Step-by-Step Proofs
Every major derivation in the course — fully worked
T1 · GDP
Proof: Expenditure = Income
We want to show that total expenditure on final goods equals total income earned in the economy.
Step 1
For any good or service $i$, define income as revenue minus cost of intermediate inputs:
$$\text{Income}_i = \text{Revenue}_i - \text{Cost}_i$$
Step 2
Sum across all $N$ goods and services in the economy:
$$\sum_{i=1}^{N}\text{Income}_i = \sum_{i=1}^{N}\text{Revenue}_i - \sum_{i=1}^{N}\text{Cost}_i$$
Step 3
Every intermediate good purchase is simultaneously a cost for the buyer and revenue for the seller. Therefore intermediate goods cancel in $\sum \text{Revenue} - \sum \text{Cost}$, leaving only final good revenues:
$$\sum_{i=1}^{N}\text{Revenue}_i - \sum_{i=1}^{N}\text{Cost}_i = \sum_{\text{final } i}\text{Revenue}_i$$
Step 4
But $\sum_{\text{final}} \text{Revenue}_i$ is exactly expenditure on final goods. Therefore:
$$\sum_{i=1}^{N}\text{Income}_i = \text{Expenditure on final goods} = \text{GDP}$$
∎ Expenditure = Income = GDP
T1 · GDP
Proof: Chain-Weighted Growth Rate is Base-Year Independent
Show that the percentage change in chain-weighted real GDP does not depend on the choice of base year $b$.
Step 1
Chain-weighted real GDP in year $t$ (with $t > b$) is defined as:
$$\text{Real GDP}_t = FI_{t,t-1} \times FI_{t-1,t-2} \times \cdots \times FI_{b+1,b} \times \text{Nominal GDP}_b$$
where each $FI_{j,j-1}$ is the Fisher Index measuring the gross change in real output from period $j-1$ to $j$.
Step 2
Compute the gross growth rate of chain-weighted real GDP from $t-1$ to $t$ by taking the ratio:
$$\frac{\text{Real GDP}_t}{\text{Real GDP}_{t-1}} = \frac{FI_{t,t-1} \times FI_{t-1,t-2} \times \cdots \times \text{Nominal GDP}_b}{FI_{t-1,t-2} \times \cdots \times \text{Nominal GDP}_b}$$
Step 3
All terms except $FI_{t,t-1}$ cancel:
$$\frac{\text{Real GDP}_t}{\text{Real GDP}_{t-1}} = FI_{t,t-1}$$
Step 4
$FI_{t,t-1}$ depends only on prices and quantities in periods $t$ and $t-1$ — it does not involve the base year $b$ at all. Therefore the growth rate:
$$\% \Delta \text{Real GDP}_t = 100 \times (FI_{t,t-1} - 1)$$
is entirely independent of $b$.
∎ This is the key advantage of chain weighting over fixed-base-year methods.
T2 · Inflation
Proof: GDP Deflator Measures the Price Level
Show that $\text{GDP Deflator} = 100 \times P_t$, where $P_t$ is the economy-wide price level.
Step 1
Recall the definitions of nominal and real GDP in year $t$ with base year $r$:
$$\text{Nominal GDP}_t = \sum_i P_{i,t} Q_{i,t}, \qquad \text{Real GDP}_t = \sum_i P_{i,r} Q_{i,t}$$
Step 2
Define the price level as the ratio of nominal to real GDP (normalized so that $P_r = 1$ in the base year):
$$P_t \equiv \frac{\text{Nominal GDP}_t}{\text{Real GDP}_t} = \frac{\sum_i P_{i,t} Q_{i,t}}{\sum_i P_{i,r} Q_{i,t}}$$
Step 3
When $t = r$ (base year), numerator = denominator, so $P_r = 1$ as required. For any $t \neq r$, $P_t$ rises with prices and is unchanged by quantities alone (if all $Q_{i,t}$ scale equally, the ratio is unchanged).
Step 4
The GDP Deflator simply multiplies by 100 for display:
$$\boxed{\text{GDP Deflator}_t = 100 \times \frac{\text{Nominal GDP}_t}{\text{Real GDP}_t}}$$
It equals 100 in the base year and rises with the overall price level.
∎ The deflator extracts the pure price change by keeping quantities fixed at their actual values each year.
T3 · Growth Rates
Proof: Constant Growth Rule
Show that if a variable grows at constant rate $g$ each period, then $y_t = y_0(1+g)^t$.
Step 1
By definition of net growth rate $g$:
$$y_{t+1} = y_t(1+g) \quad \text{for every period } t$$
Step 2
Apply recursively: since the rule holds at every $t$,
$$y_1 = y_0(1+g)$$
$$y_2 = y_1(1+g) = y_0(1+g)^2$$
$$y_3 = y_2(1+g) = y_0(1+g)^3$$
Step 3
By induction, the pattern holds for all $t \geq 0$:
$$\boxed{y_t = y_0(1+g)^t}$$
∎
T3 · Growth Rates
Proof: Rule of 70
Show that if $y$ grows at rate $g$ per year, the doubling time is approximately $70/(100g)$.
Step 1
We want the time $T$ such that $y_T = 2y_0$. Using the constant growth rule:
$$y_0(1+g)^T = 2y_0 \implies (1+g)^T = 2$$
Step 2
Take natural logs of both sides:
$$T \cdot \ln(1+g) = \ln 2$$
$$T = \frac{\ln 2}{\ln(1+g)}$$
Step 3
For small $g$, use the approximation $\ln(1+g) \approx g$. Also $\ln 2 \approx 0.693$:
$$T \approx \frac{0.693}{g} = \frac{69.3}{100g} \approx \frac{70}{100g}$$
Step 4
The approximation is exact when $g$ is expressed as a decimal. Since we typically quote $g$ as a percentage (e.g., $g = 2\%$), the formula becomes:
$$\boxed{T \approx \frac{70}{g_{\%}}}$$
where $g_\%$ is the growth rate in percent.
