A Closed-Form Optimum Under Linear Demand

Why a Closed-Form Price Is Useful

When purchase probability changes linearly with price, and the cost of each issued policy is constant, expected profit is a quadratic function of price. Its maximum can be found directly rather than by testing a long list of candidate prices. The calculation is compact, but it is only useful if the demand model is valid at the chosen price and the price lies in the range where the model applies.

The result below concerns expected underwriting profit from a single price decision. It does not claim that actual purchases or losses are certain, and it does not make a linear demand curve appropriate outside its stated range. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let $p$ be the price offered to an eligible customer, $q(p)$ the probability that the customer purchases at that price, and $c$ the expected cost per issued policy. Under linear demand, write $q(p)=\alpha-\beta p$, where $\beta>0$. Expected profit per eligible customer is $\pi(p)=q(p)(p-c)$, provided cost is incurred only when a policy is issued. The demand curve is used only on its stated feasible price interval, where $0\leq q(p)\leq 1$.

The distinction between an eligible customer and a purchaser matters. The customer produces revenue $p$ and cost $c$ only if a purchase occurs. Thus the expected profit is purchase probability multiplied by profit conditional on purchase. If there are $N$ comparable eligible customers, expected total profit is $N\pi(p)$. A fixed expense incurred whether or not a policy is issued would require a separate term in the objective; it is not included in this per-policy-cost model.

Deriving the Unconstrained Optimum

Substitute the linear demand function into expected profit. Expanding the product shows why the optimization has a closed-form solution.

\begin{aligned} \pi(p) &=(\alpha-\beta p)(p-c)\\ &=-\beta p^2+(\alpha+\beta c)p-\alpha c. \end{aligned}

The quadratic coefficient is negative because $\beta>0$. Therefore the profit curve is strictly concave: it rises up to its vertex and falls after it. Differentiating and setting the derivative equal to zero gives the vertex price.

\begin{aligned} \pi'(p)&=-2\beta p+\alpha+\beta c,\\ \pi'(p)=0 &\quad\Longrightarrow\quad p_0=\frac{\alpha+\beta c}{2\beta} =\frac{\alpha/\beta+c}{2}. \end{aligned}

The second form is often intuitive. The price $\alpha/\beta$ is the price at which the fitted linear demand curve reaches zero, sometimes called its choke price. The unconstrained profit-maximizing price is halfway between that price and the per-policy cost. That interpretation applies to the linear model; it does not mean the model is valid all the way to the choke price.

Result: For $q(p)=\alpha-\beta p$, $\beta>0$, and constant per-issued-policy cost $c$, the unconstrained profit-maximizing price is $p_0=(\alpha+\beta c)/(2\beta)$. This result applies only if the model’s assumptions hold and the price is in the model’s valid range. If the permitted interval is a closed interval $[L,U]$, the profit-maximizing feasible price is $p_0$ when $L\leq p_0\leq U$, $L$ when $p_0<L$, and $U$ when $p_0>U$, assuming the demand model is valid throughout that interval.

The endpoint rule follows from the shape of the quadratic, not from a separate approximation. Its derivative can be written as $\pi'(p)=2\beta(p_0-p)$. It is positive below $p_0$ and negative above $p_0$. Consequently, if the entire feasible interval lies below $p_0$, profit increases throughout the interval and is greatest at $U$. If it lies above $p_0$, profit decreases throughout and is greatest at $L$. If the vertex is inside, the vertex itself is feasible and is the maximum.

State the demand model and its range.
Write $q(p)=\alpha-\beta p$, identify the price interval, and check that the modeled purchase probability stays between zero and one on that interval.

Form expected profit.
Use $\pi(p)=q(p)(p-c)$ when cost is incurred per issued policy. Multiply by the number of comparable eligible customers only if total expected profit is needed.

Find and check the candidate price.
Calculate $p_0=(\alpha+\beta c)/(2\beta)$. Keep it if it is in the feasible interval; otherwise use the appropriate endpoint.

Verify the result.
Evaluate purchase probability, expected profit, and the chosen price at the final candidate. Confirm that the probability and price are within the model’s stated ranges.

Worked Applications

Worked Example: An Interior Optimum

A hypothetical insurer offers a policy to 1,000 comparable eligible customers. Its estimated purchase probability is $q(p)=0.90-0.003p$ for prices from $100 to $300. The expected cost of each issued policy is $140. Find the price that maximizes expected profit and calculate the expected total profit.

Method. Use the linear-demand vertex formula, then check that the resulting price and purchase probability are valid. The range endpoints give $q(100)=0.90-0.30=0.60$ and $q(300)=0.90-0.90=0$, so the stated interval has valid probabilities.

Here $\alpha=0.90$, $\beta=0.003$, and $c=140$. The unconstrained optimum is

p_0=\frac{0.90+(0.003)(140)}{2(0.003)} =\frac{0.90+0.42}{0.006} =\frac{1.32}{0.006} =\$220.

This price is within the $100–$300 interval. Its purchase probability is $q(220)=0.90-(0.003)(220)=0.90-0.66=0.24$. Profit conditional on purchase is $220-140=\$80$, so expected profit per eligible customer is $0.24(80)=\$19.20$. For 1,000 customers, expected total profit is $1{,}000(19.20)=\$19{,}200$.

Check. The expected number of purchases is $1{,}000(0.24)=240$; $240$ times $80 profit per policy is $19,200. The two calculations agree. The common mistake avoided here is treating the 1,000 eligible customers as 1,000 purchasers.

