Stability Of Equilibrium Under Price Adjustment

Why Equilibrium May or May Not Be Stable

Finding a price at which quantity demanded equals quantity supplied identifies an equilibrium, but it does not by itself explain what happens after a small price disturbance. If a price is just above equilibrium, for example, does the resulting market pressure tend to bring it back, or push it farther away? A stability test answers this question for a specified price-adjustment rule.

Here the rule is discrete: the price is updated once per period in response to the difference between quantity demanded and quantity supplied. The test is local, meaning it describes the behavior of prices sufficiently close to an equilibrium, not necessarily the path from any starting price.

Definition: At price \(P\), define excess demand by \(Z(P)=Q_d(P)-Q_s(P)\), where \(Q_d(P)\) and \(Q_s(P)\) are quantity demanded and quantity supplied. An equilibrium price \(P^*\) satisfies \(Z(P^*)=0\). Under the simple price-adjustment rule, \(P_{t+1}=P_t+\lambda Z(P_t)\), where \(t\) indexes periods and \(\lambda>0\) is a fixed adjustment parameter.

If \(Z(P_t)>0\), there is a shortage, so the rule raises the price. If \(Z(P_t)<0\), there is a surplus, so the rule lowers the price. The parameter \(\lambda\) controls how much the price changes for a given quantity imbalance. If price is measured in dollars per unit and quantity in units, \(\lambda\) has units of dollars per unit of quantity, making \(\lambda Z(P_t)\) a price change.

The rule is a model assumption, not a universal description of how markets adjust. The analysis below assumes a positive, fixed \(\lambda\), that the same demand and supply schedules apply during the adjustment, and that no separate price floor, ceiling, or other constraint interrupts the update. A constrained or different adjustment rule may have different stability properties.

The Local Stability Test

Suppose \(Z\) is differentiable near \(P^*\). A nearby price can be written as \(P_t=P^*+e_t\), where \(e_t\) is its deviation from equilibrium. The one-period change in that deviation is governed, to first order, by the local multiplier \(m\).

m=1+\lambda Z'(P^*)=1+\lambda\bigl(Q_d'(P^*)-Q_s'(P^*)\bigr)

Theorem: For the discrete adjustment rule, if \(|m|<1\), the equilibrium is locally stable: sufficiently small deviations shrink over successive periods. If \(|m|>1\), the equilibrium is locally unstable: sufficiently small deviations grow. When \(m=1\) or \(m=-1\), this first-order test alone is inconclusive. A negative multiplier with magnitude below one indicates convergence with alternating deviations, or damped oscillation.

For the usual case in which demand slopes downward and supply slopes upward, \(Q_d'(P^*)<0\) and \(Q_s'(P^*)>0\), so \(Z'(P^*)<0\). The price response then pushes against a small deviation, but the step size still matters. In this case, the strict stability condition can be written \(0<\lambda(-Z'(P^*))<2\). A step that is too large can overshoot and amplify deviations rather than reduce them.

Why the Multiplier Determines Stability

The test comes from approximating excess demand near equilibrium. Since \(Z(P^*)=0\), differentiability gives \(Z(P^*+e_t)=Z'(P^*)e_t+o(e_t)\), where \(o(e_t)\) denotes a remainder that becomes small relative to \(e_t\) as \(e_t\) approaches zero. Substituting this approximation into the adjustment rule gives:

\begin{aligned} e_{t+1} &=P_{t+1}-P^*\\ &=e_t+\lambda Z(P^*+e_t)\\ &=\bigl(1+\lambda Z'(P^*)\bigr)e_t+\lambda o(e_t)\\ &=m e_t+\lambda o(e_t). \end{aligned}

For sufficiently small deviations, the remainder is of smaller order than \(e_t\), so the first-order behavior is approximately \(e_{t+1}=m e_t\). If \(|m|<1\), this approximation contracts: each period leaves a smaller deviation. If \(|m|>1\), it expands. A negative \(m\) reverses the sign of the deviation each period; whether that alternation dies out depends on whether its magnitude is below one. At the boundary \(|m|=1\), the first-order term neither shrinks nor grows in magnitude, so higher-order behavior matters.

Find excess demand.
Write \(Z(P)=Q_d(P)-Q_s(P)\), keeping the sign convention clear.

