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Here’s a structured summary of core topics, concepts, principles, and equations essential for the exam, followed by Mermaid knowledge graphs to visualize relationships.
Discrete Distributions:
(a+b)/2), Variance ((n²-1)/12).p), Variance (p(1-p)).nCk p^k (1-p)^(n-k)), Mean (np), Variance (np(1-p)).(1-p)^(k-1) p), Mean (1/p), Variance ((1-p)/p²).(e^-λ λ^k)/k!), Mean (λ), Variance (λ).Continuous Distributions:
1/(b-a)), CDF, Mean ((a+b)/2), Variance ((b-a)²/12).λe^-λx), CDF (1-e^-λx), Mean (1/λ), Variance (1/λ²).(1/σ√2π) e^(-(x-μ)²/2σ²)), Mean (μ), Variance (σ²).(β^α/Γ(α)) x^(α-1) e^-βx), Mean (α/β), Variance (α/β²).Γ(α+β)/Γ(α)Γ(β) x^(α-1) (1-x)^(β-1)), Mean (α/(α+β)), Variance (αβ/((α+β)²(α+β+1))).P(X ≥ c) ≤ μ/c (for non-negative X).P(|X-μ| ≥ kσ) ≤ 1/k².P(|X̄ - μ| > δ) ≤ σ²/(nδ²) where X̄ = (X₁ + ... + Xₙ)/n.(Y - nμ)/(σ√n) ≈ N(0,1) where Y = X₁ + ... + Xₙ.L(θ) = ∏ f_X(x_i; θ).θ* = argmax_θ L(θ).f_X,Y(x,y).E[X|Y=y] = ∫ x f_X|Y(x|y) dx.Positive Definite Matrices:
A is positive definite if xᵀAx > 0 for all x ≠ 0.Properties:
A^k is positive definite for k ≥ 1.A are positive.A⁻¹ is positive definite.rA is positive definite for r > 0.Eigenvalues & Eigenvectors:
Av = λv.det(A - λI) = 0 (characteristic equation).A has n linearly independent eigenvectors, it is diagonalizable.Null(A) = {x | Ax = 0}.Convex Functions:
f(αx + (1-α)y) ≤ αf(x) + (1-α)f(y) for α ∈ [0,1].f(x) = 1/x (convex for x > 0), g(x) = 1 + |x| (convex).Composition rules:
g ∘ f is convex if g is convex and non-decreasing, and f is convex.L = (1/n) Σ (f(xᵢ) - yᵢ)² (for f(x) = w₁x₁ + w₂x₂ + ...). (Note: Question 3 uses L = - (1/n) Σ (f(xᵢ) - yᵢ)², which is negative MSE.)C = (1/n) XᵀX (for centered data).C with largest eigenvalue).Standard Form:
cᵀy subject to Ay ≤ b, y ≥ 0.Example (Question 14):
Maximize Z = 60y₁ + 45y₂
Subject to:
2y₁ + 2y₂ ≤ 200
y₁ + 3y₂ ≤ 140
y₁, y₂ ≥ 0Continuous Functions:
lim_{x→a} f(x) = f(a).Examples:
sin(x)/x (continuous at x=0 if defined as 1).x² + 3 for x ≥ 3, 3x + 3 for x < 3).Maximizing Area with Fixed Perimeter:
P, the maximum area is achieved when it’s a square (side = P/4).P = 600 ⇒ side = 150 ⇒ area = 22500.Constrained Optimization:
f(x,y) subject to g(x,y) = c, solve ∇f = λ∇g.f(x,y) = x + y subject to x² + y² = 50. Solution: x = y = 5 ⇒ f(x,y) = 10.| Topic | Equation | |||
|---|---|---|---|---|
| Markov’s Inequality | P(X ≥ c) ≤ μ/c | |||
| Chebyshev’s Inequality | `P( | X-μ | ≥ kσ) ≤ 1/k²` | |
| WLLN | `P( | X̄ - μ | > δ) ≤ σ²/(nδ²)` | |
| CLT | (Y - nμ)/(σ√n) ≈ N(0,1) | |||
| MLE | θ* = argmax_θ ∏ f_X(x_i; θ) | |||
| Conditional Expectation | `E[X | Y=y] = ∫ x f_X | Y(x | y) dx` |
| PCA Variance | Projected Variance = λ₁ (largest eigenvalue of covariance matrix) | |||
| Linear Regression Loss | L = (1/n) Σ (f(xᵢ) - yᵢ)² (or negative for Question 3) | |||
| Lagrange Multipliers | ∇f = λ∇g |