Econ 101A SP26 — Final Exam Study Guide

01

Math Toolkit

comfortable

↗ PS 1, PS 2

Key concepts

Unconstrained max (1-D): FOC $f'(x^*)=0$; SOC $f''(x^*)<0$ for local max.
Unconstrained max (2-D): $\nabla f = 0$ and Hessian negative definite — need $f_{xx}<0$ and $f_{xx}f_{yy} - f_{xy}^2 > 0$.
Lagrangian: $\mathcal{L} = f(x) - \lambda(g(x) - c)$. FOCs: $\nabla f = \lambda \nabla g$ and $g(x) = c$.
Shadow price: $\lambda^* = \partial V^*/\partial c$ — marginal value of relaxing the constraint by one unit.
Implicit Function Theorem: If $F(x, y) = 0$ and $F_y \ne 0$, then $\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$.
Concavity tests: A function is concave iff its Hessian is negative semi-definite everywhere (or equivalently for 1-D: $f''\le 0$ globally).

Lagrangian template

$$\max_{x_1,\ldots,x_n} f(x) \;\;\text{s.t.}\;\; g_k(x) = c_k \;\Longrightarrow\; \mathcal{L} = f(x) - \sum_k \lambda_k(g_k(x) - c_k)$$ $$\text{FOCs: } \partial \mathcal{L}/\partial x_i = 0 \;\;\forall i, \qquad \partial \mathcal{L}/\partial \lambda_k = 0 \;\;\forall k$$

⚡ Key insight

Almost every econ problem this semester is "set up Lagrangian → take FOCs → solve." If you can do that in your sleep, you've already gotten 30%+ of every exam.

⚠ Common pitfall

SOC in 2-D requires BOTH conditions. $f_{xx}<0$ alone is not enough — you also need $f_{xx}f_{yy} - f_{xy}^2 > 0$ (positive determinant). Skipping this lets a saddle point masquerade as a max.

✎ Practice questions

For $F(x,y) = x^2 + y^2 - 1 = 0$, find $dy/dx$ at $(x,y) = (\tfrac{1}{2}, \tfrac{\sqrt{3}}{2})$ using the IFT.
Show answer
$F_x = 2x = 1$, $F_y = 2y = \sqrt{3}$. So $dy/dx = -F_x/F_y = -1/\sqrt{3}$.
For $f(x,y) = x^2 - xy + y^2$ subject to $x + y = 10$, set up the Lagrangian and solve.
Show answer
$\mathcal{L} = x^2 - xy + y^2 - \lambda(x+y-10)$. FOCs: $2x - y = \lambda$, $2y - x = \lambda$, $x+y=10$. From first two: $x = y$, so $x = y = 5$, $\lambda = 5$.
For $f(x) = \alpha \ln x + (1-\alpha) x$ with $\alpha > 1$, find the critical point and verify it's a max.
Show answer
$f'(x) = \alpha/x + (1-\alpha) = 0 \Rightarrow x^* = \alpha/(\alpha-1)$. $f''(x) = -\alpha/x^2 < 0$ for all $x > 0$, so it's a max (also concave globally).

02

Preferences & Utility

comfortable

↗ PS 1, PS 2, PS 3

Key concepts

Rational preferences = complete + transitive. Reflexivity ($x \succeq x$) follows from completeness. Indifference $\sim$ inherits transitivity.
MRS = $\frac{\partial u/\partial x_1}{\partial u/\partial x_2}$ — units of $x_2$ you'd trade for one more $x_1$ at constant utility. Slope of indifference curve (in absolute value).
Convex preferences (averages preferred to extremes) $\Longleftrightarrow$ MRS diminishes in $x_1$ $\Longleftrightarrow$ indifference curves convex to origin.
Standard families:
  • Cobb-Douglas: $u = x_1^\alpha x_2^\beta$ → MRS $= \frac{\alpha x_2}{\beta x_1}$, expenditure shares are constants.
  • Quasi-linear: $u = \phi(x_1) + x_2$ → MRS $= \phi'(x_1)$ — depends only on $x_1$, not income.
  • Perfect substitutes: $u = ax_1 + bx_2$ → MRS $= a/b$ constant.
  • Perfect complements: $u = \min(ax_1, bx_2)$ → consume in fixed proportion $ax_1 = bx_2$.
Monotone transformations $v = g(u)$ with $g'>0$ preserve preferences. Utility is ordinal.

⚡ Key insight

Cobb-Douglas tells you everything in one shot: an agent with $u = x^\alpha y^{1-\alpha}$ and income $M$ at prices $(p_x, p_y)$ spends share $\alpha$ on $x$, share $1-\alpha$ on $y$. So $x^* = \alpha M/p_x$, $y^* = (1-\alpha)M/p_y$. Memorize this.

⚠ Common pitfall

$u^2(x)$ represents the same preferences as $u(x)$ only if $u \ge 0$ everywhere (otherwise squaring is non-monotonic and flips order on negatives). $\ln u$ requires $u > 0$. Always check the domain when applying transformations.

✎ Practice questions

Are preferences $x \succeq y \Leftrightarrow x_1 + x_2 \ge y_1 + y_2 - 1$ transitive?
Show answer
No. Counterexample: $x = (0,0)$, $y = (1,0)$, $z = (2,0)$. Then $x \succeq y$ (since $0 \ge 0$) and $y \succeq z$ (since $1 \ge 1$), but $x \not\succeq z$ (since $0 \not\ge 1$).
$u(x_1, x_2) = x_1^{1/2} x_2^{1/2}$. Compute MRS at $(x_1, x_2) = (10, 20)$.
Show answer
MRS $= \frac{(1/2)x_1^{-1/2}x_2^{1/2}}{(1/2)x_1^{1/2}x_2^{-1/2}} = \frac{x_2}{x_1} = 2$.
Why is the "MRS diminishing along an indifference curve" a hallmark of convex preferences?
Show answer
As $x_1$ rises along an indifference curve, $x_2$ falls. Diminishing MRS means each extra unit of $x_1$ is worth fewer units of $x_2$. Geometrically, this is exactly the convex-to-origin shape of indifference curves.