Worked Example: The Unconstrained Price Is Above the Feasible Range

A hypothetical product has purchase probability $q(p)=0.72-0.002p$ on the permitted price interval from $100 to $200. Its expected cost per issued policy is $120, and there are 5,000 comparable eligible customers. Find the best feasible price and expected total profit.

Method. First calculate the unconstrained vertex, then use the interval endpoint if the vertex is infeasible. The probabilities at the interval endpoints are $q(100)=0.72-0.20=0.52$ and $q(200)=0.72-0.40=0.32$, both valid.

The unconstrained price is

p_0=\frac{0.72+(0.002)(120)}{2(0.002)} =\frac{0.72+0.24}{0.004} =\$240.

Since $240 is greater than the feasible maximum of $200, the feasible optimum is $p=\$200$. At that price, $q(200)=0.32$ and the profit per purchase is $200-120=\$80$. Expected profit per eligible customer is $0.32(80)=\$25.60$, giving expected total profit of $5{,}000(25.60)=\$128{,}000$.

Check. At $100, expected profit per eligible customer is $0.52(100-120)=-\$10.40$, less than the $25.60 at $200. Because the vertex lies above the entire feasible interval, profit increases throughout that interval. The common mistake avoided is reporting the algebraic vertex, $240, as the answer even though it is outside the price range where the model may be used.

Worked Example: Converting a Centered Demand Equation

For a hypothetical coverage, an analyst models purchase probability as $q(p)=0.55-0.002(p-150)$ for prices from $100 to $300. Expected cost per issued policy is $100, and 800 eligible customers are considered. Find the profit-maximizing feasible price and expected total profit.

Method. Expand the centered equation before applying the slope-intercept formula. The intercept is not 0.55, because that value is the probability at a price of $150. Expanding gives $q(p)=0.55-0.002p+0.30=0.85-0.002p$. Thus $\alpha=0.85$ and $\beta=0.002$.

The endpoints confirm validity: $q(100)=0.85-0.20=0.65$ and $q(300)=0.85-0.60=0.25$. The unconstrained price is

p_0=\frac{0.85+(0.002)(100)}{2(0.002)} =\frac{1.05}{0.004} =\$262.50.

The price is feasible. At $262.50, the probability is $0.55-0.002(262.50-150)=0.55-0.225=0.325$. Profit per purchase is $262.50-100=\$162.50$, so expected profit per eligible customer is $0.325(162.50)=\$52.8125$. Total expected profit is $800(52.8125)=\$42{,}250$. Reported to the nearest cent, expected profit per eligible customer is $52.81; the total is calculated from the unrounded amount.

Check. The expanded form gives $0.85-(0.002)(262.50)=0.325$, the same probability as the centered form. The common mistake avoided is using 0.55 as the zero-price intercept, which would incorrectly produce a price of $187.50.

Model Checks and Common Traps

A closed-form answer is not automatically a usable price. Check the following before interpreting it:

Check the probability range. A linear formula can predict a probability below zero or above one if extrapolated too far. Use the model only over the price interval for which its probability predictions are valid and it is intended to apply.
Check the price interval separately. A price can produce a valid probability and still lie outside the permitted modeling range. In that case, maximize over the feasible interval rather than reporting the unconstrained vertex.
Use the correct cost convention. The formula assumes $c$ is incurred per issued policy. It does not subtract $c$ from every quote regardless of purchase, nor does it include fixed costs per eligible customer.
Do not confuse expected profit with conditional profit. The margin $p-c$ is profit given a purchase. Expected profit per eligible customer is $q(p)(p-c)$, which accounts for the possibility of no purchase.
Read the demand equation’s parameterization carefully. In a centered equation such as $q(p)=q_0-\beta(p-p_r)$, the intercept for the form $\alpha-\beta p$ is $\alpha=q_0+\beta p_r$, not $q_0$.

Finally, multiplying by a positive number of comparable eligible customers changes the scale of expected total profit but not the maximizing price: $N\pi(p)$ has the same maximizer as $\pi(p)$ when $N>0$. This simplification does not apply without qualification when customers have different demand curves or costs, because a single common price then affects multiple distinct profit functions.

Key takeaway: With linear purchase probability and constant cost per issued policy, expected profit is a concave quadratic. Its unconstrained maximum is $p_0=(\alpha+\beta c)/(2\beta)$. Use that price only when it lies in the valid feasible interval; otherwise, the best feasible price is the appropriate endpoint.

Check Your Understanding

For each problem, use the stated demand interval and treat cost as incurred only for a purchase.

A hypothetical policy has $q(p)=0.80-0.002p$ for prices from $100 to $300 and expected cost $100 per issued policy. Find the maximizing price and expected profit per eligible customer. (Answer: $250; $45.00.)
A product has $q(p)=0.70-0.0015p$ for prices from $100 to $400 and expected cost $100. Find the maximizing price and expected profit per eligible customer, rounding money to the nearest cent. (Answer: $283.33; $50.42.)
A coverage has $q(p)=0.90-0.003p$ on the feasible interval $150 to $250 and expected cost $80. Find the maximizing feasible price and expected profit per eligible customer. (Answer: $190; $36.30.)
A hypothetical offer has $q(p)=0.60-0.001(p-100)$ for prices from $100 to $500, with expected cost $250. Rewrite the demand equation in slope-intercept form, then find the maximizing price and expected profit per eligible customer. (Answer: $q(p)=0.70-0.001p$; $475; $50.625.)
A policy has $q(p)=0.65-0.001p$ on the permitted interval $100 to $300, expected cost $200, and 1,200 eligible customers. Find the best feasible price and total expected profit. (Answer: $300; $42,000.)

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