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The bookUnit 4 · Limit Theorems and Markov Chains74 / 83

Modes of convergence, WLLN, SLLN, CLT

Why this is asked: Know the implication diagram and one counterexample per missing arrow. The typewriter sequence and the moving-bump are the two you need.

The diagram

   L^p  ──────▶  probability  ──────▶  distribution
                     ▲
almost sure ─────────┘

No other arrows hold in general. Partial converses:

  • convergence in probability ⇒ an a.s. convergent subsequence;
  • convergence in distribution to a constant ⇒ convergence in probability;
  • probability + uniform integrability L1\Rightarrow L^{1}; P(- \sum P(|XnXX_{n} - X| >ε)<a.s.(> \varepsilon) < \infty \Rightarrow a.s. (Borel–Cantelli).

The counterexamples

Missing arrow Example
probability ⇏ a.s. typewriter: indicators of [k/2n,(k+1)/2n][k/2^{n}, (k+1)/2^{n}] sweeping [0,1]
a.s. ⇏ L1L^{1} Xn=n1(0,1/n)X_{n} = n\cdot1_{(0,1/n)}: 0a.s.,E[Xn]=1\to 0 a.s., E[X_{n}] = 1
probability ⇏ L1L^{1} same
distribution ⇏ probability X ~ N(0,1),Xn=XN(0,1), X_{n} = -X for all n
L1L^{1} ⇏ a.s. typewriter again

Laws of large numbers

  • WLLN (in probability) needs finite mean; SLLN (a.s.) also holds with finite mean (Kolmogorov).
  • Fails without a mean: the Cauchy sample mean has the same Cauchy distribution for every n.

Central limit theorem

n(Xˉμ)/σN(0,1)\sqrt{n}({\bar{X}} - \mu)/\sigma \to N(0,1) whenever σ2<\sigma^{2} < \infty. Consequences examined: P(Snnμ)- P(S_{n} \le n\mu) \to ½, so intervals with the mean as an endpoint have limiting probability ½, while intervals containing μ\mu in the interior have probability → 1.

  • Delta method: n(g(Xˉ)g(μ))N(0,σ2g(μ)2)\sqrt{n}(g({\bar{X}}) - g(\mu)) \to N(0, \sigma^{2}g'(\mu)^{2}).
  • Slutsky: XnXX_{n} \Rightarrow X and YncY_{n} \to c in probability Xn+YnX+c,XnYn\Rightarrow X_{n} + Y_{n} \Rightarrow X + c, X_{n}Y_{n} \Rightarrow cX.

Key takeaways

  • a.s. and L^p both imply probability, which implies distribution; nothing else.
  • Typewriter kills "probability ⇒ a.s."; n·1(0,1/n)1_{(0,1/n)} kills "a.s.L1a.s. \Rightarrow L^{1}".
  • CLT at the boundary gives ½ — a favourite Part-C trick.

See it move

The Central Limit Theorem, watchedinteractivefree

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