Does $X_{n}=o_{p}\left(Y_{n}\right)$ imply that $P\left(Y_{n}=0\right)=0$ for all $n$? Suppose that $X_{n}=o_{p}\left(Y_{n}\right)$. Does this imply that $P\left(Y_{n}=0\right)=0$ for all n ? Or only, e.g., that $P(Y_n = 0)\rightarrow 0$?
I guess it is a matter of definition. If $X_{n}=o_{p}\left(Y_{n}\right)$, then that means that $\frac{X_n}{Y_n}$ converges to 0 in probability. But if $Y_n$ can take value 0 with positive probability for some $n$, then I guess $\frac{X_n}{Y_n}$ is not technically a random variable for that $n$ (because it is not a map from $\Omega$ to R but rather to extended R)? Is that true? If that is true, then can we still talk about probability limit of the sequence?
 A: First, $X_n = o_p(Y_n)$, when we are not disallowing $Y_n = 0$, is probably best expressed as stating, for all $\epsilon > 0$ that
$$
P(|X_n| \le \epsilon |Y_n|) \to 1, \qquad (n \to \infty). 
$$
The "almost-sure" version would be $X_n = o(Y_n)$ with probability $1$, which is equivalent to 
$$
P(|X_n| \le \epsilon |Y_n| \text{ eventually}) = 1
$$
for all $\epsilon > 0$. 
With the definitions clear, we now see that neither $P(Y_n = 0) = 0$ nor $P(Y_n = 0) \to 0$ hold. In fact, we can have $Y_n = 0$ almost surely; we just need $X_n = 0$ almost surely as well.
To make things more interesting, let's add the stipulation that $|X_n| > 0$ almost surely. In this case it is possible for $Y_n = 0$ to occur infinitely often when $X_n = o_p(Y_n)$; just take $Y_n = n |X_n|$ with probability $(n-1)/n$ and $Y_n = 0$ otherwise. But we should have $P(Y_n = 0) \to 0$ in this case, because $Y_n = 0$ implies $|X_n| > \epsilon |Y_n|$. 
For the almost-sure version, this does imply $P(Y_n \ne 0 \text{ eventually}) = 1$, because this is a subset of $[|X_n| > 0 \text{ for all $n$, and } |X_n| \le \epsilon Y_n \text{ eventually}]$, which has probability $1$. 