03

UMP / Marshallian Demand

comfortable

↗ PS 3, PS 4 Pt A

Key concepts

UMP setup: $\max u(x)$ s.t. $p \cdot x \le M$, $x \ge 0$. Solution = Marshallian demand $x^M(p, M)$.
Interior FOC: $\nabla u = \lambda p$, i.e. $\text{MRS}_{i,j} = p_i/p_j$ for every pair. Plus budget.
$\lambda$ = marginal utility of income at optimum.
Indirect utility: $V(p, M) = u(x^M(p, M))$, the achieved utility.
Engel curve: $x^*(M)$ holding $p$ fixed. Linear in $M$ for Cobb-Douglas; quasi-linear has $x_1^*$ flat (constant) at interior.
Normal vs. inferior: $\partial x^*/\partial M > 0$ normal, $< 0$ inferior.
Labor supply is just UMP with $x = (c, l)$, prices $(p, w)$, "income" = $wT$ (total time × wage).

Marshallian demand — closed forms

$$\text{Cobb-Douglas } u = x_1^\alpha x_2^{1-\alpha}: \quad x_1^* = \frac{\alpha M}{p_1}, \quad x_2^* = \frac{(1-\alpha)M}{p_2}$$ $$\text{Quasi-linear } u = \phi(x_1) + x_2: \quad \phi'(x_1^*) = \frac{p_1}{p_2} \quad \text{(} x_1^* \text{ independent of } M \text{ at interior)}$$ $$\text{Perfect complements } u = \min(ax_1, bx_2): \quad ax_1^* = bx_2^*, \;\; p_1 x_1^* + p_2 x_2^* = M$$

⚡ Key insight

For Cobb-Douglas, you never need a Lagrangian — just apply the "share of income" rule. For quasi-linear, $\phi'(x_1^*) = p_1/p_2$ pins down $x_1^*$ without seeing M; income only affects $x_2^*$.

⚠ Common pitfall

With quasi-linear preferences, $x_1^*$ is income-independent only at an interior optimum. If $M < p_2 \cdot 0 + p_1 x_1^{interior}$ (so $x_2^* = 0$ binds), you're at a corner and $x_1^* = M/p_1$ does depend on $M$. Always check the non-negativity constraint.

✎ Practice questions

$u(x_1, x_2) = \sqrt{x_1} + x_2$. Solve for $x_1^*, x_2^*$. When does $x_2^* = 0$ corner bind?
Show answer
FOC: $\tfrac{1}{2\sqrt{x_1}} = p_1/p_2$, giving $x_1^* = p_2^2/(4p_1^2)$. Then $x_2^* = M/p_2 - p_2/(4p_1)$. Corner $x_2^* = 0$ binds when $M < p_2^2/(4p_1)$; then $x_1^* = M/p_1$.
Junyi has $u(c, l) = c^\alpha l^\beta$, time endowment $T$, wage $w$, consumption price $p$. Find optimal hours of leisure $l^*$ and consumption $c^*$.
Show answer
Budget: $pc + wl = wT$. Cobb-Douglas: $l^* = \tfrac{\beta T}{\alpha+\beta}$, $c^* = \tfrac{\alpha w T}{(\alpha+\beta)p}$. (Hours worked $= \tfrac{\alpha T}{\alpha+\beta}$.)
Cobb-Douglas demand for good 1 is $\alpha M/p_1$. What is $\partial x_1^*/\partial p_2$? Interpret.
Show answer
$\partial x_1^*/\partial p_2 = 0$. For CD, substitution effect (toward good 1 when $p_2$ rises) exactly cancels the income effect (real income falls).

04

Intertemporal Choice

comfortable

↗ PS 5, HW 4 Pt B

Key concepts

2-period budget: $c_0 + \dfrac{c_1}{1+r} = M_0 + \dfrac{M_1}{1+r}$ (present-value form).
Objective: $u(c_0) + \dfrac{1}{1+\delta} u(c_1)$, where $\delta$ = impatience rate.
Euler equation: $u'(c_0) = \dfrac{1+r}{1+\delta} u'(c_1)$.
Comparison rule:
  • $r = \delta$ → $c_0^* = c_1^*$ (perfect consumption smoothing).
  • $r > \delta$ → $c_0^* < c_1^*$ (save & defer).
  • $r < \delta$ → $c_0^* > c_1^*$ (borrow / consume now).
Log utility special case: $\partial c_0^*/\partial r = 0$ — income & substitution effects of $r$ exactly cancel.
Time inconsistency: When discounting is hyperbolic / $\beta$-$\delta$, today's "self" disagrees with tomorrow's "self". Commitment devices help.
With labor: If wage differs ($w_1 > w_0$), hours work more in high-wage period: $24 - l_t^* = (\lambda w_t)^\gamma$ (Frisch elasticity = $\gamma$).

Log utility two-period (δ = r)

$$u(c) = \ln c \;\;\Longrightarrow\;\; c_0^* = c_1^* = \frac{M_0 + M_1/(1+r)}{1 + 1/(1+r)} = \frac{(1+r)M_0 + M_1}{2+r}$$

⚡ Key insight

The Euler equation has three pieces: marginal utility today, marginal utility tomorrow, and the gross interest rate vs. the discount factor. If the ratio $(1+r)/(1+\delta) = 1$, consumption is flat. Otherwise, consumption tilts toward whichever is bigger.

⚠ Common pitfall

Don't confuse $\delta$ (impatience rate) with $\beta = 1/(1+\delta)$ (discount factor). And remember the Euler equation pairs $1+r$ with $1+\delta$ — easy to drop the "$1+$" under stress.