∎
T3 · Growth Rates
Proof: Growth Rate Algebra (Log Rules)
Show the rules $g_{x/y} \approx g_x - g_y$, $g_{xy} \approx g_x + g_y$, $g_{x^a} \approx ag_x$.
Step 1
Key approximation: for small $\epsilon$, $\ln(1+\epsilon) \approx \epsilon$. Therefore:
$$g_x = \frac{x_{t+1}-x_t}{x_t} = \frac{x_{t+1}}{x_t} - 1 \approx \ln\!\left(\frac{x_{t+1}}{x_t}\right) = \ln x_{t+1} - \ln x_t \equiv \Delta \ln x_t$$
Step 2
Product rule: If $z = xy$, then $\ln z = \ln x + \ln y$. Taking differences:
$$\Delta \ln z = \Delta \ln x + \Delta \ln y \implies \boxed{g_z \approx g_x + g_y}$$
Step 3
Ratio rule: If $z = x/y$, then $\ln z = \ln x - \ln y$. Taking differences:
$$\Delta \ln z = \Delta \ln x - \Delta \ln y \implies \boxed{g_z \approx g_x - g_y}$$
Step 4
Power rule: If $z = x^a$, then $\ln z = a\ln x$. Taking differences:
$$\Delta \ln z = a\,\Delta \ln x \implies \boxed{g_z \approx a\, g_x}$$
∎
T4 · Models
Proof: Labor Market Equilibrium Wage and Employment
Solve the linear labor supply-demand model for $w^*$ and $L^*$.
Step 1
The model has two equations and two unknowns ($w$, $L$):
$$L_s = \bar{a}w + \bar{l} \quad \text{(supply)}$$
$$L_d = \bar{f} - w \quad \text{(demand)}$$
Step 2
At equilibrium, supply equals demand: $L_s = L_d$. Set the two expressions equal:
$$\bar{a}w + \bar{l} = \bar{f} - w$$
Step 3
Collect terms in $w$ on the left side:
$$\bar{a}w + w = \bar{f} - \bar{l}$$
$$w(\bar{a} + 1) = \bar{f} - \bar{l}$$
Step 4
Divide by $(\bar{a}+1)$:
$$\boxed{w^* = \frac{\bar{f} - \bar{l}}{\bar{a} + 1}}$$
Step 5
Substitute $w^*$ back into either curve to find equilibrium employment:
$$L^* = \bar{f} - w^* = \bar{f} - \frac{\bar{f}-\bar{l}}{\bar{a}+1} = \frac{\bar{f}(\bar{a}+1) - \bar{f} + \bar{l}}{\bar{a}+1} = \frac{\bar{f}\bar{a} + \bar{l}}{\bar{a}+1}$$
$$\boxed{L^* = \frac{\bar{a}\bar{f} + \bar{l}}{\bar{a}+1}}$$
Step 6
Comparative statics: If $\bar{l}$ falls (workers retire), $w^*$ rises (↑ numerator gone, but $\bar{f}-\bar{l}$ increases so $w^*$ rises) and $L^*$ falls. If $\bar{f}$ falls (lower labour demand), $w^*$ falls and $L^*$ falls. ✓
∎ Equilibrium is the unique wage where the quantity of labor supplied equals the quantity demanded.
T5 · Production
Proof: Cobb-Douglas Has Constant Returns to Scale
Show that $F(\beta K, \beta L) = \beta F(K,L)$ for any $\beta > 0$.
Step 1
Start with the Cobb-Douglas function:
$$F(K,L) = \bar{A}K^{1/3}L^{2/3}$$
Step 2
Scale both inputs by $\beta$:
$$F(\beta K, \beta L) = \bar{A}(\beta K)^{1/3}(\beta L)^{2/3}$$
Step 3
Factor out powers of $\beta$:
$$= \bar{A}\,\beta^{1/3}K^{1/3}\cdot\beta^{2/3}L^{2/3} = \bar{A}\,\beta^{1/3+2/3}K^{1/3}L^{2/3}$$
Step 4
Since $1/3 + 2/3 = 1$:
$$= \beta^1 \cdot \bar{A}K^{1/3}L^{2/3} = \beta\, F(K,L)$$
∎ CRS holds because the exponents sum to 1.
T5 · Production
Proof: Labor Earns 2/3, Capital Earns 1/3 of GDP
Derive the factor income shares from the firm's profit maximization problem.
Step 1
The firm maximizes profit:
$$\max_{K,L} \; \bar{A}K^{1/3}L^{2/3} - rK - wL$$
Step 2
First-order condition for $K$ (set $\partial \pi / \partial K = 0$):
$$\frac{1}{3}\bar{A}K^{-2/3}L^{2/3} = r \implies r = \frac{1}{3}\frac{Y}{K}$$
Step 3
First-order condition for $L$ (set $\partial \pi / \partial L = 0$):
$$\frac{2}{3}\bar{A}K^{1/3}L^{-1/3} = w \implies w = \frac{2}{3}\frac{Y}{L}$$
Step 4
Compute total factor payments as a share of $Y$:
$$\frac{wL}{Y} = \frac{\tfrac{2}{3}(Y/L)\cdot L}{Y} = \frac{2}{3}$$
$$\frac{rK}{Y} = \frac{\tfrac{1}{3}(Y/K)\cdot K}{Y} = \frac{1}{3}$$
Step 5
Verify that shares sum to 1 (no pure profits under CRS):
$$wL + rK = \frac{2}{3}Y + \frac{1}{3}Y = Y \quad \checkmark$$
∎ Labor share = 2/3, Capital share = 1/3. This is why $\alpha = 1/3$ in the Cobb-Douglas exponent.
