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jmiller
Level I

SD of Fitted Lognormal Distribution vs SD of Fitted Normal Distribution of Log-Transformed Data

I'd like to know why the fitted lognormal distribution for a given data set provides a different standard deviation than the fitted normal distribution of the log-transformed data set. For example, suppose I have 10 datapoints: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. When I fit the lognormal distribution, I get a standard deviation of 0.6954075. When I log transform the 10 datapoints and fit a normal distribution, I get a standard deviation of 0.7330239. See snippet below. Thank you! 

 

jmiller_0-1673035658583.png

JMP Version: 16.2 

1 ACCEPTED SOLUTION

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SamGardner
Level VII

Re: SD of Fitted Lognormal Distribution vs SD of Fitted Normal Distribution of Log-Transformed Data

In v16 the documentation did not completely describe the estimation methods for the two distributions.  You can read the updated description in the v17 documentation here: https://www.jmp.com/support/help/en/17.1/#page/jmp/fit-distributions.shtml.  

 

The difference is because for the Normal Distribution fit, the unbiased estimate of the standard deviation is calculated, which divides the corrected sum of squares by (n-1), while the lognormal fit uses the maximum likelihood estimate, which divides the corrected sums of squares of the log transformed data by (n).  You can see your values have a ratio of (0.6954075)/(9.7330239)=  sqrt((n-1)/n) = sqrt(9/10).  

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3 REPLIES 3
SamGardner
Level VII

Re: SD of Fitted Lognormal Distribution vs SD of Fitted Normal Distribution of Log-Transformed Data

In v16 the documentation did not completely describe the estimation methods for the two distributions.  You can read the updated description in the v17 documentation here: https://www.jmp.com/support/help/en/17.1/#page/jmp/fit-distributions.shtml.  

 

The difference is because for the Normal Distribution fit, the unbiased estimate of the standard deviation is calculated, which divides the corrected sum of squares by (n-1), while the lognormal fit uses the maximum likelihood estimate, which divides the corrected sums of squares of the log transformed data by (n).  You can see your values have a ratio of (0.6954075)/(9.7330239)=  sqrt((n-1)/n) = sqrt(9/10).  

jmiller
Level I

Re: SD of Fitted Lognormal Distribution vs SD of Fitted Normal Distribution of Log-Transformed Data

Thank you for the quick response! Minor correction that 0.6954075 / 0.7330239 = SQRT(9/10). 

SamGardner
Level VII

Re: SD of Fitted Lognormal Distribution vs SD of Fitted Normal Distribution of Log-Transformed Data

Thanks for catching my mistake!  I corrected it in my response.  

 

This sort of question is also easy to get answered by contacting our technical support team. See www.jmp.com/support.