Dear R users, I would like to estimate mu and sigma of a log-normal distribution, where I know that the expected value is 1, as it is a normalized distribution. That means as E(x) = exp (mu + 1/2*sigma^2) = 1 that 2*mu = -sigma^2 . Therefore I only need to fit one parameter either sigma or mu. How could I do this in R? Thank you very much for your help! biophil [[alternative HTML version deleted]]
R. Michael Weylandt
2012-Aug-22 14:50 UTC
[R] log-normal distribution fitting with expected value = 1
Take a look at fitdist in the MASS package. Cheers, Michael On Wed, Aug 22, 2012 at 4:23 AM, Biophil <biophilo at gmail.com> wrote:> Dear R users, > I would like to estimate mu and sigma of a log-normal distribution, where I > know that the expected value is 1, as it is a normalized distribution. That > means as E(x) = exp (mu + 1/2*sigma^2) = 1 that 2*mu = -sigma^2 . Therefore > I only need to fit one parameter either sigma or mu. How could I do this in > R? > Thank you very much for your help! > biophil > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
David Winsemius
2012-Aug-22 17:37 UTC
[R] log-normal distribution fitting with expected value = 1
On Aug 22, 2012, at 2:23 AM, Biophil wrote:> Dear R users, > I would like to estimate mu and sigma of a log-normal distribution, > where I > know that the expected value is 1, as it is a normalized distribution.Are you sure that is what "normalized distribution" actually means in the context you are reading it? I would have assumed that it meant that the integral over its domain equaled 1, which is quite different than its mean equaling 1.> That > means as E(x) = exp (mu + 1/2*sigma^2) = 1 that 2*mu = -sigma^2 . > Therefore > I only need to fit one parameter either sigma or mu. How could I do > this in > R? > Thank you very much for your help! > biophil >-- David Winsemius, MD Alameda, CA, USA
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