Concentration Inequalities
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# Concentration Inequalities ## 🎯 Learning Objectives - State Markov's, Chebyshev's, and Chernoff's inequalities - Apply concentration bounds to analyze randomized algorithms - Derive tail bounds for sums of independent random variables - Understand the trade-off between tail decay and moment assumptions * * * ## 1...

Concentration Inequalities
🎯 Learning Objectives
- State Markov's, Chebyshev's, and Chernoff's inequalities
- Apply concentration bounds to analyze randomized algorithms
- Derive tail bounds for sums of independent random variables
- Understand the trade-off between tail decay and moment assumptions
1.1 Intuition: How Fast Does a Random Variable Concentrate?
Concentration inequalities quantify how quickly a random variable approaches its mean. They're the foundation of randomized algorithm analysis and statistical learning theory.
🔑 Key Insight: Under weak assumptions (finite variance → Chebyshev; boundedness → Hoeffding; subgaussian → Chernoff), sums of independent random variables concentrate exponentially fast around their mean.
1.2 Key Inequalities
Markov's Inequality
Pr(X≥t)≤tE[X]for X≥0Example: If average income is 50k,atmost1/4ofpeopleearn\geq 200k$.
Chebyshev's Inequality
Pr(∣X−E[X]∣≥t)≤t2Var(X)Example: For Xˉ from n i.i.d. samples: Pr(∣Xˉ−μ∣≥ϵ)≤nϵ2σ2
Chernoff Bound (Multiplicative)
For X=∑Xi where Xi∈[0,1] independent:
Hoeffding's Inequality
For Xi∈[ai,bi] independent:
1.3 Comparison
| Bound | Assumptions | Decay | Use Case |
|---|---|---|---|
| Markov | X≥0 | 1/t | Generic |
| Chebyshev | Finite variance | 1/t2 | Generic with variance |
| Chernoff | Independent, bounded | exp(−cμδ2) | Sums of bounded variables |
| Hoeffding | Independent, bounded | exp(−2nt2/(b−a)2) | Sample means |
✅ Practice Questions
Q1: Use Chebyshev to bound Pr(∣Xˉ−μ∣≥ϵ) for n i.i.d. samples with variance σ2.
SolutionVar(Xˉ)=σ2/n. By Chebyshev: Pr(∣Xˉ−μ∣≥ϵ)≤nϵ2σ2. Q2: Compare Hoeffding and Chebyshev bounds. Which is tighter for bounded variables? SolutionHoeffding decays as exp(−cnϵ2), exponentially faster than Chebyshev's 1/(nϵ2). For bounded variables, Hoeffding is exponentially tighter. Q3: How many samples n needed to estimate a mean within ±0.01 with 95% confidence for bounded [0,1] variables? SolutionHoeffding: 2exp(−2n(0.01)2)≤0.05. So exp(−0.0002n)≤0.025, −0.0002n≤ln(0.025), n≥ln(40)/0.0002≈18449. Join Discord NextRandomized SVD