Convergence in Probability and Convergent almost surely

probability theoryreal-analysisstochastic-processes

I am new to probability theory. Pardon me if question is too lame.

I came across the following theorem:

Suppose $X_n \xrightarrow P X_\infty$ and $\sum_{n=1}^\infty P( \left| X_n-X_\infty \right| > \epsilon) < \infty ~~~~~~~ \forall \epsilon >0$

Then $X_n \xrightarrow {a.s} X_\infty$

My doubts:

  1. Doesn't $\sum_{n=1}^\infty P( \left| X_n-X_\infty \right| > \epsilon) < \infty$ implies that $ \lim_{n \to \infty} P( \left| X_n-X_\infty \right| > \epsilon)=0$ and hence $X_n \xrightarrow P X_\infty$? I feel that $X_n \xrightarrow P X_\infty$ is redundant given the summation condition. Am I missing out on something? Is it to imply the existence of a limit point, $X_\infty$ ?

  2. The proof given with the theorem, uses Borel-Cantelli's lemma and then states that $ P(|X_n-X_\infty|>\epsilon, i.o.) = 0$, which implies almost sure convergence. I tried to break it down into following smaller steps.

Let $A_n = \{|X_n-X_\infty|>\epsilon\}$ be a sequence of sets.

Using Borel-Cantelli's lemma, $P( \limsup_{n \to \infty} A_n) =0$

$\implies P (\bigcap_{m=1}^ \infty \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}) =0$

Let $B_m = \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}$. This is a decreasing sequence of sets. Using continuity of probability measure,

$ P( \lim_{m \to \infty} B_m) = \lim_{m \to \infty} P(B_m)$

$\implies P( \lim_{m \to \infty} \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}) = \lim_{m \to \infty} P(\bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\})$

$ \implies P( \lim_{n \to \infty} \{|X_n-X_\infty|>\epsilon\}) = \lim_{n \to \infty} P(\{|X_n-X_\infty|>\epsilon\})$

This implies equivalence in convergence in probability and almost sure convergence under the given condition.

I am not sure if all the steps are right especially the equivalence of last two implied statement. Even if they are equivalent I am not very clear about them. It would be very helpful if you verify the proof. Thank you.

Best Answer

Your observation in 1) is correct.

For 2) let $A_k=\{\omega: |X_n(\omega) -X_{\infty}(\omega))| >\frac 1 k\}$. Then $P(A_k)=0$ for each $k$. If $A =\cup_k A_k$ then $P(A)=0$. If $\omega \notin A$ then $\omega \notin A_k$ for any $k$. Hence, for each $k$ we have $|X_n(\omega) -X_{\infty}(\omega))| \leq \frac 1 k$ whenever $n$ is sufficiently large. By deifinition of limit this says that $X_n(\omega) \to X_{\infty}(\omega)$.

The main step in 2) is to vary $\epsilon$ and get one set of probability $0$. So the equivalence you stated in your argument is not correct.

Related Question