T5 · Production
Proof: Per Capita Production Function
Derive $y = \bar{A}k^{1/3}$ from the aggregate production function using CRS.
Step 1
Start with $Y = \bar{A}K^{1/3}L^{2/3}$ and the CRS property $F(\beta K, \beta L) = \beta F(K,L)$. Set $\beta = 1/L$:
$$F\!\left(\frac{K}{L}, 1\right) = \frac{1}{L}F(K,L) = \frac{Y}{L}$$
Step 2
Define per-capita variables $y \equiv Y/L$ and $k \equiv K/L$. Then:
$$y = F(k, 1)$$
Step 3
Evaluate with the Cobb-Douglas functional form:
$$y = \bar{A}k^{1/3} \cdot 1^{2/3} = \bar{A}k^{1/3}$$
∎ $y = \bar{A}k^{1/3}$
T5 · Production
Proof: TFP as a Residual — Relative Productivity Formula
Derive the formula for country TFP relative to the US.
Step 1
The per-capita production function for any country $c$ is:
$$y_c = A_c \, k_c^{1/3}$$
and for the US:
$$y_{US} = A_{US} \, k_{US}^{1/3}$$
Step 2
Divide the two equations:
$$\frac{y_c}{y_{US}} = \frac{A_c \, k_c^{1/3}}{A_{US} \, k_{US}^{1/3}} = \frac{A_c}{A_{US}} \cdot \left(\frac{k_c}{k_{US}}\right)^{1/3}$$
Step 3
Solve for the relative TFP:
$$\boxed{\frac{A_c}{A_{US}} = \frac{y_c/y_{US}}{(k_c/k_{US})^{1/3}}}$$
Step 4
Worked example — India: $y_c/y_{US} = 0.074$, $k_c/k_{US} = 0.035$.
$$\frac{A_{India}}{A_{US}} = \frac{0.074}{(0.035)^{1/3}} = \frac{0.074}{0.327} \approx 0.23$$
India's TFP is only 23% of US TFP. Capital alone predicts $y/y_{US} = 0.327$, far above the actual 0.074 — TFP accounts for the remaining gap.
∎ TFP is a "residual" because it is computed from data, not measured directly. It captures everything (technology, institutions, human capital, allocation efficiency) not captured by $k$.
T6 · Solow
Proof: Solow Steady State $k^*$ and $y^*$
Derive the steady-state capital and output per person analytically.
Step 1
The per-capita capital accumulation equation is:
$$\Delta k_{t+1} = \bar{s}\bar{A}k_t^{1/3} - \bar{d}k_t$$
Step 2
At steady state, capital stops changing: $\Delta k = 0$. So investment equals depreciation:
$$\bar{s}\bar{A}(k^*)^{1/3} = \bar{d}k^*$$
Step 3
Divide both sides by $(k^*)^{1/3}$:
$$\bar{s}\bar{A} = \bar{d}(k^*)^{2/3}$$
Step 4
Raise both sides to the power $3/2$:
$$\boxed{k^* = \left(\frac{\bar{s}\bar{A}}{\bar{d}}\right)^{3/2}}$$
Step 5
Plug $k^*$ into the production function $y^* = \bar{A}(k^*)^{1/3}$:
$$y^* = \bar{A}\left(\frac{\bar{s}\bar{A}}{\bar{d}}\right)^{1/2} = \bar{A}\cdot\frac{(\bar{s}\bar{A})^{1/2}}{\bar{d}^{1/2}} = \frac{\bar{A}^{3/2}\bar{s}^{1/2}}{\bar{d}^{1/2}}$$
$$\boxed{y^* = \bar{A}^{3/2}\!\left(\frac{\bar{s}}{\bar{d}}\right)^{1/2}}$$
∎ Both $k^*$ and $y^*$ increase with savings rate and TFP; decrease with depreciation rate.
T6 · Solow
Proof: No Sustained Growth Without Productivity Growth
Show that in the basic Solow model, $g_y \to 0$ as $k \to k^*$.
Step 1
The economy's growth rate of capital per person satisfies:
$$g_k = \frac{\Delta k_{t+1}}{k_t} = \frac{\bar{s}\bar{A}k_t^{1/3} - \bar{d}k_t}{k_t} = \bar{s}\bar{A}k_t^{-2/3} - \bar{d}$$
Step 2
As $k_t \to k^*$, the term $k_t^{-2/3}$ converges to $(k^*)^{-2/3}$. At $k^*$, investment equals depreciation:
$$\bar{s}\bar{A}(k^*)^{1/3} = \bar{d}k^* \implies \bar{s}\bar{A}(k^*)^{-2/3} = \bar{d}$$
so $g_k \to \bar{d} - \bar{d} = 0$.
