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VEE Economics · Fiscal Multipliers and Aggregate Demand · Tutorial 729 of 998

Spending Rounds and the Geometric Series

See how each consumption round is a fixed fraction of the previous one, and use the geometric series to calculate the total output response.

Intermediate 10 min read

What You'll Learn

  • Identify the initial spending change and distinguish it from induced consumption rounds.
  • Calculate each round using the marginal propensity to consume.
  • Sum a finite number of rounds and find the response still to come.
  • Derive the infinite-round total and connect it to the spending multiplier.
  • Recognize when the geometric-series model converges and what its assumptions exclude.
  • Avoid omitting the initial government-purchases change when adding the rounds.

Why Spending Occurs in Rounds

An increase in government purchases is an initial addition to planned expenditure. In the simple expenditure model, that spending becomes income for someone else. Households then consume a fraction of the additional income, and that consumption becomes income for other households. Their consumption creates another round. The process repeats, with each induced round smaller than the one before it when the marginal propensity to consume is between zero and one.

The rounds provide a direct way to understand the government-purchases multiplier. Instead of treating the multiplier only as a ratio, we can see the total output response as the sum of the initial spending change and all the consumption generated by it. The key mathematical structure is a geometric series.

Definition: Let \(\Delta G\) be an initial change in government purchases, and let \(c\) be the marginal propensity to consume (MPC), the fraction of an additional dollar of income that households spend on consumption. In the simple rounds model, round 0 is \(\Delta G\), round 1 is \(c\Delta G\), round 2 is \(c^2\Delta G\), and round \(n\) is \(c^n\Delta G\). The total response is the sum of these rounds.

This formulation assumes a constant MPC \(c\) with \(0\leq c<1\), fixed prices, and no other changes in autonomous expenditure while the rounds occur. It also assumes that the simple expenditure model treats each round of spending as generating an equal amount of income and output. The result is a model-based change in equilibrium output, not a claim that every real-world spending change will produce that exact observed outcome. Factors such as taxes, imports, changes in prices, and responses by monetary policy are outside this simple rounds calculation.

From One Round to the Next

The initial purchase is not itself a consumption round. It is the first addition to expenditure, called round 0. The income generated by that initial purchase leads to the first induced consumption round. Each subsequent induced round is found by multiplying the previous round by \(c\).

$$ \begin{aligned} \text{Round 0} &= \Delta G,\\ \text{Round 1} &= c\Delta G,\\ \text{Round 2} &= c(c\Delta G)=c^2\Delta G,\\ \text{Round 3} &= c(c^2\Delta G)=c^3\Delta G,\\ \text{Round }n &= c^n\Delta G. \end{aligned} $$

For a positive initial increase, the rounds are positive. For an initial decrease, they are negative: reduced purchases lower income, which lowers consumption and income again. The same multiplication rule applies because each round has the same sign as the preceding round and magnitude \(c\) times as large.

Worked Example: Adding the First Five Rounds

A hypothetical city increases purchases of transit equipment by $80 million. Suppose the MPC is \(c=0.75\). Calculate rounds 0 through 4, their combined contribution, and the total response implied by all rounds.

Method. Label the initial purchase as round 0, then multiply each round by 0.75 to get the next. All amounts below are in millions of dollars.

Round 0: \(\Delta G=80\).

Round 1: \(0.75(80)=60\).

Round 2: \(0.75(60)=45\).

Round 3: \(0.75(45)=33.75\).

Round 4: \(0.75(33.75)=25.3125\).

Adding the five amounts gives \(80+60+45+33.75+25.3125=244.0625\) million. The all-round multiplier is \(1/(1-0.75)=4\), so the total response is \(80(4)=320\) million. The rounds after round 4 therefore contribute \(320-244.0625=75.9375\) million. As a check, the first omitted round is \(0.75(25.3125)=18.984375\) million, and the subsequent omitted rounds form a positive, diminishing sequence.

