Solved – Standard deviation of a ratio (percentage change)

mathematical-statisticsstandard deviationstandard error

I have 2 data sets. The first data set, let's call it $X$ has an average value of ($\bar X$) and standard deviation of ($STD_X$), the second set of data also has the average value of ($\bar Y$) and standard deviation of ($STD_Y$). I want to find out the standard error or standard deviation of a percentage change of data set 2 compared to data set 1. So I have $((\bar Y-\bar X)/\bar X)*100$. Now my question is, how do you take into account the standard deviations for this percentage value?

Best Answer

If you don't know the distribution, the usual approach would be via Taylor expansion.

e.g. see here or top of p 6 here

or

http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables

(You have to recognize that the two sample means are themselves random variables to apply it.)

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Edit:

The formula is directly relevant for your case because $Var(100(y-z)/z) = 100^2 Var(\frac{y}{z} -1) = 100^2 Var(y/z)$.

I don't know of a specific book reference off the top of my head, it feels a bit like asking for a reference for how to do long division.

It's an absolutely standard technique for approximating means and variances, based quite directly (and in a fairly obvious way) off Taylor series, which have been around for 300 years now. It's certainly mentioned in books, but I've never learned it from a book, in spite of encountering it many times - it's always 'expand this transformation in a Taylor series' (usually, but not always about the mean) and 'take expectations' or 'take variances' (or whatever, as necessary).

Once you learn how to do Taylor series (standard early-undergrad mathematics) and know a few properties of expectations and variances (standard early mathematical statistics), you're done; it's something undergrad students are given as an exercise.

I'll see if I can dig up a reference; there's sure to be something in a standard old reference like Cox and Hinkley or Kendall and Stuart or Feller or something (none of which I have to hand at the moment).

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