Step 3
Since $y_t = \bar{A}k_t^{1/3}$ and $\bar{A}$ is constant, the growth rate of output per person is:
$$g_y = \frac{1}{3}g_k \xrightarrow{k\to k^*} \frac{1}{3}\cdot 0 = 0$$
Step 4
Why? Diminishing returns: as $k$ rises, each additional unit of capital produces less output ($MPK = \frac{1}{3}\bar{A}k^{-2/3}$ falls). Investment is a constant fraction of output, so investment per unit of capital falls. Eventually it can only cover depreciation — net investment hits zero.
$$\boxed{g_y^* = 0 \text{ in the basic Solow model (no productivity growth)}}$$
∎ Only a permanent rise in $\bar{A}$ (TFP) shifts the investment curve up, establishing a new, higher steady state. Capital accumulation alone cannot sustain growth.
T7 · Solow + Growth
Proof: BGP Growth Rate $g_y^* = \tfrac{3}{2}g_A$
Derive the balanced growth path growth rate of output per person.
Step 1
From the per-capita production function $y_t = A_t k_t^{1/3}$, apply the growth rate power and product rules:
$$g_y = g_A + \frac{1}{3}g_k$$
Step 2
Definition of a balanced growth path: every endogenous variable grows at a constant rate. In particular, $g_y^* = g_k^*$ (output per person and capital per person grow at the same rate).
Step 3
Substitute $g_k^* = g_y^*$ into the equation from Step 1:
$$g_y^* = g_A + \frac{1}{3}g_y^*$$
Step 4
Rearrange and solve:
$$g_y^* - \frac{1}{3}g_y^* = g_A \implies \frac{2}{3}g_y^* = \bar{g}_A$$
$$\boxed{g_y^* = \frac{3}{2}\bar{g}_A}$$
∎ Long-run growth in output per person is entirely determined by productivity growth — amplified by a factor of 3/2.
T7 · Solow + Growth
Proof: BGP Level of Output per Person
Derive $y_t^* = A_t^{3/2}\!\left(\dfrac{\bar{s}}{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}\right)^{1/2}$.
Step 1
From the capital accumulation equation (dividing by $K_t$ and subtracting population growth):
$$g_k = \bar{s}\frac{y_t}{k_t} - \bar{d} - \bar{n}$$
Step 2
On the BGP, $g_k^* = g_y^* = \tfrac{3}{2}\bar{g}_A$. Substitute and solve for the BGP ratio $y_t^*/k_t^*$:
$$\frac{3}{2}\bar{g}_A = \bar{s}\frac{y_t^*}{k_t^*} - \bar{d} - \bar{n}$$
$$\frac{y_t^*}{k_t^*} = \frac{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}{\bar{s}}$$
Step 3
From the production function $y_t^* = A_t(k_t^*)^{1/3}$, so $k_t^* = (y_t^*/A_t)^3$. Substitute into the ratio:
$$\frac{y_t^*}{(y_t^*/A_t)^3} = \frac{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}{\bar{s}}$$
$$\frac{A_t^3}{(y_t^*)^2} = \frac{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}{\bar{s}}$$
Step 4
Solve for $y_t^*$:
$$(y_t^*)^2 = \bar{s}\cdot\frac{A_t^3}{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}$$
$$\boxed{y_t^* = A_t^{3/2}\left(\frac{\bar{s}}{\tfrac{3}{2}\bar{g}_A + \bar{d} + \bar{n}}\right)^{1/2}}$$
∎ The level grows at rate $\tfrac{3}{2}\bar{g}_A$ because $A_t$ itself grows at $\bar{g}_A$ and is raised to the power $3/2$. ✓
T8 · Labor Market
Proof: Natural Rate of Unemployment $u^* = \lambda_s/(\lambda_s + \lambda_f)$
Derive the steady-state (flow-consistent) unemployment rate from the bathtub model.
Step 1
The flow equation for the number unemployed is:
$$\Delta U_{t+1} = \lambda_s E_t - \lambda_f U_t$$
Inflows: $\lambda_s E_t$ (employed workers who lose jobs). Outflows: $\lambda_f U_t$ (unemployed who find jobs).
Step 2
Divide through by labor force size $L_t$ and use $E_t/L_t = 1 - u_t$, $U_t/L_t = u_t$:
$$\Delta u_{t+1} = \lambda_s(1-u_t) - \lambda_f u_t$$
Step 3
At the steady state (flow-consistent rate), $\Delta u_{t+1} = 0$:
$$0 = \lambda_s(1 - u^*) - \lambda_f u^*$$
Step 4
Expand and collect terms in $u^*$:
$$\lambda_s = \lambda_s u^* + \lambda_f u^* = (\lambda_s + \lambda_f)u^*$$
Step 5
Divide both sides by $(\lambda_s + \lambda_f)$:
$$\boxed{u^* = \frac{\lambda_s}{\lambda_s + \lambda_f}}$$
∎ At $u^*$: inflows $\lambda_s(1-u^*)$ exactly equal outflows $\lambda_f u^*$ — unemployment is in "flow equilibrium."
T8 · Labor Market
Proof: Higher Separation Rate Raises $u^*$
Show formally that $\partial u^*/\partial \lambda_s > 0$.
Step 1
Write $u^* = \lambda_s(\lambda_s + \lambda_f)^{-1}$ and differentiate with respect to $\lambda_s$ using the product rule:
$$\frac{\partial u^*}{\partial \lambda_s} = (\lambda_s + \lambda_f)^{-1} + \lambda_s \cdot (-1)(\lambda_s + \lambda_f)^{-2}$$
Step 2
Factor out $(\lambda_s + \lambda_f)^{-2}$:
$$= \frac{1}{(\lambda_s+\lambda_f)^2}\left[(\lambda_s+\lambda_f) - \lambda_s\right] = \frac{\lambda_f}{(\lambda_s+\lambda_f)^2}$$
Step 3
Since $\lambda_f > 0$ and $(\lambda_s + \lambda_f)^2 > 0$:
$$\frac{\partial u^*}{\partial \lambda_s} = \frac{\lambda_f}{(\lambda_s+\lambda_f)^2} > 0$$
∎ A higher job separation rate unambiguously raises the natural rate of unemployment.