Common mistake avoided: The sum of rounds 0 through 4 is only a partial total. It is not the all-round response; the remaining rounds must also be included to use the multiplier.

Summing the Geometric Series

With a positive spending increase, the total through round \(n\) is \(\Delta G+c\Delta G+\cdots+c^n\Delta G\). This is a finite geometric series: each term is the previous term multiplied by the same ratio \(c\). Factoring out the initial change gives a useful finite-round formula.

$$ \begin{aligned} \Delta Y_n &=\Delta G(1+c+c^2+\cdots+c^n)\\ &=\Delta G\frac{1-c^{n+1}}{1-c}, \qquad c\ne 1. \end{aligned} $$

Here \(\Delta Y_n\) means the cumulative response through round \(n\), including round 0. A short derivation explains the fraction: write \(S_n=1+c+\cdots+c^n\), then multiply by \(c\) to get \(cS_n=c+c^2+\cdots+c^{n+1}\). Subtracting the second equation from the first cancels all the shared middle terms, leaving \(S_n-cS_n=1-c^{n+1}\). Thus \((1-c)S_n=1-c^{n+1}\); dividing by \(1-c\) gives the finite sum formula.

When \(0\leq c<1\), the power \(c^{n+1}\) approaches zero as \(n\) grows. Therefore, the all-round sum is the limit of the finite sums:

$$ \Delta Y =\lim_{n\to\infty}\Delta G\frac{1-c^{n+1}}{1-c} =\frac{\Delta G}{1-c}. $$
Theorem: In the simple expenditure model with constant MPC \(0\leq c<1\), the rounds generated by an initial government-purchases change \(\Delta G\) sum to \(\Delta Y=\Delta G/(1-c)\). The government-purchases multiplier is the ratio of the total output response to the initial purchases change, \(\Delta Y/\Delta G=1/(1-c)\), provided \(\Delta G\ne 0\).

The formula matches the interpretation of the multiplier derived in the earlier material on the government-purchases multiplier. The geometric-series view shows why the denominator is \(1-c\): the initial change is counted once, and each later round adds a successively smaller fraction of the prior round. For example, if \(c=0.75\), the multiplier is 4 because the total of all rounds is four times the initial change.

The convergence condition matters. If \(c=1\), every round is as large as the previous one, so the sum does not approach a finite value. If \(c>1\), the rounds grow rather than shrink. Neither case gives a convergent multiplier in this model. At \(c=0\), there are no induced consumption rounds: only round 0 remains, and the multiplier is 1.

Worked Example: A Decrease in Purchases

A hypothetical regional agency reduces its purchases of maintenance services by $36 million. Assume \(c=0.60\). Find the first four rounds, the cumulative response through round 3, and the all-round response. Amounts are in millions of dollars.

Method. Use a negative value for the initial change, then multiply by the MPC for every following round.

Round 0: \(-36\) million.

Round 1: \(0.60(-36)=-21.6\) million.

Round 2: \(0.60(-21.6)=-12.96\) million.

Round 3: \(0.60(-12.96)=-7.776\) million.

The cumulative response through round 3 is \(-36-21.6-12.96-7.776=-78.336\) million. The multiplier is \(1/(1-0.60)=1/0.40=2.5\). Therefore, the all-round response is \((-36)(2.5)=-90\) million. The rounds not yet included contribute \(-90-(-78.336)=-11.664\) million, which is negative as expected for a contraction.

Common mistake avoided: A spending decrease does not make the multiplier negative. The multiplier remains positive; the negative sign comes from \(\Delta G\), so the total output response is also negative.

