0 | 0.10 |
| 100 | 80 | 0.40 |
Method. Calculate the two means, then calculate the expected product of the centered losses. The probabilities sum to \(0.40+0.10+0.10+0.40=1.00\), so the table specifies a probability distribution.
Step 1: Calculate the means.
Step 2: Calculate the expected product. Only the outcome \((100,80)\) has a nonzero product, so
Step 3: Subtract the product of the means.
Check and interpretation. The centered-product contributions are \(0.40(-50)(-40)=800\), \(0.10(-50)(40)=-200\), \(0.10(50)(-40)=-200\), and \(0.40(50)(40)=800\). Their sum is \(1{,}200\), matching the result. The positive value means outcomes in which both losses are above their means, or both below, contribute positively overall; here those contributions outweigh the opposite-direction outcomes. The common mistake this method avoids is reporting \(E[XY]=3{,}200\) as the covariance without subtracting the product of the means.
Interpreting the Sign and Magnitude
In each paired outcome, the product \((X-E[X])(Y-E[Y])\) is positive when both losses are above their means or both are below their means. It is negative when one is above its mean and the other is below. Covariance is the probability-weighted average of these contributions.
- Positive covariance: same-direction deviations contribute more, on average, than opposite-direction deviations.
- Negative covariance: opposite-direction deviations contribute more, on average.
- Zero covariance: positive and negative contributions balance in expectation. This does not mean every paired outcome has zero deviation product, nor does it by itself establish independence.
The sign gives a directional summary, not a description of every scenario. A positive covariance does not require the two losses to rise together in every outcome. Likewise, a negative covariance does not mean the losses can never both be high. The complete joint distribution may contain outcomes of all these kinds.
Worked Example: Negative Co-Movement in a Joint Loss Model
Consider two hypothetical one-year losses \(X\) and \(Y\), in dollars, with no discounting. The model assigns probabilities as follows:
| \(X\) | \(Y\) | Joint probability |
|---|---|---|
| 0 | 0 | 0.10 |
| 0 | 120 | 0.40 |
| 80 | 0 | 0.40 |
| 80 | 120 | 0.10 |
Method. Use the product-of-means formula, finding \(E[XY]\) from the joint outcomes rather than from the marginal means alone. The listed probabilities sum to \(1.00\).
Step 1: Find the means.
Step 2: Find the expected product. The only nonzero product occurs at \((80,120)\):
Step 3: Calculate and check the covariance.
Using centered deviations gives \(0.10(-40)(-60)=240\), \(0.40(-40)(60)=-960\), \(0.40(40)(-60)=-960\), and \(0.10(40)(60)=240\). The sum is \(-1{,}440\), which checks the calculation. The negative sign reflects the larger total contribution from outcomes in which one loss is below its mean while the other is above. The method avoids inferring co-movement from the marginal means, which are not enough to determine covariance.
Covariance Depends on Measurement Units
Because covariance multiplies a deviation in \(X\) by a deviation in \(Y\), its units combine the units of both losses. If each loss is recorded in dollars, the covariance is in dollars squared. If both are recorded in thousands of dollars instead, each numerical deviation is divided by \(1{,}000\), so the covariance’s numerical value is divided by \(1{,}000{,}000\). The underlying paired outcomes have not changed; only their numerical representation has.
This unit dependence is why a covariance’s magnitude should not be read like a dollar loss. Its sign has a direct directional interpretation, but its size is not a standardized scale. Always state the units and check that the calculation uses a consistent representation of the losses.
Worked Example: Reporting the Same Covariance in Thousands of Dollars
Use the first worked example’s hypothetical joint distribution, but express both annual losses in thousands of dollars. Thus \(X\) takes values \(0\) and \(0.10\), and \(Y\) takes values \(0\) and \(0.08\), with the same probabilities \(0.40, 0.10, 0.10, 0.40\) for \((0,0),(0,0.08),(0.10,0),(0.10,0.08)\), respectively. No discounting is applied.
Step 1: Calculate the means in the new units.
Step 2: Calculate the expected product.
Step 3: Subtract the product of the means and convert units.
Since one thousand dollars is \(1{,}000\) dollars, one squared thousand dollars is \(1{,}000{,}000\) dollars squared. Therefore \(0.0012\times1{,}000{,}000=1{,}200\) dollars squared, agreeing with the first example. This check catches the unit trap of converting a covariance by \(1{,}000\) instead of by \(1{,}000{,}000\) when both losses are converted from dollars to thousands.
Common Calculation Trap
A frequent error is to use the expected product \(E[XY]\) as if it were covariance. In the first example, that incorrect calculation would report \(3{,}200\) dollars squared. The correction is to subtract \(E[X]E[Y]=(50)(40)=2{,}000\) dollars squared, giving covariance \(1{,}200\) dollars squared. Another error is to multiply marginal probabilities as if the losses were independent when the problem supplies a joint distribution. Use each stated joint probability for its paired outcome; do not replace it with a product unless independence is given or established.
Check Your Understanding
For each problem, use the joint outcomes as stated and report covariance with units where specified.
- A hypothetical model has outcomes \((X,Y)=(0,0),(0,30),(50,0),(50,30)\) with probabilities \(0.30,0.20,0.20,0.30\), respectively. Calculate both means and the covariance. (Answer: \(E[X]=25\), \(E[Y]=15\), covariance \(=75\) units squared.)
- Annual losses \(X\) and \(Y\), in dollars, have outcomes \((0,0),(0,50),(100,0),(100,50)\) with probabilities \(0.20,0.30,0.30,0.20\). Find \(E[XY]\) and covariance, then interpret the sign. (Answer: \(E[XY]=1{,}000\) dollars squared; covariance \(=-250\) dollars squared, indicating negative average co-movement.)
- A joint model assigns probability \(1/3\) to each of \((20,90),(60,50),(100,10)\), where both losses are in dollars. Calculate the means and covariance, rounding only the final covariance to two decimal places. (Answer: \(E[X]=60\), \(E[Y]=50\), covariance \(=-1{,}066.67\) dollars squared.)
- In a hypothetical model, losses measured in thousands of dollars have outcomes \((0,0),(0,0.08),(0.10,0),(0.10,0.08)\) with probabilities \(0.40,0.10,0.10,0.40\). Calculate covariance in squared thousands of dollars and convert it to dollars squared. (Answer: \(0.0012\) squared thousands of dollars \(=1{,}200\) dollars squared.)
- A model has outcomes \((10,4),(10,12),(30,4)\), in dollars, with probabilities \(0.50,0.25,0.25\). Find \(E[X]\), \(E[Y]\), \(E[XY]\), and covariance. (Answer: \(E[X]=15\), \(E[Y]=6\), \(E[XY]=80\), covariance \(=-10\) dollars squared.)