✎ Practice questions

$u(c) = \ln c$, $\delta = r$, $M_0 > 0$, $M_1 = 0$. Find $c_0^*, c_1^*$.
Show answer
Smoothing: $c_0 = c_1 = c$. Budget: $c + c/(1+r) = M_0 \Rightarrow c = \tfrac{(1+r)M_0}{2+r}$.
$u(c) = \ln c$. If $r$ increases, holding $\delta$ fixed and assuming the agent is a saver, how does $c_0^*$ change? Why?
Show answer
$c_0^*$ unchanged (log utility quirk). The substitution effect (save more) and income effect (richer because saver) exactly offset for log preferences.
Why does a time-inconsistent agent (hyperbolic discounter) prefer commitment to flexibility?
Show answer
Without commitment, the period-1 self consumes too much from the period-0 self's perspective. Commitment lets period 0 lock in lower consumption, raising period-0 utility. With $\gamma > 1$ extra weight on present consumption, $c_1^c < c_1^{nc}$.

05

Producer Theory & Supply

comfortable

↗ PS 7, Midterm 2

Key concepts

Cost-min problem: $\min w \cdot z$ s.t. $f(z) \ge y$. Solution $\rightarrow$ conditional input demands and the cost function $c(w, y)$.
Marginal cost: $MC(y) = \partial c/\partial y$. Average cost: $AC(y) = c(y)/y$.
Profit-max: $\max_y p y - c(y)$. FOC: $p = MC(y^*)$.
SR supply: $p = MC$ when $p \ge AVC$; else 0.
LR supply (entry): $p = MC$ when $p \ge AC$; else 0. In LR with free entry, $\pi = 0 \Leftrightarrow p = \min AC$.
Constant returns + linear cost: $c(y) = cy$ gives MC = AC = $c$, horizontal supply.
Aggregate supply = horizontal sum of individual supplies.

⚡ Key insight

For perfectly competitive firms with $c(y) = cy + F$: variable supply pieces are flat at $p = c$, but firms only enter when revenue covers fixed cost $F$. The entry condition pins down market size in long run.

⚠ Common pitfall

$p = MC$ holds for all price-taking firms — not just PC. A monopolist faces $MR = MC$, where $MR \ne p$. Don't mix these up.

✎ Practice questions

Firm has $c(y) = y^2 + 4$. Find SR supply curve. What's $\min AC$?
Show answer
$MC = 2y$. $AVC = y$, $AC = y + 4/y$. SR supply: $y = p/2$ (since $MC \ge AVC$ always). Min $AC$ at $y = 2$ where $AC = 4$. So LR supply only at $p \ge 4$.
$f(H, K) = \min\{H, K\}$, wage $w$, capital price $r$. Find cost function and MC.
Show answer
Use inputs in fixed proportion: $H = K = y$. Cost $c(y) = (w+r)y$, $MC = w+r$ constant.

06

Perfect Competition & Taxation

comfortable

↗ PS 7, Midterm 2

Key concepts

Market eq. Set $Y^S(p) = Y^D(p)$. With linear demand $p = a - bY$ and constant MC $c$: $p^* = c$, $Y^* = (a-c)/b$.
CS = area under demand, above price. PS = area above supply, below price.
Tax incidence formula: $$\frac{\partial p^*}{\partial t} = \frac{\varepsilon_S}{\varepsilon_S - \varepsilon_D}, \qquad \frac{\partial (p^* - t)}{\partial t} = \frac{\varepsilon_D}{\varepsilon_S - \varepsilon_D}$$ (Where $\varepsilon_S > 0$ and $\varepsilon_D < 0$.) The relatively inelastic side bears more burden.
Polar cases: Horizontal supply ($\varepsilon_S \to \infty$) → consumer pays 100% of tax. Vertical supply → producer pays 100%.
DWL of tax (linear demand, horizontal supply): $DWL = \dfrac{t^2}{2b}$. Quadratic in $t$.

⚡ Key insight

Tax burden ≠ tax remittance. Even if firms write the check, consumers bear the burden if their demand is inelastic. Burden follows elasticity, not legal structure.

⚠ Common pitfall

$DWL = t^2/(2b)$ is quadratic in tax rate. Doubling $t$ quadruples DWL — taxes are increasingly distortionary at higher rates. Easy to mis-state as linear.

✎ Practice questions

Demand $p = A - bQ$, supply horizontal at $\alpha$. Tax $t$. Find new $Q$, new consumer price, new producer price, and DWL.
Show answer
Consumer pays $p = \alpha + t$. Producer keeps $\alpha$. New $Q = (A - \alpha - t)/b$. DWL = triangle: $\tfrac{1}{2} t \cdot (t/b) = t^2/(2b)$.
Why does taxing staple foods place most of the burden on consumers?
Show answer
Demand for staples is inelastic ($\varepsilon_D$ near 0). The tax-incidence formula puts the burden on the more inelastic side: consumers.

07

Monopoly

comfortable

↗ Midterm 2, Final 17 Q1, HW 9

Key concepts

Profit max: $\max_y p(y)y - c(y)$ ⇒ MR = MC.
Linear demand $p = a - by$ with $MC = c$: $$y_M^* = \frac{a-c}{2b}, \quad p_M^* = \frac{a+c}{2}, \quad \Pi_M^* = \frac{(a-c)^2}{4b}.$$
Markup rule: $\dfrac{p - MC}{p} = \dfrac{1}{|\varepsilon_D|}$. More elastic demand → smaller markup.
DWL of monopoly (vs. PC): triangle between $MC$ and demand for $y \in [y_M, y_{PC}]$. For linear demand + constant MC: $DWL = \tfrac{(a-c)^2}{8b}$.
Natural monopoly: With large fixed costs $F$, $AC$ keeps falling — only one firm can survive. PC would yield losses for all firms.

⚡ Key insight

Monopoly produces half the PC quantity at twice the markup (for linear demand). DWL = $\tfrac{1}{2}\cdot$ producer surplus. Comparing $(p_M, y_M, \Pi_M)$ to $(p_{PC}, y_{PC}, 0)$ is a standard exam routine — memorize the four numbers.

⚠ Common pitfall

Don't set $p = MC$ for a monopolist. That's the PC condition. Monopoly sets marginal revenue equal to MC — and $MR < p$ because increasing output lowers the price on all units.