Inferring the MPC and Checking Rounds

The relationship can also be used in reverse. If the initial change and the total response are known, their ratio gives the multiplier. Solving for \(c\) then reveals the implied MPC. This is valid only under the simple model assumptions: the observed total must be interpreted as the response generated by the same initial change, with a constant MPC and no other influences folded into the total.

$$ \frac{\Delta Y}{\Delta G}=\frac{1}{1-c} \quad\Longrightarrow\quad c=1-\frac{\Delta G}{\Delta Y}. $$

Worked Example: Recovering the MPC from a Total Response

A hypothetical public works purchase increases by $50 million. In a simple expenditure model, the all-round output response is $200 million. Find the implied MPC, calculate rounds 0 through 3, and determine how much of the total response remains after round 3.

Method. First use the ratio of total response to initial spending to find the multiplier. Then recover \(c\), calculate the finite rounds, and compare their sum with the stated total.

The multiplier is \(200/50=4\). Since \(4=1/(1-c)\), we have \(1-c=1/4=0.25\), so \(c=0.75\).

Round 0: $50 million.

Round 1: \(0.75(50)=37.5\) million.

Round 2: \(0.75(37.5)=28.125\) million.

Round 3: \(0.75(28.125)=21.09375\) million.

The cumulative response through round 3 is \(50+37.5+28.125+21.09375=136.71875\) million. The amount remaining is \(200-136.71875=63.28125\) million. As a check, round 4 is \(0.75(21.09375)=15.8203125\) million; the remaining rounds sum to \(15.8203125/(1-0.75)=63.28125\) million, agreeing with the subtraction.

Common mistake avoided: The ratio \(200/50=4\) is the multiplier, not the MPC. The MPC is \(1-1/4=0.75\).

Common Exam Traps and Interpretation

A frequent error is to start the series with \(c\Delta G\) and omit \(\Delta G\). That calculates only the induced consumption rounds, not the full response to government purchases. For instance, if \(c=0.80\) and \(\Delta G=10\) million, omitting the initial purchase gives \(10(0.80)/(1-0.80)=40\) million. The correct total includes round 0: \(10/(1-0.80)=50\) million. The omitted amount is exactly the initial $10 million.

A second trap is confusing a finite sum with the infinite-round multiplier. The sum through round \(n\) uses \(1-c^{n+1}\); the infinite sum uses the limit in which \(c^{n+1}\) vanishes. If a question asks for rounds 0 through 3, do not report the all-round result without also identifying the requested finite total.

The multiplier is a model-based relationship, not a statement that the MPC alone determines every actual change in output. A candidate should identify the stipulated model and the requested rounds before calculating. Within the simple model, the arithmetic is consistent: every new round equals \(c\) times the previous one, the finite sums approach a limit when \(c<1\), and the limit is the initial change multiplied by \(1/(1-c)\).

Key takeaway: Count the initial government-purchases change as round 0. Apply the MPC once for each additional round. For \(0\leq c<1\), the sum of all rounds is \(\Delta G/(1-c)\); a finite-round question instead requires the finite geometric sum.

Check Your Understanding

For each problem, treat amounts as changes in output and use the simple expenditure model unless the question asks for a finite number of rounds.

  1. A hypothetical district raises purchases by $24 million, and the MPC is 0.50. Find the cumulative response through round 3 and the amount contributed by all later rounds. (Answer: $45 million through round 3; $3 million thereafter.)
  2. A hypothetical public project is cancelled, reducing purchases by $40 thousand. If the MPC is 0.70, find the all-round output response. (Answer: a decrease of $133.33 thousand.)
  3. A $30 million increase in purchases generates a $150 million all-round response in the simple model. Find the implied MPC and the amount of round 3. (Answer: MPC \(=0.80\); round 3 \(=$15.36\) million.)
  4. A hypothetical agency increases purchases by $75 million, with MPC 0.60. Find the cumulative response through round 5 and the amount contributed by rounds after round 5. (Answer: $178.752 million through round 5; $8.748 million thereafter.)
  5. A $10 million initial purchase change is followed by rounds with MPC 0.90. What is the minimum number of rounds, counting round 0, needed for the cumulative sum to reach at least 90% of the all-round response? (Answer: 22 rounds, from round 0 through round 21.)
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