7
$\begingroup$

I am trying to understand an answer to this question (specifically Siméon's answer) Convergence in probability of the product of two random variables

Im struggling with the statement "$X_{n}$ tends to X in probability if and only if , every subsequence of $X_{n}$ has a sub sub sequence that tends to $X$ a.s"

I felt like the above statement deserved a question in itself.

If this is true I can see how the proof follows.

$\endgroup$
1
  • $\begingroup$ Any thoughts on the problem? $\endgroup$
    – saz
    Apr 11, 2018 at 18:33

1 Answer 1

19
$\begingroup$

I'll list off three main results that are useful to prove this result. If you've seen these before then the result you which to prove is pretty simple. However if you haven't, I've included proofs of the results.

Claim 1: "$X_n$ tends to $X$ in probability if every subsequence of $X_n$ has a sub-subsequence that tends to $X$ in probability"

To prove this suppose not, then there exists $\epsilon > 0$ such that $\Pr(\lvert X_n - X \rvert \geq \epsilon) \not \to 0$. Therefore there exists $\delta > 0$ such that for all $N$ there exists $n \geq N$ such that $\Pr(\lvert X_n - X \rvert \geq \epsilon) \geq \delta$. This allows us to construct a subsequence $X_{n_k}$ such that $\Pr(\lvert X_{n_k} - X \rvert \geq \epsilon) \geq \delta$ for all $k$. But then no possible sub-subsequence of $X_{n_k}$ can tend to $X$ is probability. Contradiction.

Claim 2: "If $X_n$ tends to $X$ in probability then there is a subsequence of $X_n$ which tends to $X$ almost surely"

By convergence in probability it is possible to pick a subsequence $X_{n_k}$ such that for all $k$ $$ \Pr\left(\lvert X_{n_k} - X \rvert \geq \frac{1}{k}\right) \leq \frac{1}{2^k}$$ In particular $\sum_{k = 1}^{\infty} \Pr( \lvert X_{n_k} - X \rvert \geq \frac{1}{k}) \leq 1 < \infty$ so by the first Borel-Cantelli lemma we have that $$ \Pr\left( \lvert X_{n_k} - X\rvert \geq \frac{1}{k}\ \text{for infinitely many $k$} \right) = 0 \implies \Pr \left( \lvert X_{n_k} - X\rvert < \frac{1}{k}\ \text{eventually} \right) = 1$$ Therefore almost surely there exists $K$ such that for all $k \geq K$ we have that $\lvert X_{n_k} - X \rvert < \frac{1}{k}$. This implies $X = \lim_{k \to \infty} X_{n_k}$ on this event. Therefore $X_{n_k}$ tends to $X$ almost surely.

Claim 3: "If $X_n$ tends to $X$ almost surely then $X_n$ tends to $X$ in probability"

Fix $\epsilon > 0$ and consider $\Pr(\lvert X_n - X \rvert \leq \epsilon)$. Then we use the result that for any sequence of events $(A_n)$ we have $$ \liminf \Pr(A_n) \geq \Pr(\liminf A_n)$$ (this is a version of Fatou's lemma). Applying this gives \begin{align*} \liminf \Pr(\lvert X_n - X \rvert \leq \epsilon) &\geq \Pr(\lvert X_n - X \rvert \leq \epsilon\ \text{eventually}) \\ &\geq \Pr\left(\lim_{n \to \infty} X_n = X\right) = 1 \end{align*} Therefore in fact $\lim \Pr(\lvert X_n - X \rvert \leq \epsilon) = 1$. So $\lim \Pr(\lvert X_n - X \rvert > \epsilon) = 0$ as required to convergence in probability.

Final Result: "$X_n$ tends to $X$ in probability if and only if every subsequence of $X_n$ has a sub-subsequence that tends to $X$ in almost surely"

For the forwards direction if $X_n$ tends to $X$ in probability then every subsequence of $X_n$ still tends to $X$ in probability so by Claim 2 there is a sub-subsequence which converges to $X_n$ almost surely.

Conversely if every subsequence of $X_n$ has a sub-subsequence converging to $X$ almost surely then said subsubsequence converges to $X$ in probability by Claim 3. So $X_n$ tends to $X$ in probability by Claim 1. This concludes the proof.

$\endgroup$
1
  • $\begingroup$ You have not received the credit you deserve sir. Thanks! $\endgroup$
    – user1trill
    Aug 28, 2019 at 1:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .