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In exercise 2.4.1. of Durret's Probability book, we're looking to construct a sequence of r.v. $X_n\in \{ 0,1 \}$, $X_n\rightarrow 0$ in prob., $N(n)\rightarrow \infty$ a.s., and $X_{N(n)}\rightarrow 1$ a.s.

Here's picture of the solution suggested by Durret in his solution manual.

enter image description here

However, I disagree with it. Since $X_k \rightarrow 0$ in prob. is not entirely correct. Here $k=2^n+m$. $P(|X_k-0|<\epsilon)= 1- 1/2^n $ So if I increase $m$, instead of $n$ we don't have that probability going to 1, but constant. I can always find a sufficiently big $m$ for $n=1$ where the size of the probability is 1/2.

Am I correct?

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For $2^{n}\leq k <2^{n+1}$ we have $P\{X_k >\epsilon \} =\frac 1 {2^{n}}<\frac 2 k \to 0$ as $k \to \infty$.

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  • $\begingroup$ Thanks for your answer. My mistake. ;) $\endgroup$
    – An old man in the sea.
    Aug 8, 2018 at 9:40
  • $\begingroup$ I've just noticed something in the solution. Do we really have that $N(n)\rightarrow \infty$ a.s.? Because $P(\{\omega :N(n)\rightarrow \infty \}=P(\{0\}))$ $\endgroup$
    – An old man in the sea.
    Aug 8, 2018 at 9:56
  • $\begingroup$ $N_n \geq 2^{n}$ because $m \geq 0$. $\endgroup$
    – Kavi Rama Murthy
    Aug 8, 2018 at 9:59

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