Creative Destruction in Quality-Ladder Models

Why Growth Can Replace Existing Technologies

Some productivity improvements add to an existing stock of knowledge; others replace a specific product or process with a better one. A quality-ladder model focuses on the second case. Each successful innovation moves an industry to a higher rung: output can be produced more efficiently, or a higher-quality product can be supplied. The new technology displaces the previous one, and its innovator may earn profits while holding the leading position.

This mechanism is called creative destruction. It links growth to a tension between the reward for innovating and the loss imposed on the incumbent. A larger innovation prize can encourage research, but a higher probability that another innovation will arrive shortens the incumbent’s expected time earning profits. The same force that advances productivity can therefore reduce the private value of an existing innovation.

Definition: A quality-ladder model represents technological progress as a sequence of successful innovations. Each success raises productivity by a fixed factor $ \gamma>1 $ and replaces the previous technology. The process of displacing an existing technology with a superior one is creative destruction.

To make the mechanism precise, consider a discrete-time model with periods measured in years. Let $A_t$ denote productivity at the start of year $t$. In each year, at most one innovation succeeds, with probability $ \lambda $, independently across years. If it succeeds, productivity is multiplied by $ \gamma $; otherwise, productivity is unchanged. This is a simplified model: it treats the innovation probability as constant and does not model strategic competition over the timing of research.

A_{t+1} = \begin{cases} \gamma A_t, & \text{with probability } \lambda,\\ A_t, & \text{with probability } 1-\lambda. \end{cases}

From Innovation Probability to Productivity Growth

In a single year, productivity either stays at $A_t$ or rises to $\gamma A_t$. Taking the expectation across these two outcomes gives

E[A_{t+1}\mid A_t] = \lambda\gamma A_t+(1-\lambda)A_t = A_t\bigl[1+\lambda(\gamma-1)\bigr].

Thus the expected one-year growth rate of the level of productivity is $ \lambda(\gamma-1) $. This is not generally the same as expected log growth. The log change is $ \ln\gamma $ if an innovation occurs and zero otherwise, so expected log growth is $ \lambda\ln\gamma $. These are different measures, especially when the quality step is large. Under the stated independent, constant-probability assumptions, the expected productivity level after $n$ years is $A_0[1+\lambda(\gamma-1)]^n$, while expected cumulative log growth is $n\lambda\ln\gamma$.

Worked Example: Separating Level Growth From Log Growth

A hypothetical equipment industry has productivity $A_0=200$ index points. Each year, one successful innovation raises productivity by 25%, and the probability of success is $0.2000$. Assume at most one success per year, with independent outcomes and a constant success probability. Find the expected productivity after three years, the expected annual growth rate of the productivity level, and expected annual log growth.

Method. Set $ \gamma=1.25 $ and $ \lambda=0.2000 $. The expected one-year productivity factor is

1+\lambda(\gamma-1) =1+0.2000(1.25-1) =1+0.2000(0.25) =1.05.

The expected level grows by $1.05-1=0.05$, or 5.00% per year. Applying the factor for three years,

E[A_3]=200(1.05)^3 =200(1.157625) =231.525 \text{ index points}.

For log growth, the annual expected change in $ \ln A $ is

\lambda\ln\gamma =0.2000\ln(1.25) =0.2000(0.223144) =0.044629,

or about 4.463% expected log growth per year. The two growth figures differ because one averages productivity levels, while the other averages log changes. As a check, the expected number of innovations over three years is $3(0.2000)=0.6000$, and expected cumulative log growth is $0.6000\ln(1.25)=0.133886$, which is three times the annual expected log growth. This calculation avoids treating the innovation probability as if every year’s productivity rose by the full 25%.

The Incumbent’s Profit Prize

In a simple quality-ladder model, the leading firm earns a flow of operating profit while its technology remains the best available. Let $ \pi $ be that profit per year, $r$ the effective annual discount rate, and $ \beta=1/(1+r) $ the one-year discount factor. Let $ \lambda $ be the probability that an innovation displaces the incumbent at the end of each year. Assume profits are received for the current year, the innovation outcome is independent across years, and displacement ends the firm’s profit stream with no salvage value.

