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I have the following issue: I would like to do a power analysis (find the right sample size) for a ratio metric ($Z = \frac{X}{Y}$). The in-house statistical software I inherited uses a delta correction or delta method for that and before blindly using that I would rather like to understand what this method does. I understand it is somehow related to the variable $Z$ not being normally distributed and/or $X$ and $Y$ being correlated.

More concretely I would like to understand what happens if I don't use the delta correction, in what way the sample size estimation may be off if I don't use it and in how far it compensates for the (presumed) violation of the prerequisites. Also, I would be interested in whether there is something I can test the distribution against which would allow me to omit the delta correction.

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    $\begingroup$ That's going to depend on the ranges of $X$ and $Y$. For example, if $X,Y \in [100,150]$, you may not have much of a problem, whereas if $Y \in [1,100]$ you probably will. $\endgroup$
    – jbowman
    Commented Jan 5, 2023 at 18:08

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The $\delta$-method is an elementary statistical result for calculating the asymptotic standard error of a smooth function of random variables. A ratio such as $X/Y$ is smooth provided 0 is not in the sample space of $Y$ ($Y$ must be strictly positive or strictly negative). The $\delta$-method has nothing to do with the actual normality of the variables since it is an asymptotic result. If $\bar{X}$ is asymptotically normal and $\bar{Y}$ is asymptotically normal (with a very low probability of achieving 0) then their ratio $\overline{X/Y}$ is asymptotically normal with a variance given by the quadratic form defined by the covariance matrix of $X,Y$ and the Jacobian.

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  • $\begingroup$ +1 Should the answer maybe say "approximate asymptotic standard error", since typically only the first order term, and rarely the second order term are actually calculated out of an infinite series? $\endgroup$
    – Alexis
    Commented Jan 5, 2023 at 18:34
  • $\begingroup$ @Alexis The asymptotic standard error should be correct. The later order terms shouldn't matter because the estimator converges to its expectation by the SLLN. In finite samples, "approximation" should be used very loosely, because the rate of convergence can be abysmal. $\endgroup$
    – AdamO
    Commented Jan 5, 2023 at 18:41
  • $\begingroup$ TY! So the higher-order terms in the $\delta$-method do what? Accelerate the rate of asymptotic convergence? Incorporating them often produces results which are different—even if only by a small amount—from the first order term only, so what does that difference (i.e. 1st order only, or 1st order plus higher order terms) imply? $\endgroup$
    – Alexis
    Commented Jan 5, 2023 at 18:45
  • $\begingroup$ @Alexis you seem to have some intuition about the purpose of higher order terms, since you said yourself "the second order term is rarely calculated out of an infinite series". Incorporating higher order terms in the form of tensors will improve the local approximation of the standard error. $\endgroup$
    – AdamO
    Commented Jan 5, 2023 at 18:59
  • $\begingroup$ Yes. Sorry... taking advantage of you to learn stuffs. :D $\endgroup$
    – Alexis
    Commented Jan 5, 2023 at 19:04

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