T8 · Labor Market
Proof: Bathtub Model Converges to $u^*$ from Any Starting Point
Show that $|u_t - u^*|$ shrinks each period so $u_t \to u^*$.
Step 1
Define the deviation from the natural rate: $\varepsilon_t \equiv u_t - u^*$. We want to show $|\varepsilon_{t+1}| < |\varepsilon_t|$.
Step 2
From the flow equation $\Delta u_{t+1} = \lambda_s(1-u_t) - \lambda_f u_t$, write $u_{t+1}$:
$$u_{t+1} = u_t + \lambda_s(1-u_t) - \lambda_f u_t = u_t(1 - \lambda_s - \lambda_f) + \lambda_s$$
Step 3
Subtract $u^* = \lambda_s/(\lambda_s + \lambda_f)$ from both sides. Note that $\lambda_s = u^*(\lambda_s + \lambda_f)$:
$$\varepsilon_{t+1} = \varepsilon_t(1 - \lambda_s - \lambda_f)$$
Step 4
Therefore:
$$\varepsilon_t = (1 - \lambda_s - \lambda_f)^t \, \varepsilon_0$$
Step 5
Since $\lambda_s, \lambda_f \in (0,1)$ and $\lambda_s + \lambda_f < 1$ in practice:
$$0 < 1 - \lambda_s - \lambda_f < 1 \implies |1 - \lambda_s - \lambda_f|^t \to 0$$
$$\boxed{u_t = u^* + (1 - \lambda_s - \lambda_f)^t(u_0 - u^*) \longrightarrow u^*}$$
∎ Convergence is geometric at rate $(1-\lambda_s - \lambda_f)$ per period. Higher $\lambda_s + \lambda_f$ means faster convergence — a more dynamic labor market returns to its natural rate more quickly.
T9 · Short Run
Proof: Decomposing Actual Output into Trend and Gap
Derive $Y_t = \bar{Y}_t(1 + \tilde{Y}_t)$ from the definition of the output gap.
Step 1
Define the output gap (short-run output) as the percentage deviation of actual from potential:
$$\tilde{Y}_t \equiv \frac{Y_t - \bar{Y}_t}{\bar{Y}_t}$$
Step 2
Multiply both sides by $\bar{Y}_t$:
$$\tilde{Y}_t \cdot \bar{Y}_t = Y_t - \bar{Y}_t$$
Step 3
Add $\bar{Y}_t$ to both sides:
$$Y_t = \bar{Y}_t + \tilde{Y}_t \cdot \bar{Y}_t = \bar{Y}_t(1 + \tilde{Y}_t)$$
$$\boxed{Y_t = \bar{Y}_t(1 + \tilde{Y}_t)}$$
∎ Actual output equals potential output scaled up or down by $(1 + \tilde{Y}_t)$. A recession means $\tilde{Y}_t < 0$, so $Y_t < \bar{Y}_t$.
T10 · Money
Proof: Fisher Equation $R_t \approx i_t - \pi_t$
Derive the relationship between real and nominal interest rates.
Step 1
If you save $\$S_t$ at nominal rate $i_t$, your nominal savings next period are:
$$S_{t+1}^{\text{nom}} = (1+i_t)S_t$$
Step 2
The real value of savings (purchasing power) adjusts for inflation $\pi_t$. Prices rise by factor $(1+\pi_t)$, so:
$$S_{t+1}^{\text{real}} = \frac{(1+i_t)}{(1+\pi_t)}S_t$$
Step 3
The real interest rate is the net growth in real savings:
$$1 + R_t = \frac{1+i_t}{1+\pi_t}$$
Step 4
This is the exact Fisher equation. For small $i_t$ and $\pi_t$, expand the denominator using $(1+\pi_t)^{-1} \approx 1 - \pi_t$:
$$1 + R_t \approx (1+i_t)(1-\pi_t) = 1 + i_t - \pi_t - i_t\pi_t \approx 1 + i_t - \pi_t$$
Step 5
Subtract 1 from both sides:
$$\boxed{R_t \approx i_t - \pi_t}$$
(The cross-term $i_t\pi_t$ is negligible when both rates are small.)
∎ The real rate equals the nominal rate minus inflation. With $i_t = 5\%$ and $\pi_t = 3\%$, the real return is only $\approx 2\%$.
T11 · IS Curve
Proof: Deriving the IS Curve from the Short-Run Model
Derive $\tilde{Y}_t = \bar{a}_t - \bar{b}(R_t - \bar{r})$ from the national income identity and model equations for each expenditure component.
