Mathematics for Economics — Econ 503 · Example-heavy Notes अर्थशास्त्रका लागि गणित — Econ 503 · उदाहरण-केन्द्रित नोट

Notation

Set: a well-defined collection of distinct objects. Element: $x \in A$ means $x$ belongs to $A$. Empty set: $\varnothing$ or $\{\}$. Subset: $A \subseteq B$ means every $x \in A$ is also in $B$.

Operations: Union $A \cup B$; Intersection $A \cap B$; Difference $A \setminus B$; Complement $A^c$; Cartesian product $A \times B = \{(a, b) : a \in A, b \in B\}$.

Hierarchy

$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$.

• $\mathbb{N}$: natural numbers $\{1, 2, 3, \ldots\}$.
• $\mathbb{Z}$: integers $\{\ldots, -2, -1, 0, 1, 2, \ldots\}$.
• $\mathbb{Q}$: rationals $\{p/q : p, q \in \mathbb{Z}, q \neq 0\}$.
• $\mathbb{R}$: reals — fills the gaps left by $\mathbb{Q}$ (irrationals like $\sqrt 2$, $\pi$, $e$).
• $\mathbb{R}^n$: $n$-dimensional Euclidean space $\{(x_1, x_2, \ldots, x_n) : x_i \in \mathbb{R}\}$.

Definitions

A function $f: X \to Y$ assigns each $x \in X$ exactly one $y \in Y$. Notation: $y = f(x)$.

• Domain $\text{Dom}(f) = X$. Codomain $= Y$. Range = $\{f(x) : x \in X\} \subseteq Y$.
• Injective (one-to-one): $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
• Surjective (onto): Range = Codomain.
• Bijective: both injective and surjective. Then $f^{-1}$ exists.
• Composite: $(g \circ f)(x) = g(f(x))$.

Sequence convergence

$\{x_n\}$ converges to $L$, written $\lim_{n \to \infty} x_n = L$, if for every $\varepsilon > 0$ there exists $N$ such that $n \geq N \Rightarrow |x_n - L| < \varepsilon$.

A sequence is bounded if $\exists M$ with $|x_n| \leq M$ for all $n$. Monotone if non-decreasing or non-increasing throughout.

Monotone Convergence Theorem: every bounded monotone sequence converges (in $\mathbb{R}$).

Geometric series

$$\sum_{n=0}^\infty r^n = 1 + r + r^2 + \cdots = \frac{1}{1 - r}, \quad |r| < 1.$$

If $|r| \geq 1$, the series diverges. This is the workhorse formula behind the Keynesian multiplier ($1/(1-b)$ where $b$ is MPC).

$\varepsilon$-$\delta$ definition

$\lim_{x \to x_0} f(x) = L$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that $0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < \varepsilon$.

$f$ is continuous at $x_0$ iff $\lim_{x \to x_0} f(x) = f(x_0)$ (limit exists AND equals function value AND function is defined there).

Basic rules

Partial derivative

For $z = f(x, y)$, holding $y$ constant and differentiating w.r.t. $x$:

Total differential:

This is the foundation of comparative statics.

Chain rule (multivariable)

If $z = f(x, y)$ and both $x = x(t)$, $y = y(t)$:

If $z = f(x_1, x_2, \ldots, x_n)$ where each $x_i = x_i(t_1, \ldots, t_m)$:

Single-variable

To maximize/minimize $f(x)$:

1. FOC: $f'(x^*) = 0$.
2. SOC: $f''(x^*) < 0$ ⇒ local max; $f''(x^*) > 0$ ⇒ local min; $f''(x^*) = 0$ ⇒ inconclusive (use higher derivatives).

Multi-variable

For $f(x_1, x_2, \ldots, x_n)$:

1. FOC: all partials zero: $\nabla f(x^*) = 0$.
2. SOC: Hessian $H = [f_{ij}]$ must be (locally):
   • Negative definite (ND) for strict max: all leading principal minors alternate in sign starting negative.
   • Positive definite (PD) for strict min: all leading principal minors positive.

For 2 variables: $\det H = f_{11} f_{22} - f_{12}^2$. Then:

• $f_{11} < 0$ AND $\det H > 0$ ⇒ max.
• $f_{11} > 0$ AND $\det H > 0$ ⇒ min.
• $\det H < 0$ ⇒ saddle point.

Test for concavity via Hessian

A $C^2$ function $f$ on a convex open set is:

• Concave iff Hessian $H(x)$ is negative semi-definite (NSD) for all $x$.
• Strictly concave if $H(x)$ is negative definite (ND) for all $x$.
• Likewise convex iff PSD, strictly convex if PD.

2-variable: $f$ concave iff $f_{xx} \leq 0$, $f_{yy} \leq 0$, $\det H \geq 0$ everywhere.

Method

To solve $\max f(x, y)$ subject to $g(x, y) = c$:

1. Form the Lagrangian: $\mathcal{L}(x, y, \lambda) = f(x, y) + \lambda [c - g(x, y)]$.
2. Set partials to zero:
   $$\mathcal{L}_x = 0, \quad \mathcal{L}_y = 0, \quad \mathcal{L}_\lambda = 0.$$
3. Solve for $x^*, y^*, \lambda^*$.
4. (Optional) Verify SOC via bordered Hessian.

Economic interpretation of $\lambda^*$: shadow price of the constraint. $\dfrac{df^*}{dc} = \lambda^*$ (envelope theorem).

SOC for 2-variable constrained max

Define the bordered Hessian:

where $g_i = \partial g/\partial x_i$ at the optimum. Then:

• $\det \bar H > 0$ ⇒ constrained maximum.
• $\det \bar H < 0$ ⇒ constrained minimum.