Write $V$ for the value of holding the leading position at the start of a year, just before earning that year’s profit. The firm receives $ \pi $ now. With probability $1-\lambda$, it remains incumbent next year and the continuation value is discounted by $ \beta $. With probability $ \lambda $, it is displaced and receives no continuation value. Therefore,

V=\pi+\beta(1-\lambda)V, \qquad V=\frac{\pi}{1-\beta(1-\lambda)} =\frac{\pi}{1-\frac{1-\lambda}{1+r}}.

The denominator is positive for $r>0$ and $0\leq\lambda\leq1$. This value is lower when the displacement probability rises: more rapid creative destruction shortens the expected stream of incumbent profits. The formula also makes the timing convention important. Here the first profit is received immediately at the start of the valuation period; a model in which the first profit arrives one year later would discount that first payment.

Worked Example: Valuing an Incumbent Position

A hypothetical industrial sensor firm earns operating profit of $12$ thousand dollars at the start of each year while its design remains the industry leader. A rival innovation displaces it at the end of each year with probability $0.2000$. The annual effective discount rate is $10\%$. Under the assumptions above, calculate the value of the incumbent position.

Method. First calculate the discount factor: $ \beta=1/(1+0.10)=10/11 $. The probability of remaining incumbent through the year is $1-0.2000=0.8000$. The value equation is

V=12+\frac{10}{11}(0.8000)V =12+\frac{8}{11}V.

Move the continuation term to the left:

\left(1-\frac{8}{11}\right)V=12, \qquad \frac{3}{11}V=12, \qquad V=12\left(\frac{11}{3}\right)=44.

The incumbent position is worth \$44,000. A substitution check gives $12+(8/11)(44)=12+32=44$, so the value satisfies the original equation. This calculation avoids valuing the profit stream as a perpetuity with no displacement risk.

Research Incentives and the Innovation Prize

A prospective innovator compares the expected value of success with the cost of research. Let $R$ denote research effort, let $p(R)$ be the probability that the project succeeds, and let $C(R)$ be its cost paid at the start of the period. In a stripped-down setting, success grants the firm a leading position worth $V$; failure grants no prize. Taking $V$ as given, the firm’s expected net return is

\Pi(R)=p(R)V-C(R).

For an interior choice of effort, the first-order condition is $p'(R)V=C'(R)$: the firm increases research until the marginal expected prize equals marginal research cost. This condition assumes differentiable functions, a feasible interior choice, and that the firm takes the value of the prize as given. If effort is constrained, the optimum may instead be at a boundary. The condition describes a private decision; it does not by itself establish that the chosen effort is socially optimal.

Value the prize.
Determine the value of becoming the leader, taking account of the profit flow and the chance of later displacement.

Compute expected private return.
Multiply the success probability by the prize value, then subtract the research cost.

Compare the alternatives.
Research is worthwhile relative to doing nothing only when the best feasible expected net return is positive.

Worked Example: Choosing Research Effort

A hypothetical medical-device start-up can obtain the leading position if its design project succeeds. The position is worth $44$ thousand dollars. Research effort $R$ has cost $C(R)=50R^2$ thousand dollars and success probability $p(R)=0.5000R$, for $0\leq R\leq2$. Find the effort that maximizes expected net return and determine whether the project is worthwhile.

Method. The expected net return, in thousands of dollars, is

\Pi(R)=(0.5000R)(44)-50R^2 =22R-50R^2.

Differentiate with respect to effort:

\Pi'(R)=22-100R.

