Why Model Continuous Demand?
An insurer may observe a continuous measure of customer demand, such as the coverage amount a customer requests after receiving a quote. A useful first model relates that quantity directly to the quoted price. In a level-on-level regression, both demand and price enter in their original units: the model predicts a change in demand units for a one-unit change in price.
That price slope is not itself an elasticity. It is a marginal effect with units, such as thousands of dollars of requested coverage per dollar of premium. To express responsiveness as a unit-free percentage comparison, evaluate the model’s implied elasticity at a particular price and demand level. This connection is especially useful when interpreting a fitted demand relationship for pricing analysis. It is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
For example, if demand is measured in thousands of dollars of requested coverage and price is measured in dollars of annual premium, then the slope is in thousands of dollars of coverage per premium dollar. Its units matter: a slope of \(-0.20\) means the fitted demand measure falls by \(0.20\) demand units for each one-dollar increase in price. The interpretation is a model-based change in the conditional mean, not necessarily a causal effect.
Fitting the OLS Line
Ordinary least squares (OLS) chooses the intercept and slope that minimize the sum of squared residuals, where a residual is observed demand minus fitted demand. For paired observations \((P_i,Q_i)\), define \(\overline{P}\) and \(\overline{Q}\) as their sample means. Provided the observed prices are not all identical, the OLS estimates are
The slope is the sample price-demand cross-product divided by the price sum of squares. A negative numerator, with positive price variation, produces a negative fitted slope. That sign describes the direction of the fitted linear association; it does not by itself establish that changing the price causes demand to change.
Worked Example: Estimate the Price Slope From Quote Data
A hypothetical insurer records requested coverage, in thousands of dollars, for five comparable quotes. Premium is in dollars per year. Fit the OLS line, interpret its price slope, and calculate the fitted-demand elasticity at a premium of $120.
| Premium \(P_i\) ($) | Requested coverage \(Q_i\) ($000s) |
|---|---|
| 80 | 52 |
| 100 | 46 |
| 120 | 43 |
| 140 | 35 |
| 160 | 34 |
Step 1: Calculate the means. The price mean is \(\overline{P}=(80+100+120+140+160)/5=600/5=120\). The demand mean is \(\overline{Q}=(52+46+43+35+34)/5=210/5=42\).
Step 2: Calculate the slope. The deviations from the means are \((-40,-20,0,20,40)\) for price and \((10,4,1,-7,-8)\) for demand. The cross-products are \(-400,-80,0,-140,-320\), summing to \(-940\). The squared price deviations sum to \(1{,}600+400+0+400+1{,}600=4{,}000\). Therefore,
Step 3: Calculate the intercept. Using the means, \(\widehat{\beta}_0=42-(-0.235)(120)=42+28.2=70.2\). The fitted line is \(\widehat{Q}(P)=70.2-0.235P\).
Step 4: Interpret and check the fit. The slope says that each additional dollar of premium is associated with \(0.235\) thousand dollars, or $235, less fitted requested coverage. At the five observed prices, fitted demands are \(51.4,46.7,42.0,37.3,32.6\). Subtracting these fitted values from the observations gives residuals \(0.6,-0.7,1.0,-2.3,1.4\), which sum to zero, as expected for a fitted OLS line with an intercept.
Step 5: Calculate elasticity at $120. The fitted demand there is \(70.2-0.235(120)=42\) thousand dollars. The model-implied point elasticity is \((-0.235)(120)/42=-28.2/42=-0.6714\), rounded to four decimal places. At that point, a 1% increase in premium corresponds approximately to a 0.6714% decrease in fitted demand, under the model’s local linear approximation.
Method and trap avoided. This is a direct OLS fit followed by a point-elasticity calculation. The slope is not the elasticity: \(-0.235\) has demand-per-dollar units, while elasticity also depends on the price and demand at which it is evaluated.
From Marginal Effect to Elasticity
For a differentiable demand function \(Q(P)\), point price elasticity is the percentage change in demand per percentage change in price, evaluated at a specified price. In differential form,
For the fitted level-on-level line, \(d\widehat{Q}/dP=\widehat{\beta}_1\). Consequently, at a price \(P\) where fitted demand is positive,
This follows from the ordinary percentage-change rule. For a small price change \(dP\), the fitted demand change is approximately \(d\widehat{Q}=\widehat{\beta}_1dP\). Dividing by fitted demand gives the proportional demand change \(d\widehat{Q}/\widehat{Q}\); dividing that by the proportional price change \(dP/P\) yields \(\widehat{\beta}_1P/\widehat{Q}\). The units cancel, so elasticity is dimensionless.
The slope is constant along a level-on-level line, but its elasticity generally is not. Both \(P\) and fitted demand vary along the line. Elasticity is undefined when fitted demand is zero, and a negative fitted demand has no sensible interpretation as a demand quantity. The linear model should therefore be used only over a range where its predictions and assumptions are reasonable.
Worked Example: Convert a Demand Slope Into Elasticities
A hypothetical coverage-selection study estimates \(\widehat{Q}(P)=90-0.40P\), where \(P\) is the annual premium in dollars and \(Q\) is requested coverage in thousands of dollars. Find fitted demand and point elasticity at premiums of $100 and $125. Also state the predicted demand change for a $10 premium increase.