Step 1
Start with the national income accounting identity, dividing both sides by potential output $\bar{Y}_t$:
$$\frac{Y_t}{\bar{Y}_t} = \frac{C_t}{\bar{Y}_t} + \frac{I_t}{\bar{Y}_t} + \frac{G_t}{\bar{Y}_t} + \frac{EX_t}{\bar{Y}_t} - \frac{IM_t}{\bar{Y}_t}$$
Step 2
Substitute the five model equations for each expenditure share:
$$\frac{Y_t}{\bar{Y}_t} = \bar{a}_{c,t} + \left[\bar{a}_{i,t} - \bar{b}(R_t - \bar{r})\right] + \bar{a}_{g,t} + \bar{a}_{ex,t} - \bar{a}_{im,t}$$
Step 3
Subtract 1 from both sides, using $\tilde{Y}_t \equiv Y_t/\bar{Y}_t - 1$:
$$\tilde{Y}_t = \bar{a}_{c,t} + \bar{a}_{i,t} + \bar{a}_{g,t} + \bar{a}_{ex,t} - \bar{a}_{im,t} - 1 - \bar{b}(R_t - \bar{r})$$
Step 4
Define the aggregate demand shock as the sum of all exogenous expenditure components relative to potential:
$$\bar{a}_t \equiv \bar{a}_{c,t} + \bar{a}_{i,t} + \bar{a}_{g,t} + \bar{a}_{ex,t} - \bar{a}_{im,t} - 1$$
Step 5
Substitute to obtain the IS curve:
$$\boxed{\tilde{Y}_t = \bar{a}_t - \bar{b}(R_t - \bar{r})}$$
∎ The IS curve is downward-sloping in $(R_t, \tilde{Y}_t)$ space. When $R_t = \bar{r}$ (interest rate at trend), output equals the demand shock: $\tilde{Y}_t = \bar{a}_t$. In the long run, $\bar{a}_t = 0$ so $\tilde{Y}_t = 0$.
T11 · IS Curve
Proof: Multiplier Effect on the IS Curve
Derive the multiplied IS curve when consumption also depends on current income ($\tilde{Y}_t$) with MPC $\bar{x} \in (0,1)$.
Step 1
Modified consumption equation:
$$\frac{C_t}{\bar{Y}_t} = \bar{a}_{c,t} + \bar{x}\tilde{Y}_t$$
where $\bar{x}$ (the MPC) $\in (0,1)$ captures the fraction of extra income spent on consumption.
Step 2
Substitute into the income identity (as in the baseline derivation), now with the additional $\bar{x}\tilde{Y}_t$ term on the right:
$$\tilde{Y}_t = \bar{a}_t + \bar{x}\tilde{Y}_t - \bar{b}(R_t - \bar{r})$$
Step 3
Move the $\bar{x}\tilde{Y}_t$ term to the left side:
$$(1 - \bar{x})\tilde{Y}_t = \bar{a}_t - \bar{b}(R_t - \bar{r})$$
Step 4
Divide both sides by $(1 - \bar{x}) > 0$:
$$\boxed{\tilde{Y}_t = \frac{1}{1-\bar{x}}\left[\bar{a}_t - \bar{b}(R_t - \bar{r})\right]}$$
The multiplier $\frac{1}{1-\bar{x}} > 1$ amplifies any initial shock.
∎ Because $0 < \bar{x} < 1$, the multiplier exceeds 1. A $1\%$ demand shock produces more than $1\%$ output change — each round of higher income induces further consumption, which raises income again, and so on.
T12 · Phillips Curve
Proof: Accelerationist Phillips Curve from Adaptive Expectations
Show that combining the expectations-augmented PC with adaptive expectations yields $\Delta\pi_t = \bar{\nu}\tilde{Y}_t + \bar{o}_t$.
Step 1
Start with the expectations-augmented Phillips Curve:
$$\pi_t - \pi_t^e = \bar{\nu}\tilde{Y}_t + \bar{o}_t$$
Inflation exceeds expected inflation when output is above potential, plus cost shocks.
Step 2
Apply adaptive expectations: agents set this period's expected inflation equal to last period's actual inflation:
$$\pi_t^e = \pi_{t-1}$$
Step 3
Substitute $\pi_t^e = \pi_{t-1}$ into Step 1:
$$\pi_t - \pi_{t-1} = \bar{\nu}\tilde{Y}_t + \bar{o}_t$$
Step 4
By definition, $\Delta\pi_t \equiv \pi_t - \pi_{t-1}$. Therefore:
$$\boxed{\Delta\pi_t = \bar{\nu}\tilde{Y}_t + \bar{o}_t}$$
The change in inflation (not its level) depends on the output gap and cost shocks.
∎ With adaptive expectations, any output boom ($\tilde{Y}_t > 0$) accelerates inflation, while a recession decelerates it. There is no stable long-run tradeoff between $\pi$ and $\tilde{Y}$.
T12 · Phillips Curve
Proof: Long-Run Phillips Curve is Vertical
Show that in the long run (steady state), the output gap is zero regardless of the inflation rate.
Step 1
In the long run, inflation is constant: $\pi_t = \pi_{t-1} = \pi^*$. This means $\Delta\pi_t = 0$.
Step 2
Substitute $\Delta\pi_t = 0$ and $\bar{o}_t = 0$ (no long-run shocks) into the accelerationist PC:
$$0 = \bar{\nu}\tilde{Y}^* + 0$$
Step 3
Since $\bar{\nu} > 0$, this implies:
$$\boxed{\tilde{Y}^* = 0}$$
regardless of what $\pi^*$ is.
∎ In the long run output always returns to potential. Monetary policy cannot permanently raise output above potential — attempting to do so only causes accelerating inflation (Friedman-Phelps result).
T13 · IS-MP Model
Proof: IS-MP Equilibrium — Solving for Short-Run Output
Given the IS curve and the central bank's choice of $R_t$, derive the equilibrium $\tilde{Y}_t$.
Step 1
The IS curve gives output as a function of the real interest rate:
$$\tilde{Y}_t = \bar{a}_t - \bar{b}(R_t - \bar{r})$$
Step 2
The central bank sets $R_t$ via the MP curve (a horizontal line at the chosen rate). Substituting $R_t$ directly into the IS equation gives equilibrium output immediately — there is no further solving needed since $R_t$ is set exogenously by the Fed.
