# Divergence in probability implies almost sure divergence

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?

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 $$1$$.
• 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. – stat Aug 15 '19 at 10:02
• 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 $\log (n) >\epsilon$. – Kavi Rama Murthy Aug 15 '19 at 10:06
• But does it at least follow that $log(n)(X_n+Y_n) \to \infty$ in probability? – stat Aug 15 '19 at 10:11