Setup

Maximize $f(x)$ subject to $g_i(x) \leq b_i$ for $i = 1, \ldots, m$ and $x \geq 0$.

Form Lagrangian $\mathcal{L} = f(x) + \sum_i \lambda_i [b_i - g_i(x)]$.

KKT conditions (necessary; sufficient under constraint qualifications + concave $f$, convex $g$):

Complementary slackness: either a constraint is binding ($g_i(x^*) = b_i$, $\lambda_i > 0$) or it is slack ($g_i(x^*) < b_i$, $\lambda_i = 0$).

Envelope theorem (unconstrained)

If $x^*(a)$ solves $\max_x f(x, a)$, then

Only the direct effect of $a$ on $f$ matters; the indirect effect via $x^*$ is zero at the optimum (because of FOC).

Envelope theorem (constrained)

If $\max f(x; a)$ s.t. $g(x; a) = b$ has Lagrangian $\mathcal{L}$:

Form and solution

General solution = complementary function (CF) + particular integral (PI):

If initial condition $y(0) = y_0$, then $A = y_0 - b/a$:

Stability: $a > 0$ ⇒ $e^{-at} \to 0$, so $y(t) \to b/a$ as $t \to \infty$ — stable. $a < 0$ ⇒ diverges — unstable.

Integrating factor method

Form: $\dot y + a(t) y = b(t)$.

Multiply both sides by integrating factor $\mu(t) = e^{\int a(t) dt}$:

Integrate: $\mu y = \int \mu b \, dt + C$, hence

Form and solution

Particular integral: $y_p = b/a_2$ (when $a_2 \neq 0$).

Complementary function: solve the homogeneous version $\ddot y + a_1 \dot y + a_2 y = 0$ via characteristic equation:

Three cases:

• Case 1: distinct real roots $r_1 \neq r_2$: $$y_c = A_1 e^{r_1 t} + A_2 e^{r_2 t}.$$
• Case 2: repeated real root $r_1 = r_2 = r$: $$y_c = (A_1 + A_2 t) e^{r t}.$$
• Case 3: complex roots $r = \alpha \pm i\beta$: $$y_c = e^{\alpha t} (A_1 \cos \beta t + A_2 \sin \beta t).$$ Oscillatory; damped if $\alpha < 0$, explosive if $\alpha > 0$.

Form and solution

Steady state $y^* = b/(1-a)$ (when $a \neq 1$).

General solution:

Stability conditions:

• $0 < a < 1$: monotone convergence to $y^*$ — stable.
• $-1 < a < 0$: oscillating convergence — stable.
• $a > 1$: monotone divergence — unstable.
• $a < -1$: explosive oscillation — unstable.
• $a = 1$: uniform path (no steady state if $b \neq 0$).
• $a = -1$: bounded oscillation around average (neutrally stable).

In short: stable iff $|a| < 1$.

Form and solution

Particular: $y_p = b/(1 + a_1 + a_2)$ (when denominator $\neq 0$).

Characteristic: $r^2 + a_1 r + a_2 = 0$. Three cases parallel to differential case but use $r^t$ instead of $e^{rt}$:

• Distinct real $r_1 \neq r_2$: $y_c = A_1 r_1^t + A_2 r_2^t$.
• Repeated $r$: $y_c = (A_1 + A_2 t) r^t$.
• Complex $r = \alpha \pm i\beta$: $y_c = R^t (A_1 \cos \theta t + A_2 \sin \theta t)$ where $R = \sqrt{\alpha^2 + \beta^2}$ and $\theta = \arctan(\beta/\alpha)$.

Stability: $|r_i| < 1$ for all roots.

Schur conditions (quick test for stability of $r^2 + a_1 r + a_2 = 0$):

Setup

• $x(t)$: state variable — what the system is (capital, inventory, debt).
• $u(t)$: control variable — what the decision-maker chooses (investment, consumption, hiring).
• $F$: per-period objective (e.g., utility, profit).
• $g$: equation of motion for state.

Hamiltonian and necessary conditions

Define the Hamiltonian:

$\mu(t)$ = co-state variable = shadow price of $x$ at time $t$.

Necessary conditions for optimum:

1. $\dfrac{\partial H}{\partial u} = 0$ — Hamiltonian maximized w.r.t. control.
2. $\dot \mu = -\dfrac{\partial H}{\partial x}$ — co-state equation of motion.
3. $\dot x = \dfrac{\partial H}{\partial \mu} = g(x, u, t)$ — state equation (recovers original).
4. Transversality: $\mu(T) = 0$ if $x(T)$ is free; or $x(T) = x_T$ if endpoint fixed.

Maximization form

In matrix notation: $\max c^\top x$ s.t. $A x \leq b$, $x \geq 0$.

Steps

1. Convert inequality constraints to equalities by adding slack variables: $a_{i1} x_1 + \cdots + a_{in} x_n + s_i = b_i$.
2. Initial basic feasible solution: all $x_j = 0$, $s_i = b_i$. $Z = 0$.
3. Check optimality: look at the row-0 ("$Z$-row") coefficients of non-basic variables. If all $\geq 0$ (for max), we're done.
4. Otherwise, choose entering variable: the non-basic with the most negative row-0 coefficient.
5. Leaving variable: minimum-ratio test — for each row, compute $b_i / a_{i,\text{entering}}$ (only positive coefficients); the smallest ratio identifies the leaving variable.
6. Pivot: row operations to make the entering variable basic.
7. Repeat from step 3.

Primal-dual pair

Strong duality theorem: if either has an optimum, both do, and their values coincide.

Economic interpretation: dual variable $y_i$ = shadow price of primal resource $i$ — the increase in optimal $Z$ from relaxing $b_i$ by one unit.