At a $100 premium: Fitted demand is \(90-0.40(100)=90-40=50\) thousand dollars. The marginal effect is \(-0.40\) thousand dollars per premium dollar, or \(-$400\) of requested coverage per premium dollar. Elasticity is \(\widehat{\epsilon}(100)=(-0.40)(100)/50=-40/50=-0.8000\).
At a $125 premium: Fitted demand is \(90-0.40(125)=90-50=40\) thousand dollars. Elasticity is \(\widehat{\epsilon}(125)=(-0.40)(125)/40=-50/40=-1.2500\). The same slope gives different elasticities because the price-demand combination differs.
For a $10 increase: The fitted change is \(\Delta\widehat{Q}=\widehat{\beta}_1\Delta P=(-0.40)(10)=-4\) thousand dollars. Thus the model predicts $4,000 less requested coverage. This is a level change, not a percentage change.
Check. The fitted values fall from 50 to 46 thousand dollars between $100 and $110, exactly the four-thousand-dollar change calculated from the slope. The stated elasticities are local values at their respective prices, not one constant elasticity for the entire line.
What the Regression Assumptions Support
The OLS formulas compute a best-fitting line for the observed data, but stronger claims require assumptions. For a useful conditional-mean interpretation, the model assumes the mean demand at each price follows a straight line: \(E[Q_i\mid P_i]=\beta_0+\beta_1P_i\). The observations must have price variation to estimate a slope. In standard regression inference, errors are also assumed to have conditional mean zero; independence and constant error variance support conventional standard-error calculations. If those variance conditions are not appropriate, conventional uncertainty estimates may not be reliable.
A fitted association does not automatically measure the effect of changing a quote. For example, customers who receive different prices might also differ in characteristics related to demand. Without a basis for treating price variation as exogenous, interpret the slope as an association in the modeled data, not a causal response. Likewise, extrapolating the fitted line far beyond the observed price range can produce implausible results.
Worked Example: Evaluate a Linear Model Within and Beyond Its Range
A hypothetical insurer fits \(\widehat{Q}(P)=72-0.24P\), with price \(P\) in dollars and demand \(Q\) in requested coverage units. The observed premiums ranged from $100 to $250. Calculate predicted demand and elasticity at $150 and $200, then assess the prediction at $350.
At $150: \(\widehat{Q}(150)=72-0.24(150)=72-36=36\). The elasticity is \((-0.24)(150)/36=-36/36=-1.0000\).
At $200: \(\widehat{Q}(200)=72-0.24(200)=72-48=24\). The elasticity is \((-0.24)(200)/24=-48/24=-2.0000\). The line predicts fewer demand units at the higher price, while the magnitude of point elasticity is larger there.
At $350: The algebraic prediction is \(72-0.24(350)=72-84=-12\). A negative demand quantity is not meaningful here. Since $350 is also beyond the observed premium range, this is a warning against using the line as an unrestricted demand function. The elasticity formula at this point does not repair the implausible demand prediction.
Method and trap avoided. Evaluate both the fitted level and its elasticity at each requested price. A mathematically computable prediction is not automatically a plausible business quantity, particularly when it is negative or extrapolated.
Common Mistakes
- Calling the slope an elasticity. The slope is demand units per price unit. Elasticity is \(\widehat{\beta}_1P/\widehat{Q}(P)\), evaluated at a stated price and demand level.
- Dropping the demand denominator. The incorrect calculation \(\widehat{\beta}_1P\) is not an elasticity. Divide by the demand level, usually the fitted conditional mean when reporting a model-implied elasticity.
- Mixing units. If demand is recorded in thousands, a prediction of 40 means 40 thousand demand units, not 40 individual dollars. Keep price and demand units consistent through the calculation.
- Assuming a constant elasticity. A level-on-level line has a constant marginal effect, but its elasticity changes with price and fitted demand.
- Claiming causation from a fitted slope alone. OLS describes the fitted relationship; a causal interpretation requires appropriate assumptions about how prices were assigned or varied.
Check Your Understanding
Work each problem using the units stated, then compare your result with the answer.
- A fitted model is \(\widehat{Q}=64-0.20P\), with \(P\) in dollars. Find fitted demand and elasticity at \(P=120\). (Answer: fitted demand \(=40\); elasticity \(=-0.6000\))
- A fitted model is \(\widehat{Q}=100-0.50P\). Calculate demand at \(P=120\) and \(P=140\), the predicted change, and elasticity at \(P=140\). (Answer: demands \(40\) and \(30\); change \(=-10\); elasticity \(=-2.3333\))
- For a data set, \(\overline{P}=150\), \(\overline{Q}=38\), \(\sum_i(P_i-\overline{P})^2=5{,}000\), and \(\sum_i(P_i-\overline{P})(Q_i-\overline{Q})=-1{,}250\). Find the OLS line, prediction at \(P=140\), and elasticity there. (Answer: \(\widehat{Q}=75-0.25P\); prediction \(=40\); elasticity \(=-0.8750\))
- A fitted model is \(\widehat{Q}=84-0.30P\). At \(P=180\), calculate fitted demand, the prediction after a $5 increase, and the elasticity at the original price. (Answer: demand \(=30\); new prediction \(=28.5\); elasticity \(=-1.8000\))
- A fitted model is \(\widehat{Q}=110-0.20P\). Find fitted demand and elasticity at \(P=250\), and state whether the model implies a positive demand there. (Answer: demand \(=60\); elasticity \(=-0.8333\); yes, the prediction is positive)