✎ Practice questions

Demand $p = 100 - q$, $MC = 20$. Monopoly $q^*, p^*, \Pi^*, DWL$?
Show answer
$q^* = 40$, $p^* = 60$, $\Pi^* = 40 \cdot 40 = 1600$. $q_{PC} = 80$. $DWL = \tfrac{1}{2}(80-40)(60-20) = 800$.
When is a natural monopoly socially preferred to PC?
Show answer
When fixed costs $F$ are so large that PC firms would all earn negative profits and the market would shut down. A regulated monopoly producing at $AC$ (zero profit) restores trade and is better than nothing.

08

Walrasian Equilibrium & Welfare Theorems

comfortable

↗ PS 8, Midterm 2

Key concepts

Edgeworth box: 2 agents, 2 goods, fixed total endowment. Each point = allocation $(x^A, x^B)$ with $x^A + x^B = \omega$.
Walrasian equilibrium = prices $p^*$ and allocation $(x^{A*}, x^{B*})$ s.t. (i) each agent optimizes given $p^*$ from their endowment $\omega^i$; (ii) markets clear.
Demand from endowment (CD with share $\alpha$ on good 1): $x_1^{A*} = \dfrac{\alpha(p_1 \omega_1^A + p_2 \omega_2^A)}{p_1}$.
Pareto optimum: No feasible reallocation makes someone better off without making another worse. Tangency: $\text{MRS}_A = \text{MRS}_B$.
Contract curve: Pareto-optimal allocations that Pareto-improve on the endowment.
FWT (First Welfare Theorem): Every Walrasian equilibrium is Pareto optimal.
SWT (Second Welfare Theorem): Every Pareto optimum can be supported as a WE with appropriate lump-sum transfers.
Walras' Law: If $n-1$ markets clear, the $n$th one does too — only need to check one fewer.

⚡ Key insight

Normalizing $p_2 = 1$ is always fine — only relative prices matter. So a "2-good 2-agent" problem really has only one price to find: $p_1^*$. Use market clearing in good 1, then Walras's law gives you good 2 for free.

⚠ Common pitfall

An allocation can be Pareto optimal but not a Pareto improvement on the endowment (it might make someone worse than their starting point). The contract curve is the intersection of "PO" and "individually rational."

✎ Practice questions

Both A and B have $u = \log x_1 + \log x_2$. Endowments $\omega^A = \omega^B = (5, 5)$. Find the WE price ratio.
Show answer
Symmetric: $p_1^*/p_2^* = 1$. The endowment is already on the contract curve (Pareto-optimal) and is itself the WE allocation.
How does Walras' Law help in solving 2-good equilibria?
Show answer
You only need to impose market clearing in one market and solve for $p_1^*/p_2^*$. The other market clears automatically.
What does the Second Welfare Theorem buy us beyond the First?
Show answer
FWT says markets are efficient. SWT says markets can also implement any efficient distribution we want, via lump-sum transfers — separating efficiency from equity.

09

EMP / Hicksian Demand / Duality

★ harder

↗ PS 3

Key concepts

Expenditure-minimization problem (EMP): $\min p \cdot x$ s.t. $u(x) \ge \bar u$. The "dual" of UMP.
Hicksian demand: $h(p, \bar u)$ = solution to EMP. Aka compensated demand.
Expenditure function: $e(p, \bar u) = p \cdot h(p, \bar u)$ = minimum cost of reaching $\bar u$.
Shephard's Lemma: $\dfrac{\partial e}{\partial p_i} = h_i(p, \bar u)$.
Slutsky decomposition: $$\frac{\partial x_i^M}{\partial p_j} = \underbrace{\frac{\partial h_i}{\partial p_j}}_{\text{substitution (≤0 own)}} \;-\; \underbrace{x_j^M \cdot \frac{\partial x_i^M}{\partial M}}_{\text{income effect}}$$
Roy's identity: $x_i^M = -\dfrac{\partial V/\partial p_i}{\partial V/\partial M}$ — recover Marshallian from indirect utility.
Duality: $x^M(p, M) = h(p, V(p, M))$. Same answer, different angle.

Hicksian for perfect complements

$$u = \min(2x_1, x_2),\;\; \bar u: \quad h_1 = \bar u/2, \quad h_2 = \bar u$$ $$\text{(no price dependence — fixed proportion at kink)}$$

⚡ Key insight

The substitution effect is always non-positive on own price ($\partial h_i/\partial p_i \le 0$). Marshallian demand can slope up (Giffen) only because the income effect overwhelms substitution — but Hicksian never does.

⚠ Common pitfall

For non-convex preferences (e.g., $u = 2x_1^2 + x_2^2$, isoquant is an ellipse), expenditure minimization can put you at a corner, not a tangency. Solve by comparing corner costs, not by setting MRS = price ratio.

✎ Practice questions

$u = \min(2x_1, x_2)$, find Hicksian. Why is there no price dependence?
Show answer
$h_1 = \bar u/2$, $h_2 = \bar u$. Perfect complements consume at the kink $2x_1 = x_2 = \bar u$, regardless of prices.
$u = 2x_1^2 + x_2^2$, $\bar u$. Find Hicksian.
Show answer
Isoquant is an ellipse — corner solutions. Compare $(x_1, x_2) = (\sqrt{\bar u/2}, 0)$ at cost $p_1\sqrt{\bar u/2}$ vs. $(0, \sqrt{\bar u})$ at cost $p_2\sqrt{\bar u}$. Pick the cheaper. Switch at $p_1 = \sqrt{2}p_2$.
Why is it that for Cobb-Douglas, $\partial x_1^M/\partial p_2 = 0$?
Show answer
By Slutsky: substitution effect toward good 1 (when $p_2$ rises) exactly cancels income effect (real income falls, so consume less of normal good 1). CD has a knife-edge cancellation.

