Almost sure convergence
Xₙ→X almost surely means P({ω: Xₙ(ω)→X(ω)})=1. Variables are defined on a common probability space.
Find a concept
REFERENCE SHEET
Modes of convergence
Distinguish pathwise behavior, small error probabilities, average error, and convergence of laws.
Before this Random variables & moments, Convergence theorems & iterated integrals
Lᵖ ⇒ probability ⇒ distribution; almost surely ⇒ probability
Definitions
Convergence in probability and Lᵖ
Convergence in probability means P(|Xₙ−X|>ε)→0 for every ε>0. For 1≤p<∞, Lᵖ convergence means E|Xₙ−X|ᵖ→0.
Convergence in distribution
For real random variables, Xₙ converges in distribution to X if their distribution functions converge at every continuity point of the distribution function of X. A common probability space is not required.
Key theorems
Implications
Almost sure convergence implies convergence in probability. Lᵖ convergence implies convergence in probability by Markov’s inequality. Convergence in probability implies convergence in distribution.
A constant limit
Convergence in distribution to a constant c implies convergence in probability to c.
Dominated convergence in Lᵖ
If Xₙ→X almost surely and |Xₙ|≤Y for all n, with E[Yᵖ]<∞, then Xₙ→X in Lᵖ. Apply dominated convergence to |Xₙ−X|ᵖ≤2ᵖYᵖ.
Notable applications
Limit theorems
The strong law concerns almost sure convergence of averages; the central limit theorem concerns convergence in distribution of rescaled errors.
Interchanging limits and expectations
L¹ convergence guarantees E[Xₙ]→E[X]. Convergence in probability alone does not control rare, large values.
Work it out
Try before revealing01 / Counterexample
On (0,1) with uniform probability, let Xₙ(ω)=n when ω<1/n and 0 otherwise. Find the almost sure limit and E[Xₙ].
Hint
Fix a positive ω first, then compute the area under the spike.
Solution
For any ω>0, eventually n>1/ω, so Xₙ(ω)=0. Thus Xₙ→0 almost surely, but E[Xₙ]=n·(1/n)=1. There is no convergence in L¹.
02 / Prove
Show Lᵖ convergence implies convergence in probability.
Hint
Apply Markov’s inequality to |Xₙ−X|ᵖ.
Solution
For ε>0, P(|Xₙ−X|>ε)≤E|Xₙ−X|ᵖ/εᵖ→0. The bound controls each fixed error threshold.
03 / Counterexample
Let X take values −1 and 1 equally likely, and set Xₙ=−X for every n. Does Xₙ converge in distribution to X? In probability?
Hint
Compare the distributions and then |Xₙ−X|.
Solution
Every Xₙ has the same distribution as X, so convergence in distribution holds. But |Xₙ−X|=2 surely, so convergence in probability fails.
Connect the ideas
Go deeper
Further reading and full-length treatments: