# When to use sample median as an estimator for the median of a lognormal distribution?

I myself would always use geometric mean to estimate a lognormal median. However, in the industry world, sometimes using the sample median gives better results. The question thus is, is there a cutoff range/point starting from which the sample median can be used reliably as an estimator for the population median?

Also, the sample geometric mean is MLE for median, but not unbiased. An unbiased estimator would be $\hat{\beta}_{\mbox{CGM0}}=\exp(\hat{\mu}-\sigma^2/2N)$ if $\sigma$ is known. In practice, a biased corrected estimator $\hat{\beta}_{\mbox{CGM}}$ (see below) is used since $\sigma$ is always unknown. There are papers saying that this bias-corrected geomean estimator is better because of smaller MSE and unbiasedness. However, in reality, when we only have a sample size of 4 to 6, can I argue that the bias correction makes no sense since

1. Unbiasedness means the estimator is centered around the true population parameter, neither under nor over-estimate the parameter. For positively skewed distribution, the center is the median not the mean.
2. Invariant to transformation is important property in my current area(transformation between DT50 and degradation rate k, k=log(2)/DT50). You will get different results based on original data and on the transformed data.
3. For limited sample size, mean unbiasedness is potentially misleading. Bias is not error, an unbiased estimator can give bigger error. From a Bayesian point of view, the data is known and fixed, the MLE maximizes the probability of observing the data, while the bias correction is based on fixed parameters.

The sample geometric mean estimator is MLE, median-unbiased, invariant to transformations. I think it should be preferred to the bias-corrected geomean estimator. Am I right?

Assumming $X_1,X_2,...,X_N \sim \mbox{LN}(\mu,\sigma^2)$

$\beta = \exp(\mu)$

$\hat{\beta}_{\mbox{GM}}= \exp(\hat{\mu})= \exp{(\sum\frac{\log(X_i)}{N})} \sim \mbox{LN}(\mu,\sigma^2/N)$

$\hat{\beta}_{\mbox{SM}}= \mbox{median}(X_1,X_2,...,X_N)$

$\hat{\beta}_{\mbox{CGM}}= \exp(\hat{\mu}-\hat\sigma^2/2N)$

where, $\mu$ and $\sigma$ are the log-mean and log-sd, $\hat\mu$ and $\hat\sigma$ are the MLEs for $\mu$ and $\sigma$.

A related question: for the variance of the sample median, there is an approximate formula $\frac{1}{4Nf(m)^2}$; what is a big enough sample size to use this formula?

• Your expression for $\hat{\beta}_{\mbox{CGM}}$ doesn't have a hat on the $\sigma^2$. Does that mean it assumes $\sigma^2$ is known? That would seem to make it not very useful. Jul 17, 2013 at 13:29
• sorry, it should be $\hat\sigma^2$ Jul 17, 2013 at 13:31
• It is unclear what your estimators are because you have not defined $\hat{\mu}$ or $\hat{\sigma}$. The main concern about lognormal models and small samples is that the lognormal-based estimators are sensitive to the lognormal assumption, so unless you have good evidence that this assumption is correct, it is usually better to use alternative robust estimators.
– whuber
Jul 17, 2013 at 14:07
• @whuber, $\hat\mu$ and $\hat\sigma$ are the MLEs. I agree with the concern of the lognormal assumption. In my current working area, lognormal assumption is standard practice and is accepted by the authorities. So all my questions are based on the lognormal assumption being correct. Jul 17, 2013 at 14:30
• no, the $\mu$ and $\sigma$ are the log-mean and log-sd, not the mean and sd for the lognormal. I will edit the question to make it clear. Jul 17, 2013 at 14:34