Let $X_1,X_2, \dots$ be iid RVs with mean $\alpha$ and variance $\sigma^2$ , and let $Y_1,Y_2, \dots$ be iid RVs with mean $\beta(\neq 0)$ and variance $\tau^2$. Find the limiting distribution of $$Z_n=\frac{\sqrt{n}( \bar X_n - \alpha)}{\bar Y_n}$$ ,where $\bar X=\frac{1}{n}\sum_{i=1}^{n} X_i$ and $\bar Y=\frac{1}{n}\sum_{i=1}^{n} Y_i$

Trial: By Lindeberg_Levy CLT , $$ \bar X_n \text{~} N(\alpha,\frac{\sigma^2}{n})$$ and $$ \bar Y_n \text{~} N(\beta,\frac{\tau^2}{n})$$ I know Slutsky's theorem but unable to use. Please help.

From Max's hint I think my answer will be :

Since $ \bar X_n \text{~} N(\alpha,\frac{\sigma^2}{n})$ we have $ \sqrt{n}(\bar X_n -\alpha)\text{~} N(0,\sigma^2)$

and here $ \bar Y_n \text{~} N(\beta,\frac{\tau^2}{n})$ where $\beta \neq 0$

So now by using Slutsky's theorem we have

$ Z_n =\frac{\sqrt{n}( \bar X_n - \alpha)}{\bar Y_n}\text{~} N(\frac{0}{\beta},\frac{\sigma^2}{\tau^2/n})=N(0,\frac{n\sigma^2}{\tau^2}) $

Here my variance comes out $\frac{n\sigma^2}{\tau^2}$.

Please tell me where I make mistake? Thank you in advance.

  • 3
    $\begingroup$ This is just a shot in the dark so I'm sure someone else will give a better answer. I believe $\sqrt{n}\left(\bar{X}_n-\alpha\right)$ is normally distributed with zero mean and variance $\sigma^2$. As $n\rightarrow\infty$, $\bar{Y}_n$ goes to the constant $\beta$. I would expect that in the limit, $Z_n$ is normally distributed with zero mean and variance $\left(\frac{\sigma}{\beta}\right)^2$. $\endgroup$ Commented Jan 22, 2013 at 18:07
  • 1
    $\begingroup$ @Max: If that is a shot in the dark, it's a pretty good one... :) $\endgroup$
    – cardinal
    Commented Jan 22, 2013 at 20:44
  • $\begingroup$ Is a better title available for this question? $Z_n$ could refer to almost everything; a description of the ratio we are finding the limit of might be helpful. $\endgroup$
    – Silverfish
    Commented Nov 24, 2014 at 21:33
  • $\begingroup$ @Silverfish Changed the title to suit the question in the post itslef. $\endgroup$ Commented Nov 25, 2014 at 3:47

1 Answer 1


Since user Max never turned his comment into an answer, to lay this one to officially rest:

By Lindeberg-Levy CLT indeed

$$\sqrt{n}( \bar X_n - \alpha) \xrightarrow{d} N(0,\sigma^2)$$

By the Law of Large Numbers

$$\bar Y_n \xrightarrow{p} \beta$$

Then we can apply Slutsky's theorem

$$Z_n=\frac{\sqrt{n}( \bar X_n - \alpha)}{\bar Y_n} \xrightarrow{d}\frac 1{\beta}N(0,\sigma^2) = N(0, \sigma^2/{\beta^2})$$


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