Asymptotic Expectation of Ratio of Sample Averages I have two random variables: $X$ and $Y$. I know that:
\begin{equation}
E[X]=E[Y]=\mu>0
\end{equation}
I know that variance of both can be bounded:
\begin{equation}
\operatorname{Var}[X]<k, \quad
\operatorname{Var}[Y]<k
\end{equation}
The variables might be correlated. 
Suppose I create a statistic:
\begin{equation}
Z_n=\frac{ \overline{x}-\overline{y}}{ \overline{y}}=\frac{\frac{1}{n}\sum x_{i}-\frac{1}{n}\sum y_{i}}{\frac{1}{n}\sum y_{i}}.
\end{equation}
Clearly:
\begin{equation}
\overline{x}-\overline{y}\rightarrow0 \quad and \quad \overline{y} \rightarrow\mu \quad as \quad n\rightarrow\infty \quad
\end{equation}
My suspicion is that:
\begin{equation}
Z_{n}\rightarrow0 \quad as \quad n\rightarrow\infty \quad?
\end{equation}
(in some statistical sense - in probability, almost surely, etc.)
I am an economist and reasonably decent at statistics, but this is stretching my abilities. It seems obvious in some sense, but I need to provide a proof for paper. My intuition is that E[x-y] is going to 0 and E[y] is always bounded from 0, and it seems as if the averages are getting tighter and tighter so that correlations between X and Y can't matter much as n rises. But, I worry that there are weird cases I need to exclude for X and Y (which are pretty simple variables in actuality). 
I was thinking that one way forward is to do a Taylor Expansion (http://www.stat.cmu.edu/~hseltman/files/ratio.pdf) and then show all of the extra terms must go to zero. But, it seems like there must be an easier way? Like, maybe this is a very obvious and simple? 
 A: By the Strong Law of Large Numbers, $\bar{X}_n$ and $\bar{Y}_n$ converge almost surely to $\mu$. For $\mu \neq 0$, $g(x,y) = \frac{x - y}{y}$ is continuous at $(\mu, \mu)$ hence by the Continuous Mapping Theorem, $g(X_n, Y_n) \xrightarrow[a.s.]{}g(\mu, \mu) = 0$.
A: Slutsky's theorem will do the trick (expanding on comment by hejseb).
Let 
$$Z_n = \frac{\bar{x}_n - \bar{y}_n}{\bar{y}_n}$$
By the Weak Law of Large Numbers, we have
\begin{align*}
\bar x_n &\stackrel{p}{\rightarrow} \mu \\
\bar y_n &\stackrel{p}{\rightarrow} \mu 
\end{align*}
Since convergence in probability is stronger than in distribution, we trivially have $\bar x_n \stackrel{d}{\rightarrow} \mu$. Also define $c_n = \bar x_n - \bar y_n$. You state that "clearly"
$$c_n \stackrel{d}{\rightarrow} 0$$
This is true, but it is actually an application of Slutsky's theorem. A second application of Slutsky's theorem gives you 
$$Z_n = \frac{c_n}{\bar y_n} \stackrel{d}{\rightarrow} \frac{0}{\mu} = 0$$
Convergence in probability implies convergence and distribution. Although the converse is not true in general, it is true when we have convergence to a constant. Thus $Z_n \stackrel{p}{\rightarrow} 0$ as desired. 