Locate the equilibrium of interest.
Use a price \(P^*\) satisfying \(Z(P^*)=0\).

Calculate the local slope and multiplier.
Evaluate \(Z'(P^*)\), then calculate \(m=1+\lambda Z'(P^*)\).

Classify the local behavior.
Compare \(|m|\) with one. If stable, use the sign of \(m\) to distinguish nonalternating convergence from damped oscillation.

Worked Stability Assessments

Worked Example: A Stable Price Path Without Overshooting

A market has \(Q_d(P)=100-2P\) and \(Q_s(P)=10+P\). Under the rule \(P_{t+1}=P_t+0.2Z(P_t)\), determine whether the equilibrium is locally stable. Starting from \(P_0=34\), calculate two updates and check whether they agree with the stability diagnosis.

Method. Find the equilibrium, derive the excess-demand slope, and apply the multiplier test. Then use the stated rule directly to check the predicted behavior. This avoids mistaking the sign of excess supply for excess demand.

Equilibrium. Equating the schedules gives \(100-2P=10+P\), so \(90=3P\), \(P^*=30\), and \(Q^*=100-2(30)=40\). The excess-demand function is \(Z(P)=(100-2P)-(10+P)=90-3P\), and \(Z'(P)=-3\).

Stability test. Here \(\lambda=0.2\), so \(m=1+0.2(-3)=0.4\). Since \(|0.4|<1\) and \(m>0\), the equilibrium is locally stable with nonalternating convergence.

Update check. At \(P_0=34\), \(Z(34)=90-3(34)=90-102=-12\), so \(P_1=34+0.2(-12)=34-2.4=31.6\). Then \(Z(31.6)=90-94.8=-4.8\), so \(P_2=31.6+0.2(-4.8)=31.6-0.96=30.64\). The deviations from equilibrium are \(4\), \(1.6\), and \(0.64\). Each is \(0.4\) times the preceding deviation, as the multiplier predicts.

Worked Example: Stable Adjustment With Damped Oscillations

Suppose \(Q_d(P)=100-P^2\) and \(Q_s(P)=40+4P\), for prices near \(P=6\). Use \(P_{t+1}=P_t+0.1Z(P_t)\). Verify the equilibrium, test its local stability, and calculate three updates starting from \(P_0=5\).

Method. Differentiate excess demand at the equilibrium before assessing stability, then apply the update period by period. This prevents treating a price overshoot on the first update as evidence of instability.

Equilibrium and slope. At \(P=6\), demand is \(100-6^2=64\), and supply is \(40+4(6)=64\). Thus \(P^*=6\) and \(Z(P)=60-P^2-4P\). Its derivative is \(Z'(P)=-2P-4\), so \(Z'(6)=-16\). The multiplier is \(m=1+0.1(-16)=-0.6\). Because \(|-0.6|<1\), the equilibrium is locally stable, and the negative multiplier predicts alternating deviations.

First update. At \(P_0=5\), demand is \(100-25=75\), supply is \(40+20=60\), and \(Z(5)=15\). Therefore \(P_1=5+0.1(15)=6.5\), above equilibrium.

Second update. At \(P_1=6.5\), demand is \(100-42.25=57.75\), supply is \(40+26=66\), and \(Z(6.5)=57.75-66=-8.25\). Therefore \(P_2=6.5+0.1(-8.25)=5.675\).

Third update. At \(P_2=5.675\), demand is \(100-(5.675)^2=100-32.205625=67.794375\), supply is \(40+4(5.675)=62.7\), and \(Z(5.675)=5.094375\). Therefore \(P_3=5.675+0.1(5.094375)=6.1844375\). The deviations are \(P_0-6=-1\), \(P_1-6=0.5\), \(P_2-6=-0.325\), and \(P_3-6=0.1844375\). They alternate and shrink, consistent with local stability; the exact ratios need not equal \(-0.6\) away from equilibrium because the schedules are nonlinear.

Worked Example: An Adjustment Step That Is Too Large

Use \(Q_d(P)=100-2P\) and \(Q_s(P)=10+P\), with \(P_{t+1}=P_t+0.8Z(P_t)\). The equilibrium is \(P^*=30\), as found by solving \(100-2P=10+P\). Determine local stability and calculate two updates from \(P_0=32\).

