A few questions about the statistics of a lognormal distribution?

Question

If Y = ln(X) and Y is normally distributed the distribution of X is Lognormal.

How do you account for observations of X that are 0 since there is no such thing as ln(0)?

Also, once the variance and mean of Y are computed, do standard z-table based confidence intervals work?

Answer 1

Yes, if Y=ln(x) and Y is normally distributed, then X is log-normally distributed. An assumption of using this transformation is that X cannot be 0;  as you say, it can't be computed.  Are these missing datapoints, or does your distribution of X include meaningful 0s?

No, once the variance and mean of Y are computed, the z-table is not the appropriate table for calculating confidence intervals.  A z-table is used when you know the population parameters (mean and standard deviation).  Since you are calculating the standard deviation of the sample, and using that as an estimate of the population parameter, you need to use a student's t distribution.

Good luck!

Answer 2

Are you trying to force a log transform?  Remember that the actual definition for a log normal random variable is: "If X is a random variable whose logarithm is normally distributed (that is, log X ~ N(μ, σ²)), then X has a lognormal distribution.  

Because you have a data set X, with values of zero, it is likely that X does not follow the lognormal.

I wouldn't remap the data points, the remapping you have done will likely skew the data, how would you see the difference between a true "1" and a mapped "0"?

the expectation of Y = logX is E[Y] = exp(μ+σ²/2) and the variance is: Var[Y] = exp(2 * (μ + σ²) - exp(2μ + σ²)

the distribution function of X is:

f(x | μ, σ²) = (2π)^(-1/2) * 1/x * exp( (log(x) - μ)² / (2σ²))  

for 0 < x < ∞ ; - ∞ < μ < ∞ ; and σ > 0

as you can see for X to have the log normal the X must be strickly positive, so in the case you have noted it is not possible for X to follow the lognormal.