10

Comparative Statics & IFT

★ harder

↗ PS 1, PS 7, HW 4 Pt B, Final 21 Q4

Key concepts

Single-variable IFT: If $F(x, \theta) = 0$ defines $x^*(\theta)$ and $F_x \ne 0$, then $\dfrac{dx^*}{d\theta} = -\dfrac{F_\theta}{F_x}$.
Recipe: Write the FOC (or equilibrium condition) as $F = 0$. Compute $F_\theta$ (how $F$ shifts with the parameter). Compute $F_x$ (typically the SOC). Then divide and apply the minus sign.
Sign analysis: SOC implies $F_x < 0$ (or has known sign), so $\operatorname{sign}(dx^*/d\theta) = \operatorname{sign}(F_\theta)$. Most exam problems reduce to: "what's the sign of $F_\theta$?"
Common applications:
  • Tax incidence: $\partial p^*/\partial t = \varepsilon_S/(\varepsilon_S - \varepsilon_D)$.
  • Insurance demand: $\partial \alpha^*/\partial w$ depends on whether ARA rises or falls in $w$.
  • Equilibrium wage: $\partial w^*/\partial N < 0$ when labor demand fixed and supply rises.

IFT — single-variable

$$F(x^*, \theta) = 0 \;\;\Longrightarrow\;\; \frac{dx^*}{d\theta} = -\frac{F_\theta(x^*, \theta)}{F_x(x^*, \theta)}$$

⚡ Key insight

The IFT lets you do comparative statics without solving the model explicitly. For implicit equations like $\frac{2}{3}(100 - w - r) = N[T - (3w)^{-3/2}]$, you'd never close-form $w^*(N)$ — but you can still sign $\partial w^*/\partial N$.

⚠ Common pitfall

The minus sign in $-F_\theta/F_x$ is easy to forget. A safer mental check: differentiate $F(x^*(\theta), \theta) = 0$ totally, get $F_x \cdot (dx^*/d\theta) + F_\theta = 0$, then solve. The minus follows automatically.

✎ Practice questions

$G(w^*, N) = \tfrac{2}{3}(100 - w - r) - N[T - (3w)^{-3/2}] = 0$ defines equilibrium wage. Sign of $\partial w^*/\partial N$?
Show answer
$G_N = -[T - (3w)^{-3/2}] < 0$. $G_w = -\tfrac{2}{3} - \tfrac{3}{2}N(3w)^{-5/2} < 0$. So $\partial w^*/\partial N = -G_N/G_w = -(\text{neg})/(\text{neg}) < 0$. More workers → lower wage.
For insurance: $F(\alpha^*, w) = p(1-q)u'(w_2) - (1-p)q u'(w_1) = 0$. Under DARA, is $\partial \alpha^*/\partial w$ positive or negative?
Show answer
Numerator $F_w$ proportional to $r_A(w_1) - r_A(w_2)$. Under DARA & $w_2 < w_1$: $r_A(w_2) > r_A(w_1)$, so $F_w < 0$. SOC: $F_\alpha < 0$. So $\partial \alpha^*/\partial w = -F_w/F_\alpha < 0$. Richer = less insurance.

11

Uncertainty & Expected Utility

★ harder

↗ PS 5, HW 4 Pt 2

Key concepts

Expected utility: $EU = \sum_s p_s\, u(w_s)$ for state-dependent wealth.
Risk aversion $\Leftrightarrow$ $u'' < 0$ (concave). Risk neutral: $u$ linear. Risk loving: $u'' > 0$.
Risk premium $\pi$: $u(\bar w - \pi) = E[u(w)]$ — what you'd pay to avoid the gamble.
Absolute Risk Aversion (ARA): $r_A(w) = -u''(w)/u'(w)$.
Relative Risk Aversion (RRA): $r_R(w) = w \cdot r_A(w)$.
CRRA utility: $u(w) = \dfrac{w^{1-\rho}}{1-\rho}$. Gives $r_R = \rho$ constant, $r_A = \rho/w$ (DARA).
CARA utility: $u(w) = -e^{-aw}/a$. Gives $r_A = a$ constant, $r_R = aw$ (IRRA).
Insurance FOC: $\dfrac{u'(w_2)}{u'(w_1)} = \dfrac{(1-p)q}{p(1-q)}$. Fair pricing ($q = p$) ⇒ full insurance $\alpha^* = L$.

Insurance — key results

$$\text{Fair: }q=p \;\Longrightarrow\; \alpha^* = L \text{ (full coverage) for any } u'' < 0$$ $$\text{Unfair: }q > p \;\Longrightarrow\; 0 < \alpha^* < L,\;\; \operatorname{sign}\!\left(\tfrac{\partial \alpha^*}{\partial w}\right) = \operatorname{sign}\!\bigl(r_A(w_1) - r_A(w_2)\bigr)$$

⚡ Key insight

Under fair insurance, any risk-averse agent buys full coverage — full stop. The interesting case is unfair pricing, where the choice depends on whether risk aversion grows or shrinks with wealth. DARA (CRRA, CARA→no — wait — CARA has constant $r_A$) is the most common assumption: rich people are less risk averse in dollar terms.

⚠ Common pitfall

$r_A$ vs. $r_R$ confusion is the #1 error here. CRRA has $r_R = \rho$ constant but $r_A = \rho/w$ decreasing in $w$. CARA has $r_A$ constant but $r_R$ increasing. Always check which is constant by name.

✎ Practice questions

$u(w) = -e^{-aw}/a$. Compute $r_A$ and $r_R$.
Show answer
$u'(w) = e^{-aw}$, $u''(w) = -a e^{-aw}$. So $r_A = a$ (constant). $r_R = aw$ (linear in $w$, increasing). This is CARA.
Stock returns +10% or −5% (prob 1/2 each). Bond return = 0%. Linear utility. What share in stocks?
Show answer
Risk neutral → maximize expected return. $E[\text{stock}] = 0.025 > 0$. Invest 100% in stocks.
Under CRRA, why does insurance demand scale with wealth?
Show answer
CRRA is homothetic — preferences over fractional gambles don't depend on absolute wealth. Doubling $(w, L)$ doubles optimal $\alpha$, so $\alpha^*/L$ depends only on $w/L$ — invariant to proportional scaling.

