# Lognormal Standard Error

What is the Standard Error of the Lognormal distribution? I am particularly interested in comparing two probabilities from the distribution. I have the two proportions based on experiments, in general these proportions follow a lognormal fit.

The standard error of differences from a normal distribution is straightforward (I've seen one other version of this which I think can see on wikipedia):

$SE = \sqrt{\frac{p (1-p)}{n} + \frac{q (1-q)}{m}}$

where $p, q$ are probabilities and $n, m$ sample sizes, expected to be from a normally distributed sample.

How would this be extended to the lognormal?

Edit:

The basic idea is to learn if the difference between two probabilities would be significant or not. I don't think I'm expressing my question correctly as I'm not a strict statistician.

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When you mention standard error are you talking about the standard deviation of the sample mean from lognormal data or for the difference in means from two samples from possibly different lognormal distirbutions? What you presented above is the standard error for the difference of two independent estimates of proportion for two BINOMIAL distributions and not the difference of sample means for two NORMAL distributions, – Michael Chernick Sep 18 '12 at 16:34
Standard error of the differences between the two independently sampled probabilities. I've used the SE equation above to represent survey data, which is assumed to be normal, I'm noting that I need a Standard Error calculation for a lognormal. If the standard error for the normal distribution is different than above, feel free to share. This is just what was taught to me as part of survey analysis. – Lillian Milagros Carrasquillo Sep 18 '12 at 16:38
What, precisely, do you mean by "probabilities from the distribution" (in your question) and "sampled probabilities" (in your comment)? When we sample things from a lognormal distribution, we observe (nonnegative) numbers, not probabilities; and when we estimate values, they are usually parameters like the geometric mean and geometric standard deviation (not probabilities). – whuber Sep 18 '12 at 16:50
To go along with what whuber said the formula that you learned is for a proportion and is the standard error for the diffference between two independent binomial proportions not two normal means. However the confusion could be that when this was discussed n and m are large and the normal approximation to the binomial is used. Then the binomial variables which are sums of independent Bernoulli random variables can be taken to be approximately normal and the difference of their sample means will have the expression you gave for the standard error of the mean difference. – Michael Chernick Sep 18 '12 at 17:04

The variance of a lognormal is [exp(σ$^2$)-1] exp(2(σ+μ)) where μ and σ are the mean and standard deviation parameters of the related normal distribution. Now the maximum likelihood estimate of the lognormal variance would be obtained by plugging in the maximum likelihood estimates of μ and σ into the expression above for the variance. Call that estimate V.

The mean of the lognormal is M = exp(μ+σ$^2$/2) and its mle would be obtained by plugging in the mles for μ and σ in its expression. Call that estimate E

The estimate of the standard error for the difference of the means would depend on what estimate you use for the mean. If you use E I think it would be complicated.

However if the ordinary sample means were used to estimate the population means for the lognormal distirbutions then standard error for those means would be the population standard deviation divided by √n where n is the sample size for the lognormal. This could be estimated by √(V/n). So if we compared means for the lognormal by their sample mean difference the standard error of the estimate would be √(V$_1$/n$_1$+V$_2$/n$_2$) where V$_1$ and V$_2$ are the respective mles for the lognormal variances for populations 1 and 2 respectively and n$_1$ and n$_2$ are the corresponding sample sizes taken from populations 1 and 2.

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