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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?

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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, $g(X_n, Y_n) \xrightarrow[a.s.]{}g(\mu, \mu) = 0$.

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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.

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