12

Oligopoly — Cournot & Bertrand

★ harder

↗ HW 9, Final 17 Q1, Final 21 Q4

Key concepts

Cournot = firms choose quantities simultaneously. Bertrand = firms choose prices.
Cournot best response (linear demand $p = a - bY$, MC $c$): $y_i = \dfrac{a-c}{2b} - \dfrac{y_{-i}}{2}$.
Symmetric Cournot NE (I firms): $$y_i^* = \frac{a-c}{b(I+1)}, \quad p^* = c + \frac{a-c}{I+1}, \quad \pi_i^* = \frac{(a-c)^2}{b(I+1)^2}.$$
Identity at NE: $\pi_i^* = (y_i^*)^2$ (for linear demand, identical firms, constant MC).
Bertrand (identical products, $I \ge 2$): $p = MC$, profits zero. Just 2 firms suffice to replicate PC.
Limit: Cournot $\xrightarrow{I \to \infty}$ PC. Markup decays as $1/(I+1)$.
Stackelberg (sequential quantities, F1 leader): leader produces more, gets more profit; follower produces less. $y_1^{Sb} > y_1^{Cb}$.

Cournot ↔ Bertrand summary (linear demand)

$$\text{Monopoly: } Y = \tfrac{a-c}{2b}, \quad p = \tfrac{a+c}{2}$$ $$\text{Cournot I-firm: } Y = \tfrac{I(a-c)}{b(I+1)}, \quad p = c + \tfrac{a-c}{I+1}$$ $$\text{Bertrand (I≥2, identical): } Y = \tfrac{a-c}{b}, \quad p = c$$ $$\text{Stackelberg: } y_1 = \tfrac{a-c}{2b}, \; y_2 = \tfrac{a-c}{4b}, \; Y = \tfrac{3(a-c)}{4b}, \; p = \tfrac{a+3c}{4}$$

⚡ Key insight

Stackelberg leader gets more profit than Cournot — commitment value. Follower gets less. Total industry quantity is higher than Cournot, so consumers prefer Stackelberg (higher CS).

⚠ Common pitfall

Bertrand ≠ "Cournot with prices." With identical products, Bertrand collapses to PC even with 2 firms — the strategic variable is what drives the equilibrium, not the number of firms. Differentiated Bertrand is a different story (positive margins) — but the homework version assumes identical products.

✎ Practice questions

Cournot duopoly: $p = 100 - Q$, MC = 20. Compute $y_1^*, y_2^*, p^*, \pi_i^*$.
Show answer
$y_i^* = 80/3 \approx 26.7$. $Y = 160/3$. $p^* = 100 - 160/3 \approx 46.7$. $\pi_i^* = (80/3)^2 \approx 711$.
Same demand and MC. 5-firm symmetric Cournot. What's $p^*$?
Show answer
$p^* = c + (a-c)/(I+1) = 20 + 80/6 \approx 33.3$. As $I$ grows, $p \to c = 20$.
Stackelberg (same demand, MC = 20). Leader and follower quantities, and total output. Do consumers prefer Stackelberg or Cournot?
Show answer
$y_1 = 40$, $y_2 = 20$, $Y = 60$. Cournot total was $53.3$. Stackelberg has higher $Y$, lower $p$ → higher CS, so consumers prefer Stackelberg.

13

Game Theory — NE, Mixed, SPNE

★ harder

↗ HW 9, Final 17 Q2, Final 21 Q3, Midterm 2

Key concepts

Pure NE: $(s_1^*, s_2^*)$ such that no player has a profitable unilateral deviation. Check cell-by-cell.
Dominant strategy: best response to any opponent action. Dominated: always worse than some other strategy.
Mixed NE: each player randomizes over actions; equilibrium probabilities make the opponent indifferent across pure strategies.
Mixed indifference trick: if player 1 plays mixed $(p, 1-p)$, then $p$ is set so player 2 is indifferent. The constraint comes from 2's payoffs, not 1's.
Best response correspondence: $BR_i(p_{-i})$ — for 2×2 games, a vertical/horizontal step function.
SPNE (Subgame Perfect NE): NE in every subgame. Found by backward induction. Rules out non-credible threats.
Stackelberg = SPNE of sequential quantity game. Leader internalizes follower's reaction.
Repeated games: cooperation can be sustained in long-run if discount factor $\beta$ is high (Tit-for-Tat, grim trigger).

Mixed NE recipe (2 × 2)

$$\text{Player 1 mixes } (p, 1-p) \text{ s.t. } \mathbb{E}[\pi_2|A] = \mathbb{E}[\pi_2|B]$$ $$\text{Player 2 mixes } (q, 1-q) \text{ s.t. } \mathbb{E}[\pi_1|A] = \mathbb{E}[\pi_1|B]$$

⚡ Key insight

For backward induction in a finite tree: start at the last decision node, choose the best action there, replace that node with its payoff, repeat. The remaining "strategy" at non-last nodes is the SPNE strategy. Non-credible threats — actions that wouldn't actually be best at the moment of decision — get ruled out.

⚠ Common pitfall

Mixed-strategy probabilities are set using the opponent's payoffs. Many students compute "what makes me indifferent" instead — that gives the opponent's mixing probability, not yours. Swap the perspective.

✎ Practice questions

Game: $\begin{pmatrix}(-2,-2) & (4,0)\\(0,6) & (0,0)\end{pmatrix}$ (rows = F1's actions Enter/Don't, cols = F2's). Find all NE.
Show answer
Pure: (Enter, Don't) and (Don't, Enter). Mixed: $p_1 = 3/4$ (makes F2 indifferent: $-2p_1 + 6(1-p_1) = 0$), $p_2 = 2/3$ (makes F1 indifferent: $-2p_2 + 4(1-p_2) = 0$).
In Bat's dilemma with payoffs $(7,7)$, $(5,0)$, $(0,5)$, $(6,6)$ for (Share, Share), (Share, Don't), (Don't, Share), (Don't, Don't): is there a dominant strategy?
Show answer
No. Bat A's best response to "Share" is Share ($7>0$). To "Don't" is Don't ($6>5$). No dominant strategy. Two pure NE: (Share, Share) and (Don't, Don't).
Why does backward induction rule out non-credible threats?
Show answer
Backward induction requires each player to play optimally at every decision node. A threat to "punish off the equilibrium path" is non-credible if it wouldn't actually maximize the threatener's payoff at that node — backward induction discards such threats.

