Convergence in Probability and Convergence with Probability 1


For a random vector X and a sequence of random vectors X_{1},\cdots X_{n},\cdots defined on a probability space (\Omega,\mathcal{F},P).

1. Convergence in Probability or Convergence of probability measures, is kind of convergence in measure, in concept of the measure theory.
X_{n}\stackrel{P}{\to}X: P\{||X_{n}-X||>\epsilon\}\to0 for any fixed \epsilon>0.

2. Convergence with Probability 1, is kind of pointwise convergence in real analysis, it is also called almost surely convergence in the measure theory.
X_{n}\stackrel{a.s.}{\to}X: if P\{X_{n}\to X\}=1 or P\{X_{n}\nrightarrow X\}=0.

Convergence in Probability is weaker than Convergence with Probability 1, as stated below:

Theorem: X_{n}\stackrel{a.s.}{\to}X\Rightarrow X_{n}\stackrel{P}{\to}X

Proof: \{X_{n}\to X\}={\omega: for each \epsilon>0 there exists an N(\omega)>0 such that ||X_{k}(\omega)-X(\omega)||\leq\epsilon for all k>N(\omega)}

={\cap_{\epsilon>0}{\omega: there exists an N(\omega)>0 such that ||X_{k}(\omega)-X(\omega)||\leq\epsilon for all k>N(\omega)}}

={\cap_{\epsilon>0}\cup_{n=1}^{\infty}{\omega: ||X_{k}(\omega)-X(\omega)||\leq\epsilon for all k>N(\omega)}}

={\cap_{\epsilon>0}\cup_{n=1}^{\infty}\cap_{k>n}{\omega: ||X_{k}(\omega)-X(\omega)||\leq\epsilon}}

So X_{n}\stackrel{a.s.}{\to}X\Longleftrightarrow P\{X_{n}\to X\}=1

\Longleftrightarrow P\{\cap_{\epsilon>0}\cup_{n=1}^{\infty}\cap_{k>n}\{\omega:||X_{k}(\omega)-X(\omega)||\leq\epsilon\}\}=1

\Longleftrightarrow P\{\cup_{n=1}^{\infty}\cap_{k>n}\{\omega:||X_{k}(\omega)-X(\omega)||\leq\epsilon\}\}=1 for all \epsilon>0

\Longleftrightarrow\lim_{n\to\infty}P\{\cap_{k>n}\{\omega:||X_{k}(\omega)-X(\omega)||\leq\epsilon\}\}=1 for all \epsilon>0

\Longleftrightarrow\lim_{n\to\infty}P\{\cup_{k>n}\{\omega:||X_{k}(\omega)-X(\omega)||>\epsilon\}\}=0 for all \epsilon>0

\Rightarrow\lim_{n\to\infty}P\{||X_{n}-X||>\epsilon\}=0 for all \epsilon>0, or P\{||X_{n}-X||>\epsilon\}\to0 for any fixed \epsilon>0, that is X_{n}\stackrel{P}{\to}X.

if X_{n}\stackrel{a.s.}{\to}X, let the set of unconvergence points as U_{a.s.}=\{\omega:X_{n}(\omega)\nrightarrow X(\omega)\} then P\{U_{a.s.}\}=0.
if X_{n}\stackrel{P}{\to}X, let the set of unconvergence points as U_{P}=\{\omega:X_{n}(\omega)\nrightarrow X(\omega)\} then we donot have P\{U_{P}\}=0.

Convergence in Probability do not require P\{U_{P}\}=0 but release the condition to \lim_{n\to\infty}P\{U_{P}(n)\}=0, where U_{P}(n) is a set which will go smaller as n\to\infty.

From the proof above we can see that U_{P}=\{\cup_{\epsilon}\cap_{n=1}^{\infty}\cup_{k}\{\omega:||X_{k}(\omega)-X(\omega)||>\epsilon\}\} and U_{P}(n)=\{\omega:||X_{n}-X||>\epsilon\}, that is to say U_{P} is bigger than U_{a.s.}

A Course in Large Sample Theory(Lecture notes), Xianyi Wu
A Course in Large Sample Theory, Thomas S. Ferguson

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