lognormal-to-normal-to-lognormal distribution transformation

[lelvira]

Hello,
I have a lot of doubts in a procedure I am trying to create.
The idea is that I have a mu+6sigma limit defined based on a normal distribution.
However, one of the parameters is not normally distributed but lognormal. Then what I try to do is to transform the lognormal distribution into a normal one, take the limit, and do the transformation back. So, I would like to calculate the equivalent to the mu+6sigma limit in a normal distribution but in a lognormal one.
I guess it has to be an equivalent. For example if mu+6sigma is 20 units in the normal distribution, it has to be an equivalent of X units in a lognormal one, hasn't it?

The procedure I am doing is as follow:
1.Lognormal distribution. From such distribution I get the lognormal mu and sigma.

2.Transformation of the lognormal mu and sigma to a normal distribution based on the formulas for the lognormal distribution

mu=log(mulog^2/SQRT(mulog^2+sigmalog^2))
sigma^2=log((sigmalog/mulog)^2+1)

3.Take the mu + 6sigma limit based on the normal distribution defined as:
limit=mu+6sigma

4.Transformation of the limit to the lognormal distribution. How?????
limitlog=exp(limit) I do not think this last step is correct at all.

Suggestions/comments are welcomed.

Thank you very much in advance

[Old Mac User]

>From your description I'm not sure your are doing the first step
correctly. Begin by taking Ln(Data). Hopefully the "logged data" is
approx. normally distributed. Calculate the Avg. and SD of the
"logged data". Calculate Upper Limit = Avg + 6 *SD (remember... Avg
and SD were calculated from "logged data". Calculate anti-log of
Limit. OMU

[lelvira]

If I understood correct what you propose is as follow:
1. Take the log of all the samples and calculate the mu and sigma. Those are normally distributed.
Avg=1/N sum(log(xi))
SD = 1/N sum(log(xi)-mu)^2
2. Calculate the limit Avg+6*SD
3. Take the ani-log:
exp(Avg+6*SD)

I have my doubts about this procedure since in the first step you are doing a sum of log's to take the average ( log(x1)+log(x2)+....+log(xN)). Then in the last step you take the exp of such sum of logs + SD which also includes log's. So it seems to me like this:

exp(log(x1)+log(x2)+....+log(xN) + SD)

Is this really the conversion back step?. I have the impression this last step is not correct but of course I can be wrong.

Thank you.

[Jack Tomsky]

OMU is right. If x is log-normal, then ln(x) is normal. Calculate your limits from the normal and then take the anti-log (i.e., the exponential) to get back to the original units.

Your alternative approach of transforming the mean and variance will yield slightly different results.

Jack