14

Externalities & Pigouvian Tax

★ harder

↗ HW 10, Final 21 Q2

Key concepts

Externality: one agent's action affects another's payoff not via prices — e.g., pollution, congestion, network effects.
Decentralized = inefficient: agents don't see external cost/benefit. Negative externalities → too much activity; positive → too little.
Social planner internalizes all effects: maximize joint utility / total surplus.
Pigouvian tax = marginal external cost evaluated at the social optimum. Imposed on the producer of the externality, restoring the first best via decentralized choice.
Common-pool / open access (e.g., hunting in shared forest): each user's entry imposes congestion on others. Restore efficiency via quota, permit, or entry fee.
Coase theorem: If property rights are clear and transaction costs are zero, parties can bargain to efficient outcome regardless of who holds the right. (Heroic assumptions.)

Pigouvian tax recipe

$$\text{1. Solve private optimum } x_p^*. \;\;\;\;\;\; \text{2. Solve social optimum } x_s^* < x_p^*.$$ $$\text{3. Tax } t \text{ s.t. private FOC with tax matches social FOC.}$$ $$\text{HW10 example: } b_p^*(t) = 9/(1+t),\;\; b_s^* = 5 \Rightarrow t = 0.8.$$

⚡ Key insight

Pigouvian taxation is the magic of public economics: a single corrective price restores first-best efficiency without dictating quantities. Agents still choose, but they choose the right thing because they now face true social marginal cost.

⚠ Common pitfall

The Pigouvian tax equals MEC at the social optimum, not at the private optimum. Many students use the private-optimum MEC and over-tax. Always compute the tax at the desired (efficient) quantity.

✎ Practice questions

150 hunters, forest yields $\tfrac{1}{2}x^{-1/2}$ per hunter ($x$ in forest), plains yield 0.05 per hunter. Efficient $x$?
Show answer
Maximize total: $\tfrac{1}{2}x^{1/2} + 0.05(150-x)$. FOC: $\tfrac{1}{4}x^{-1/2} = 0.05 \Rightarrow x = 25$.
Free-entry allocation in same problem: how many hunters in forest?
Show answer
Indifference: $\tfrac{1}{2}x^{-1/2} = 0.05 \Rightarrow x = 100$. Way too many (negative externality of crowding).
Pigouvian permit price (game sold at $5/ton) to achieve $x = 25$?
Show answer
In equilibrium, $5 \cdot \tfrac{1}{2}\cdot 25^{-1/2} - p = 5 \cdot 0.05 \Rightarrow p = 0.25$.

15

Adverse Selection / Lemons

★ harder

↗ HW 10, Final 17 Q4, Final 21 Q1

Key concepts

Asymmetric information: one side of the market knows more than the other (sellers know quality, buyers don't).
Adverse selection: only the worst types want to participate at the pooling price, dragging down expected quality.
Market unraveling: if the buyer's value premium isn't large enough, no price clears the market. Trade collapses despite gains from trade existing for every type.
Lemons condition (linear values): sellers $\sim q$, buyers $\sim \kappa q$, $q \sim U[0,1]$. Market survives iff $\kappa > 2$. For $\kappa = 3/2$: unravels.
Signaling (Spence): informed party takes a costly action (education, warranty) cheaper for high types than low types — separating equilibrium.
Screening (Rothschild-Stiglitz): uninformed party offers a menu; types self-select.
Minimum quality standard: exclude $q < \bar q$. Pro: restores some trade. Con: excludes efficient low-q trades. Welfare improvement over no-trade but not first best.

Unraveling — uniform model

$$E[q | q \le p] = p/2, \quad \text{buyer's WTP} = \kappa \cdot p/2$$ $$\text{Trade iff } \kappa \ge 2. \;\; (\kappa = 3/2 \Rightarrow \text{ unravels.})$$ $$\text{With min quality } \bar q: \quad p \in [\bar q, \tfrac{3\bar q}{2-\kappa}\text{-style range}]$$

⚡ Key insight

Lemons unraveling is not about buyers being suspicious — it's a coordination failure. Even with full common knowledge of the model, there's no positive price where the average car offered for sale is worth what the buyer pays.

⚠ Common pitfall

A minimum quality standard is Kaldor-Hicks improving, not Pareto improving. Sellers below $\bar q$ are forced out of the market and lose option value. Don't claim "everyone is better off."

✎ Practice questions

$q \sim U[0,1]$, sellers value at $q$, buyers value at $\tfrac{3}{2}q$. Does the market unravel?
Show answer
At price $p$, sellers $q \le p$ participate, $E[q|q\le p] = p/2$. Buyer's expected value: $\tfrac{3}{4}p < p$ for any $p > 0$. Yes, market unravels.
Same model, but a minimum quality standard $\bar q$. For what $p$ does trade occur?
Show answer
$E[q | \bar q \le q \le p] = (\bar q + p)/2$. Buyer participates iff $\tfrac{3}{2} \cdot \tfrac{\bar q + p}{2} \ge p \Leftrightarrow p \le 3\bar q$. So $\bar q \le p \le 3\bar q$.
Restaurants: 30% good (value $30 to you, cost $20 to chef), 70% bad ($15 to you, cost $10 to chef). Pooling pricing?
Show answer
WTP $= 0.3 \cdot 30 + 0.7 \cdot 15 = 19.5$. Good chef sells iff $19.5 \ge 20$ — no. So only bad meals at price $\le 19.5$. WTP drops to $15$. Market for good restaurants unravels.

16

Moral Hazard / Principal-Agent

★ harder

↗ Final 17 Q3

Key concepts

Moral hazard: one party (agent) takes a costly action (effort) the other (principal) can't observe.
Linear contract: $w = a + by$ where $y$ = output, $a$ = base, $b$ = sensitivity.
Output process: $y = e + \varepsilon$, $\varepsilon \sim N(0, \sigma^2)$.
Agent's CARA-flavored EU: $a + be - \tfrac{1}{2}b^2\sigma^2 - \tfrac{1}{2}e^2$. FOC in $e$: $e^* = b$.
Two constraints on principal's problem:
• Incentive Compatibility (IC): $e^* = b$.
• Individual Rationality (IR): $EU(e^*) \ge \bar u$.
Principal maximizes $E[y] - E[w] = e - a - be$ subject to IC + IR.
Optimal $b$: $$b^* = \frac{1}{1 + \sigma^2}.$$
Noise raises insurance value (lower $b$ shifts risk back to principal). $\sigma = 0$: $b^* = 1$, full incentive. $\sigma \to \infty$: $b^* \to 0$, no incentive.
Fixed-wage contract ($b = 0$): agent's optimal effort $= 0$, principal's profit $= 0$. Need contingent pay.
Observable effort: pay directly on effort, no risk borne by agent → first-best.