Step 3
The Fed's stabilization problem: choose $R_t$ to achieve $\tilde{Y}_t = 0$.
$$0 = \bar{a}_t - \bar{b}(R_t^* - \bar{r}) \;\Rightarrow\; R_t^* - \bar{r} = \frac{\bar{a}_t}{\bar{b}}$$
$$\boxed{R_t^* = \bar{r} + \frac{\bar{a}_t}{\bar{b}}}$$
Step 4
For a negative shock $\bar{a}_t < 0$: $R_t^* < \bar{r}$ — the Fed cuts the rate below trend to offset the contractionary shock. For a positive shock $\bar{a}_t > 0$: $R_t^* > \bar{r}$ — the Fed raises the rate to prevent overheating.
∎ Perfect stabilization requires the Fed to move $R_t$ by exactly $\bar{a}_t / \bar{b}$ relative to the trend rate — offsetting any demand shock dollar-for-dollar through the interest rate channel.
T13 · IS-MP Model
Proof: Why Controlling $i_t$ Controls $R_t$ in the Short Run
Show that with sticky inflation, a change in the nominal rate $i_t$ translates one-for-one to a change in the real rate $R_t$.
Step 1
Fisher equation: $R_t = i_t - \pi_t$. The real rate equals the nominal rate minus inflation.
Step 2
In the short run, $\pi_t$ is predetermined (sticky) — firms and workers set prices/wages based on past conditions and cannot instantly adjust. Thus, at any point in time, $\pi_t$ is approximately fixed.
Step 3
Differentiate with respect to $i_t$ holding $\pi_t$ fixed:
$$\frac{\partial R_t}{\partial i_t} = 1 \quad (\pi_t \text{ held constant})$$
A 1 percentage point increase in $i_t$ raises $R_t$ by exactly 1 percentage point.
Step 4
This breaks down in the long run when inflation adjusts. If the Fed permanently raises $i_t$ and inflation adjusts upward by the same amount, $R_t$ returns to $\bar{r}$ — money is neutral in the long run.
∎ Monetary non-neutrality is a short-run phenomenon. The sticky inflation assumption is what gives the central bank real power over the economy through conventional policy.
T14 · Unconventional Policy
Proof: ELB Constraint on the Real Interest Rate
Show that the ELB limits how low the real interest rate can go, and that higher inflation targets expand policy room.
Step 1
By definition, the ELB constrains the nominal rate: $i_t \geq \text{ELB}$.
Step 2
Apply the Fisher equation $R_t = i_t - \pi_t$:
$$R_t = i_t - \pi_t \geq \text{ELB} - \pi_t$$
Step 3
Define the real lower bound: $R_t^{\min} = \text{ELB} - \pi_t$. The Fed cannot reduce $R_t$ below $R_t^{\min}$.
$$\boxed{R_t \;\geq\; \text{ELB} - \pi_t}$$
Step 4
Policy room = how far below $\bar{r}$ the Fed can push $R_t$:
$$\text{Policy room} = \bar{r} - R_t^{\min} = \bar{r} - \text{ELB} + \pi_t$$
Higher $\pi_t$ (and thus higher $\pi_t^e \approx \bar{\pi}$) directly increases policy room. A 1 pp higher inflation target expands the real lower bound by 1 pp.
∎ This is the core argument for why central banks target 2% rather than 0% inflation — it maintains ~2pp of additional policy room beyond the ELB, reducing the frequency and severity of hitting the constraint.
T14 · Unconventional Policy
Proof: Bond Prices and Yields Move Inversely
Show formally that if the market price of a bond rises, its YTM must fall.
Step 1
YTM is defined as the $y$ that solves:
$$P = \sum_{k=1}^N \frac{C_k}{(1+y)^k} + \frac{FV}{(1+y)^N}$$
where $P$ is the price, $C_k$ are coupon payments, and $FV$ is face value. All cash flows are fixed.
Step 2
Define $f(y) = \sum_{k=1}^N \frac{C_k}{(1+y)^k} + \frac{FV}{(1+y)^N}$. Take the derivative:
$$f'(y) = -\sum_{k=1}^N \frac{k \cdot C_k}{(1+y)^{k+1}} - \frac{N \cdot FV}{(1+y)^{N+1}} < 0$$
Every term is strictly negative (since $C_k, FV, k, N > 0$). Therefore $f$ is strictly decreasing in $y$.
Step 3
Since $P = f(y)$ and $f$ is strictly decreasing: if $P$ rises to $P' > P$, then the new YTM $y'$ satisfying $P' = f(y')$ must be smaller: $y' < y$.
$$\boxed{P \uparrow \;\Leftrightarrow\; \text{YTM} \downarrow}$$
∎ The inverse relationship holds for any coupon bond with positive cash flows. This is the mechanism through which forward guidance and QE — by raising bond demand and prices — lower long-term yields and stimulate the economy.
T15 · AS-AD
Proof: Deriving the Aggregate Demand Curve
Derive the AD equation $\tilde{Y}_t = \bar{a}_t - \bar{b}\bar{m}(\pi_t - \bar{\pi})$ by combining the IS curve and monetary policy rule.
