Assume we have a positive sequence of random variables $X_{n}$ such that $$ X_{n} \overset{a.s.}{\to} 0 $$ and also it is known that $$ X_{n} = O_{p}\left(\frac{1}{n^{\delta}}\right), $$ which means that for any $\varepsilon >0$ there exists finite $M$ and $N$ s.t. $P(n^{\delta}X_{n} > M) < \varepsilon$ for all $n > N$.
What is the limit of $n^{\delta}P(X_{n} > 0)$? Does it go to zero or which condition could be for going to zero?
In general no: let $X_n=n^{-\delta}X$, where $\delta>0$ and $X$ is a random variable such that $\mathbb P(X>0)=1$. Then $X_n\to 0$ almost surely and $(n^\delta X_n)$ is tight, since the random variable in this sequence are all equal to $X$. Then $n^\delta \mathbb P(X>0)=n^\delta$ and does not tend to zero.