Suppose $X{\sim}N(\mu_1,\sigma_1^2)$ and $Y{\sim}N(\mu_2,\sigma_2^2)$, $x_1,...x_n$ and $y_1,...,y_n$ are i.i.d samples from X and Y, respectively. Consider the estimator $r=\frac{\sum_{i=1}^{n}(x_i-\overline{x})^2}{\sum_{i=1}^{n}(y_i-\overline{y})^2}$. What does r converge to when sample size $n$ is large? Intuitively, this should be $\frac{\sigma_1^2}{\sigma_2^2}$. But what are steps to reach this conclusion and the basis for each step? any help is highly appreciated. In particular, is the following true? On the right hand side, I treat $X_i$ and $Y_i$ as random variables. \begin{equation*} lim_{n\rightarrow{\infty}}\frac{\sum_{i=1}^{n}(x_i-\overline{x})^2}{\sum_{i=1}^{n}(y_i-\overline{y})^2}=\frac{E[\sum_{i=1}^{n}(X_i-\mu_1)^2]}{E[\sum_{i=1}^{n}(Y_i-\mu_2)^2]} \end{equation*}

  • $\begingroup$ Are you assuming X and Y are independent? $\endgroup$ – Adrian Dec 14 '15 at 8:04
  • $\begingroup$ @Adrian Yes, they are. $\endgroup$ – Ruth Dec 14 '15 at 8:15

There is something not quite right about the equation at the end of your question. On the left you have $n\to\infty$, but on the right $n$ still appears as a finite number. Fortunately, your question can be answered in a straightforward way using the Strong Law of Large Numbers.

First, though, you have to understand what you mean by "iid samples". These samples are actually random variables. Formally, we think of iid samples as being multiple independent copies of $X$, which we denote by $X_1,X_2,...$, all with the same distribution. Likewise, we also consider independent identically distributed copies $Y_1,Y_2,...$ of $Y$. Then the claim would be that $r=\frac{\sum_{i=1}^{n}(X_i-\overline{X}_n)^2}{\sum_{i=1}^{n}(Y_i-\overline{Y}_n)^2}\to \frac{\sigma_1^2}{\sigma_2^2}$ almost surely i.e. everywhere except perhaps on a set of probability zero.

To prove this, you apply the Strong Law of Large Numbers to the sequence $X_i^2$, giving $\sum_{i=1}^{n} X_i^2/n\to E(X^2)=\mu_1^2+\sigma_1^2$ almost surely. Next apply the Strong Law to $X_i$, telling you that $\sum_{i=1}^{n} X_i/n\to E(X)=\mu_1$. Therefore $\sum_{i=1}^{n}(X_i-\overline{X}_n)^2/n\to\sigma_1^2$ almost surely. You can apply the same argument to the $Y$ random variables as well. Then divide, to get the conclusion you were looking for.

Note that you don't need independence between $X$ and $Y$ for this conclusion to hold. And you don't need normality, only that the expectation and variance of $X$ and $Y$ are finite.

  • $\begingroup$ Where have you justified interchanging the limit and the ratio, i.e., that the limit of the ratio = the ratio of the limits? $\endgroup$ – Mark L. Stone Dec 16 '15 at 23:54
  • $\begingroup$ @MarkL.Stone This is a well known property of almost sure convergence and it follows from the Continuous mapping theorem - see en.wikipedia.org/wiki/Continuous_mapping_theorem. $\endgroup$ – S. Catterall Dec 17 '15 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.