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Let $X_n$, $Y_n$ be dependent random variables

with $X_n \to c$ almost surely with $c>0$ and $\log(n) Y_n \to 0$ in probability.

Can I state the following:

$\log(n) (X_n + Y_n) \to \infty$ almost surely, i.e. the term diverges almost surely?

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No. Let $\{Y_n\}$ be independent random variables such that $P(Y_n=-1)=\frac 1 n$ and $P(Y_n=0)=1-\frac 1 n$. Let $X_n=1$ for all $n$. Note that $P(|\log(n)Y_n|>\epsilon)\leq \frac 1 n \to 0$ for any $\epsilon >0$. Since $\sum P(Y_n=-1)=\sum \frac 1 n=\infty$ it follows by Borel Cantelli Lemma that $Y_n=-1$ infinitely often with probability $1$. Hence $log(n)(X_n+Y_n)=0$ infinitely often with probability $1$.

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  • $\begingroup$ does $P(|log(n)Y_n|>\epsilon) = 1/n$ really hold for any $\epsilon>0$ and any n? probably for any $\epsilon$ we can find a suitable $n$? E.g. if $epsilon = 1$ and $n = 2$ it would not hold, if I am right. $\endgroup$ – stat Aug 15 '19 at 10:02
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    $\begingroup$ If $|\log (n)Y_n| >\epsilon$ the $Y_n$ cannot be $0$. Since $Y_n$ takes only the values $0$ and $-1$ it must take the value $-1$ for which the probability is $\frac 1 n$. Equality is not important but equality hold as long as $\log (n) >\epsilon$. $\endgroup$ – Kavi Rama Murthy Aug 15 '19 at 10:06
  • $\begingroup$ But does it at least follow that $log(n)(X_n+Y_n) \to \infty$ in probability? $\endgroup$ – stat Aug 15 '19 at 10:11
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    $\begingroup$ Yes, that is true. $\endgroup$ – Kavi Rama Murthy Aug 15 '19 at 10:12

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