Method. Compare the multiplier with the stability threshold before interpreting the numerical updates. The common mistake is to assume that because excess demand falls with price, every positive adjustment step must be stable.

Multiplier. Excess demand is \(Z(P)=90-3P\), so \(Z'(30)=-3\). With \(\lambda=0.8\), \(m=1+0.8(-3)=-1.4\). Since \(|m|=1.4>1\), the equilibrium is locally unstable under this step size.

First update. At \(P_0=32\), demand is \(100-64=36\), supply is \(10+32=42\), and \(Z(32)=-6\). Thus \(P_1=32+0.8(-6)=27.2\), a deviation of \(27.2-30=-2.8\).

Second update. At \(P_1=27.2\), demand is \(100-54.4=45.6\), supply is \(10+27.2=37.2\), and \(Z(27.2)=8.4\). Thus \(P_2=27.2+0.8(8.4)=33.92\), a deviation of \(3.92\). The deviations have changed from \(2\) to \(-2.8\) to \(3.92\): their signs alternate, and their magnitudes are multiplied by \(1.4\). This exactly matches the linear model's multiplier.

Common Mistakes and Practical Limits

Reversing the excess-demand sign. With \(Z=Q_d-Q_s\), a shortage is positive and calls for an increase in price. Replacing \(Z\) with \(Q_s-Q_d\) without also changing the rule reverses the adjustment direction.
Ignoring the size of the step. A negative excess-demand slope is not enough to establish stability. The product of the adjustment parameter and the slope determines the multiplier.
Calling every stable path monotone. A stable negative multiplier produces alternating deviations. Convergence and nonalternation are different properties.
Extending a local result too far. The derivative test concerns prices near \(P^*\). It does not prove that a distant starting price converges, particularly if schedules are nonlinear or the adjustment rule changes.
Treating the boundary as stable or unstable without further analysis. When \(m=1\) or \(m=-1\), the first-order multiplier test does not decide the result. Higher-order terms or the exact adjustment equation may be needed.

Key takeaway: For \(P_{t+1}=P_t+\lambda(Q_d(P_t)-Q_s(P_t))\), calculate \(m=1+\lambda(Q_d'(P^*)-Q_s'(P^*))\). The equilibrium is locally stable when \(|m|<1\), unstable when \(|m|>1\), and the first-order test is inconclusive when \(|m|=1\).

Check Your Understanding

For each problem, use the stated adjustment rule and assess only the equilibrium or local behavior requested.

A market has \(Q_d(P)=84-3P\), \(Q_s(P)=12+P\), and \(\lambda=0.3\). Find the equilibrium price and quantity, calculate the local multiplier, and classify the adjustment. (Answer: \(P^*=18\), \(Q^*=30\), \(m=-0.2\); locally stable with damped oscillations.)
Let \(Q_d(P)=150-5P\), \(Q_s(P)=30+P\), and \(\lambda=0.25\). Starting at \(P_0=18\), find the equilibrium and the next two prices, then classify local stability. (Answer: \(P^*=20\), \(Q^*=50\), \(P_1=21\), \(P_2=19.5\), \(m=-0.5\); locally stable with damped oscillations.)
For \(Q_d(P)=90-P\) and \(Q_s(P)=10+P\), find the equilibrium and multiplier if \(\lambda=1.2\). Is the equilibrium locally stable? (Answer: \(P^*=40\), \(Q^*=50\), \(m=-1.4\); locally unstable.)
Suppose \(Q_d(P)=100-P^2\), \(Q_s(P)=25+10P\), and \(\lambda=0.04\), for prices near \(P=5\). Verify the equilibrium at \(P=5\) and determine the local behavior. (Answer: At \(P=5\), both quantities are \(75\); \(Z'(5)=-20\), \(m=0.2\); locally stable with nonalternating convergence.)
A market has \(Q_d(P)=70-2P\) and \(Q_s(P)=10+P\), with adjustment rule \(P_{t+1}=P_t+0.1(Q_d(P_t)-Q_s(P_t))\). Find the equilibrium and multiplier. What multiplier would result if the excess-supply expression \(Q_s-Q_d\) were substituted into the rule without changing its plus sign? (Answer: \(P^*=20\), \(Q^*=30\); with excess demand \(m=0.7\), locally stable; using excess supply with the unchanged plus sign gives multiplier \(1.3\), locally unstable.)

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