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Recenly I am reading "Mathematical statistics and data analysis" written by Rice myself. At page 207, theorem A said: With simple random sampling, approimxate variance of $R=\frac{\overline{Y}}{\overline{X}}$ is $Var(R)\approx\dfrac{r^2\sigma^2_{\overline{X}}+\sigma^2_{\overline{Y}}-2r\sigma^2_{\overline{XY}}}{\mu^2_{x}}$ where $r=\dfrac{\sum^N_{i=1}{y_i}}{\sum^N_{i=1}{x_i}}$. Would anyone give me some hint about the procedure of the proof? Thanks a lot!

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    $\begingroup$ If you could divine the correct terms to use, a search of this site would give you abundant answers! Try "delta method". The answer will depend on whether $x_i$ are viewed as random or not. $\endgroup$ – whuber Sep 18 '12 at 18:23
  • $\begingroup$ If I had better premise on "delta method" of this problem, it would be solved without posting it on stackexchange... It is appreciated to give some hits on the problem if I struggle on it for few days. $\endgroup$ – Randolph Chou Sep 19 '12 at 18:40
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The delta method is often used to approximate variances of functions of random variables. The variances for the original random variables are known and go into he expression.

Here R is a ratio of U/V where U and V are sample means. It is just gotten by applying the first order Taylor series terms to the function about a central point and taking the square and then the expectation.

So for a differentiable function f(x) we take Var[f(X)] =E[f(X)-f(a)]$^2$ where f(a) is the mean for the random variable f(X).

Expanding the Taylor series about a gives f(X)≈f(a) +f'(a)(X-a) or

f(X)-f(a)≈f'(a)(X-a)

Then E[f(X)-f(a)]$^2$≈ [f'(a)]$^2$ E[X-a]$^2$ This gives the approximation

Var[f(X)]≈f'(a)]$^2$ Var(X).

In your case we can apply the bivariate extension of the delta method as follows:

here.

Note that they give the approximation for your exact example but include a finite population correction for the case where the sample means were gotten from simple random sampling from a finite population. It is similar but not exactly expressed in the same form as your formula from Rice. But they are probably mathematically equivalent (under some added assumptions about the means of X and Y that Rice must give but you have not mentioned). It is clear that the delta method was used.

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  • $\begingroup$ Thank you Michael! I use second order Taylor expression of $g(r,s)=r/s$ and set approached point $(r,s)=(E[R],E[S])$ then get the result. $\endgroup$ – Randolph Chou Sep 18 '12 at 18:49
  • $\begingroup$ Suppose that U and V are normally distributed, as we would expect if they are sample means. Do we need to worry about the ratio of two standard normals not having the first two moments? In other words, do we need to worry about existence? $\endgroup$ – Charlie Sep 18 '12 at 19:38
  • $\begingroup$ See this article, for example: jstatsoft.org/v16/i04/paper $\endgroup$ – Charlie Sep 18 '12 at 19:46
  • $\begingroup$ @Charlie If U and V are standard normal and independent the ratio is Cauchy and its first and second moments would not exist and clearly the delta method could not be applied. The same difficult occurs if V the variable in the denominator has a significant probability mass in a small interval about zero. Having the denominator very small can make the ratio very large and hence not have a second moment and possibly not even a first. $\endgroup$ – Michael R. Chernick Sep 18 '12 at 19:58
  • $\begingroup$ But this will not be a problem if the Xs and Ys must all be nonnegative. Then both U and V will be bounded away from zero. We really only need this to be the case for V. $\endgroup$ – Michael R. Chernick Sep 18 '12 at 19:59

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