Optimal linear contract

$$y = e + \varepsilon, \;\; \varepsilon \sim N(0, \sigma^2)$$ $$\text{Agent: } \max_e a + be - \tfrac{1}{2}b^2\sigma^2 - \tfrac{1}{2}e^2 \;\;\Longrightarrow\;\; e^* = b$$ $$\text{Principal: } \max_{a,b} e - a - be \;\;\text{s.t. IR, IC} \;\;\Longrightarrow\;\; b^* = \frac{1}{1+\sigma^2}$$

⚡ Key insight

$b^* = 1/(1+\sigma^2)$ captures the risk-incentive tradeoff. With no noise ($\sigma^2 = 0$), pay 100% on output ($b=1$) — the agent's effort fully determines output, no insurance needed. With huge noise, paying-on-output mostly transfers risk, not incentive — so $b \to 0$.

⚠ Common pitfall

The IR constraint binds at the optimum (firm pulls agent down to outside option). Don't forget to use the IR equality to substitute out $a$. Many students leave $a$ as a free variable and get stuck.

✎ Practice questions

CEO chooses $e$ to max $a + be - \tfrac{b^2\sigma^2}{2} - \tfrac{e^2}{2}$. Find $e^*$.
Show answer
FOC: $b - e = 0$, so $e^* = b$.
If $\sigma^2 = 0$, what's $b^*$? Why does this make sense?
Show answer
$b^* = 1$. With no noise, output perfectly reveals effort — no risk-sharing needed, just pay 100% on output (effectively, pay on effort).
Under a fixed-wage contract $w = a$, what's the firm's profit? Why?
Show answer
Agent has no incentive to work: $e^* = 0$. Output $= 0 + \varepsilon$, expected output $= 0$. So expected profit $= -a < 0$. The firm would set $a = 0$ if IR doesn't bind, but then nobody accepts. Either way, zero profit.

★

Formulas at a Glance

cheatsheet

Consumer choice

$$\text{MRS}_{1,2} = \frac{u_1}{u_2}, \quad \text{interior FOC: } \text{MRS}_{1,2} = \frac{p_1}{p_2}$$ $$\text{Cobb-Douglas: } x_i^* = \frac{\alpha_i M}{p_i}$$ $$\text{Quasi-linear: } \phi'(x_1^*) = p_1/p_2 \text{ (income-independent at interior)}$$

Intertemporal

$$c_0 + \frac{c_1}{1+r} = M_0 + \frac{M_1}{1+r}$$ $$\text{Euler: } u'(c_0) = \frac{1+r}{1+\delta} u'(c_1); \;\; r=\delta \Rightarrow c_0 = c_1$$

Uncertainty

$$r_A(w) = -\frac{u''(w)}{u'(w)}, \quad r_R(w) = w \cdot r_A(w)$$ $$\text{CRRA: } u = \frac{w^{1-\rho}}{1-\rho}, \;\; r_R = \rho, \;\; r_A = \rho/w$$ $$\text{Insurance FOC: } \frac{u'(w_2)}{u'(w_1)} = \frac{(1-p)q}{p(1-q)}$$

Producer / market structure

$$\text{PC: } p = c, \quad Y_{PC} = (a-c)/b, \quad \Pi = 0$$ $$\text{Monopoly: } y_M = (a-c)/(2b), \;\; p_M = (a+c)/2, \;\; \Pi_M = (a-c)^2/(4b)$$ $$\text{Cournot } I: y_O = \tfrac{a-c}{b(I+1)}, \;\; p_O = c + \tfrac{a-c}{I+1}, \;\; \pi_O = \tfrac{(a-c)^2}{b(I+1)^2}$$ $$\text{Bertrand (} I \ge 2\text{): } p = c, \;\; \Pi = 0$$ $$\text{Stackelberg: } y_1 = \tfrac{a-c}{2b}, \;\; y_2 = \tfrac{a-c}{4b}$$

Taxation & DWL

$$\frac{\partial p^*}{\partial t} = \frac{\varepsilon_S}{\varepsilon_S - \varepsilon_D}, \quad DWL_\text{linear} = \frac{t^2}{2b}$$

Game theory

$$\text{Mixed NE (2}\times\text{2): probability of player } i \text{ set by player } -i\text{'s indifference}$$ $$\text{SPNE: backward induction; rules out non-credible threats}$$

Lemons (linear)

$$E[q|q\le p] = p/2; \;\; \text{trade iff buyer multiplier} \ge 2$$

Moral hazard (linear contract)

$$e^* = b, \;\; b^* = \frac{1}{1+\sigma^2}, \;\; \text{(fixed wage } b=0 \Rightarrow e^* = 0\text{)}$$

IFT

$$F(x^*, \theta) = 0 \;\;\Longrightarrow\;\; \frac{dx^*}{d\theta} = -\frac{F_\theta}{F_x}$$

Econ 101A — Final Exam Study Guide

Contents

Math Toolkit

Preferences & Utility

UMP / Marshallian Demand

Intertemporal Choice

Producer Theory & Supply

Perfect Competition & Taxation

Monopoly

Walrasian Equilibrium & Welfare Theorems

EMP / Hicksian Demand / Duality

Comparative Statics & IFT

Uncertainty & Expected Utility

Oligopoly — Cournot & Bertrand

Game Theory — NE, Mixed, SPNE

Externalities & Pigouvian Tax

Adverse Selection / Lemons

Moral Hazard / Principal-Agent

Formulas at a Glance