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I have a dataset, always considered log-normally distributed, from which I calculate a median value (0.735). In a book, I have seen someone using similar data but calculating related median value by taking the antilog of the mean of log10 of these data. I get 0.69 instead with this method.

What is the point of using this type of antilog method in general?

At the end, I would like to create a cumulative distribution curve based on the median value to calculated the chances (P25, P50 and P75) of having certain values of my dataset.

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  • $\begingroup$ do you have an even number of observations in the data? Any quantile should be invariance under monotonic transformations of the data (e.g. antilogs) but if you have an even number and you need to pick a point between the two middle values the middle number of original values will be different than the middle number of transformed values except maybe if the transformation is affine. $\endgroup$
    – Lucas Roberts
    Commented Apr 2, 2018 at 13:25
  • $\begingroup$ That's a good point, but I don't have an even number here (neither does the book's author). $\endgroup$
    – JrCaspian
    Commented Apr 2, 2018 at 13:45

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I could be wrong (and I welcome the correction if so), but I believe the rationale is driven by an attempt to adhere to some normality distribution assumption. That is, if I have a normal distribution, I know $P_{25} \approx -\frac{2}{3}$. So, if my distribution is not normal to start, and if I can transform it to a reasonably normal distribution, then I can use this approximation.

Of course, as you noted, the final results will not be the same. A quick example would be the average of 1 and 9 is 5, but if you log the values (0, 0.95), average (0.477), and then antilog, you get the average as 3.

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    $\begingroup$ You're correct, but it's a little more detailed than that. For the lognormal, taking logs allows you to use the sample mean as your maximum likelihood estimate of the centrality parameter of the distribution, which corresponds to the median on both the log and, when exponentiated, on the original scale. So there's some MLE / asymptotic optimality principle at work there as well. $\endgroup$
    – jbowman
    Commented Apr 2, 2018 at 14:23

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