Yes, I'm sure. It's about normalized citation scores, where the individual
raw citation score is divided by the average of the whole set of
publications.

On Aug 22, 2012 7:37 PM, "David Winsemius" <dwinsemius@comcast.net> wrote:

>
> 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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