Set the derivative equal to zero to find the interior candidate: $22-100R=0$, so $R=0.2200$. This is feasible, and $ \Pi''(R)=-100<0 $, confirming a maximum. The success probability at this effort is $p(0.2200)=0.1100$. Expected prize value is $0.1100(44)=4.84$ thousand dollars, while research cost is $50(0.2200)^2=50(0.0484)=2.42$ thousand dollars. Expected net return is therefore $4.84-2.42=2.42$ thousand dollars, which is positive. The firm prefers this research choice to doing nothing, which has net return zero. This calculation avoids comparing the full prize value with research cost as if success were certain.

Creative Destruction’s Two-Sided Effect

The innovation probability has different effects depending on whose position is being considered. A higher probability of a successful quality step increases expected productivity growth, holding the step size fixed. But it also reduces the expected duration of the incumbent’s profit stream. The latter effect lowers the value of becoming an incumbent and can weaken a prospective innovator’s private incentive to spend on research.

This trade-off can be seen directly from the incumbent-value formula. Write $D=1-\beta(1-\lambda)$, so $V=\pi/D$. Since $dD/d\lambda=\beta$,

\frac{dV}{d\lambda} =-\frac{\pi\beta}{[1-\beta(1-\lambda)]^2}<0 \qquad \text{when } \pi>0,\ \beta>0.

The inequality shows that an increase in the displacement probability reduces incumbent value. It does not imply that more innovation lowers productivity growth: under the ladder assumptions, a higher probability still raises expected growth. Nor does it by itself determine the total effect on research, because changing the innovation environment may also change costs, prizes, or other firms’ behavior. Those elements are held fixed in the simple private-choice calculation.

Key takeaway: Quality-ladder growth comes from repeated productivity steps, but each step displaces the previous leader. The expected growth calculation depends on the step size and innovation probability; the private research incentive depends additionally on the value and cost of the innovation prize.

Common Exam Trap

A frequent error is to apply the quality improvement in every period, then describe the result as expected growth. Suppose $ \gamma=1.25 $ and $ \lambda=0.2000 $. The incorrect calculation is $1.25-1=25\%$ expected annual growth, which assumes an innovation occurs every year. The correct expected productivity factor is

(0.2000)(1.25)+(0.8000)(1.00)=1.05, \qquad \text{expected level growth}=5.00\%.

Another related trap is to confuse the expected level growth rate $ \lambda(\gamma-1) $ with expected log growth $ \lambda\ln\gamma $. State which measure the question asks for, and use its corresponding formula. When valuing profits, also check whether the problem gives an innovation probability that displaces the incumbent, rather than a probability that the incumbent survives; these probabilities are complements in the model above.

Check Your Understanding

Use the stated discrete-time quality-ladder assumptions where applicable. Each problem is independent.

A software-tool industry has a 10% productivity step and an annual success probability of $0.2500$. Find the expected one-year growth rate of the productivity level. (Answer: 2.50%)
An incumbent receives $9$ thousand dollars in profit at the start of each year. The annual effective discount rate is 5%, and displacement probability is $0.2000$ per year. Find the value of the incumbent position. (Answer: \$37,800)
A device’s quality rises by a factor of $1.20$ if an innovation occurs. The annual innovation probability is $0.1000$. Find expected annual log growth, to four decimal places as a percentage. (Answer: 1.8232%)
A research prize is worth $30$ thousand dollars. Research effort $R$ gives success probability $p(R)=0.4000R$, and costs $C(R)=20R^2$ thousand dollars, with feasible effort including $R=0.3000$. Find the effort maximizing expected net return, its success probability, and the maximum expected net return. (Answer: $R=0.3000$, success probability $0.1200$, maximum net return $1.8$ thousand dollars)
An incumbent earns $7$ thousand dollars at the start of each year. The annual effective discount rate is 8%, and the displacement probability is $0.2000$. A research project costs $4$ thousand dollars and succeeds with probability $0.2000$, producing the same value as an incumbent position. Calculate the position value and the project’s expected net return. (Answer: Position value $27$ thousand dollars; expected net return $1.4$ thousand dollars)

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