Step 1
Write the IS curve:
$$\tilde{Y}_t = \bar{a}_t - \bar{b}(R_t - \bar{r}) \tag{IS}$$
Step 2
Write the monetary policy rule:
$$R_t - \bar{r} = \bar{m}(\pi_t - \bar{\pi}) \tag{MP rule}$$
Step 3
Substitute the MP rule into (IS) — replace $(R_t - \bar{r})$:
$$\tilde{Y}_t = \bar{a}_t - \bar{b}\cdot\bar{m}(\pi_t - \bar{\pi})$$
Step 4
This is the AD curve:
$$\boxed{\tilde{Y}_t = \bar{a}_t - \bar{b}\bar{m}(\pi_t - \bar{\pi})}$$
Slope in $(\tilde{Y}, \pi)$ space: $\frac{d\pi}{d\tilde{Y}} = \frac{-1}{\bar{b}\bar{m}} < 0$ — downward sloping.
Y-intercept (at $\tilde{Y}=0$): $\pi = \bar{\pi} + \frac{\bar{a}_t}{\bar{b}\bar{m}}$.
Higher $\bar{m}$ → flatter slope → more aggressive response to inflation deviations.
∎ The AD curve is not a behavioral equation on its own — it is a reduced form combining two structural equations (IS + policy rule). A demand shock $\bar{a}_t$ shifts it; a change in $\bar{\pi}$ also shifts it.
T15 · AS-AD
Proof: Steady State of the AS-AD Model
Show that the unique steady state is $(\tilde{Y}^*, \pi^*) = (0, \bar{\pi})$.
Step 1
In steady state: all variables are constant over time, so $\pi_t = \pi_{t-1} = \pi^*$, and there are no shocks: $\bar{a}_t = \bar{o}_t = 0$.
Step 2
Substitute into the AS curve $\pi_t = \pi_{t-1} + \bar{\nu}\tilde{Y}_t + \bar{o}_t$:
$$\pi^* = \pi^* + \bar{\nu}\tilde{Y}^* + 0 \;\Rightarrow\; \bar{\nu}\tilde{Y}^* = 0$$
Since $\bar{\nu} > 0$: $\;\boxed{\tilde{Y}^* = 0}$.
Step 3
Substitute $\tilde{Y}^* = 0$ and $\bar{a}_t = 0$ into the AD curve:
$$0 = 0 - \bar{b}\bar{m}(\pi^* - \bar{\pi}) \;\Rightarrow\; \bar{b}\bar{m}(\pi^* - \bar{\pi}) = 0$$
Since $\bar{b}, \bar{m} > 0$: $\;\boxed{\pi^* = \bar{\pi}}$.
Step 4
Uniqueness: the AS is upward sloping and AD is downward sloping in $(\tilde{Y}, \pi)$ space. Two lines with opposite slopes intersect in exactly one point. That point is $(0, \bar{\pi})$.
∎ In the long run, the economy always returns to potential output with inflation at the target. The central bank can influence the long-run inflation rate (by changing $\bar{\pi}$) but cannot permanently raise output above potential.
T15 · AS-AD
Proof: Inflation Convergence After a One-Period Supply Shock
Show that after a one-period shock $\bar{o}_1 > 0$, the economy converges back to $(0, \bar{\pi})$ under adaptive expectations, and that convergence is monotone.
Step 1
Solve AS and AD simultaneously in period $t$ (with $\bar{a}_t = 0$, $\bar{o}_t = 0$ for $t \geq 2$):
$$\text{AD: } \tilde{Y}_t = -\bar{b}\bar{m}(\pi_t - \bar{\pi})$$
$$\text{AS: } \pi_t = \pi_{t-1} + \bar{\nu}\tilde{Y}_t$$
Substitute AD into AS:
$$\pi_t = \pi_{t-1} + \bar{\nu}\cdot(-\bar{b}\bar{m})(\pi_t - \bar{\pi})$$
$$\pi_t = \pi_{t-1} - \bar{\nu}\bar{b}\bar{m}(\pi_t - \bar{\pi})$$
Step 2
Let $\lambda \equiv \bar{\nu}\bar{b}\bar{m} > 0$. Rearrange to solve for $\pi_t$:
$$\pi_t + \lambda\pi_t = \pi_{t-1} + \lambda\bar{\pi}$$
$$(1+\lambda)\pi_t = \pi_{t-1} + \lambda\bar{\pi}$$
$$\pi_t = \frac{\pi_{t-1}}{1+\lambda} + \frac{\lambda\bar{\pi}}{1+\lambda} = \bar{\pi} + \frac{\pi_{t-1} - \bar{\pi}}{1+\lambda}$$
Step 3
Define the gap $g_t \equiv \pi_t - \bar{\pi}$. Then:
$$g_t = \frac{g_{t-1}}{1+\lambda}$$
This is a first-order linear difference equation with convergence factor $\frac{1}{1+\lambda} \in (0,1)$. Starting from $g_1 > 0$ (due to the shock), each period the gap shrinks by factor $\frac{1}{1+\lambda}$:
$$g_t = g_1 \cdot \left(\frac{1}{1+\lambda}\right)^{t-1} \to 0 \text{ as } t \to \infty$$
Step 4
Convergence is faster when $\lambda = \bar{\nu}\bar{b}\bar{m}$ is large — i.e., when the PC slope ($\bar{\nu}$), IS sensitivity ($\bar{b}$), or policy aggressiveness ($\bar{m}$) are larger. Output gap each period:
$$\tilde{Y}_t = -\bar{b}\bar{m}\cdot g_t < 0 \quad \text{(recession throughout adjustment)}$$
∎ After a one-period supply shock, the inflation gap decays geometrically at rate $1/(1+\lambda)$. The output gap is negative throughout — the central bank sustains a mild recession to wring out elevated inflation. Convergence is fastest when the economy is farthest from steady state.