# Asymptotic Expectation of Ratio of Sample Averages

I have two random variables: $$X$$ and $$Y$$. I know that: $$$$E[X]=E[Y]=\mu>0$$$$ I know that variance of both can be bounded: $$$$\operatorname{Var}[X] The variables might be correlated.

Suppose I create a statistic:

$$$$Z_n=\frac{ \overline{x}-\overline{y}}{ \overline{y}}=\frac{\frac{1}{n}\sum x_{i}-\frac{1}{n}\sum y_{i}}{\frac{1}{n}\sum y_{i}}.$$$$ Clearly: $$$$\overline{x}-\overline{y}\rightarrow0 \quad and \quad \overline{y} \rightarrow\mu \quad as \quad n\rightarrow\infty \quad$$$$ My suspicion is that: $$$$Z_{n}\rightarrow0 \quad as \quad n\rightarrow\infty \quad?$$$$ (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?

• Commented Feb 21, 2019 at 20:09

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, $$g(X_n, Y_n) \xrightarrow[a.s.]{}g(\mu, \mu) = 0$$.
$$Z_n = \frac{\bar{x}_n - \bar{y}_n}{\bar{y}_n}$$
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.