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I am trying to understand a proof of the following theorem:

$X_n$ is a sequence of random variables.

$X_n \to X$ in probability $\implies$ that each sub-sequence of $X_n$ has a sub-sequence which almost surely converges to $X$.

The proof goes as follows: $X_{n_l}$ is a sub-sequence of $X_n$, thus $\forall j$ $P(|X_{n_l}-X|>\frac{1}{j}) \to 0$ as $l \to \infty$ (because $X_n \to X$ in probability).

Then it follows that $\exists X_{n_{l_j}}$ such that $P(|X_{n_{l_j}}-X|>\frac{1}{j}) \le \frac{1}{j^2}$. I do not really see why such a sub-sub-sequence necessarily must exist. Any hints?

zesy
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1 Answers1

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After sleeping this over I guess I see the answer myself (also thanks to the commentator of the question).

$\forall j$ $\exists l_0(j)$ such that $\forall l \ge l_0(j)$ holds $P(|X_{n_l}-X|>\frac{1}{j}) \le \frac{1}{j^2}$ because $X_{n_l}$ converges in probability.

So to construct $X_{n_{l_j}}$ we firstly find $l_0(1)$ and take $X_{n_{l_0(1)}}$ as the first element of the sub-sub-sequence $X_{n_{l_j}}$.

Then we find $l_0(2)$, if $l_0(2) \le l_0(1)$ We take $X_{n_{l_0(1)+1}}$ as the second element of the sub-sub-sequence (note $P(|X_{n_{l_0(1)+1}}-X|>\frac{1}{2}) \le \frac{1}{2^2}$), otherwise $X_{n_{l_0(2)}}$ is the second element.

We proceed this way and find further elements for $X_{n_{l_j}}$. Apparently this process will never stop because $\forall j$ $\exists l_0(j)$.

Thus what we construct is indeed a sub-sub-sequence satisfying $P(|X_{n_{l_j}}-X|>\frac{1}{j}) \le \frac{1}{j^2}$.

zesy
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