# Standard error on median for exponential distribution

I am trying to find the standard error on the median, $\sigma_\tilde{x}$, for a sample, $X_i$, of a population whose pdf could be modelled as $\lambda e^{-\lambda x}$ if normalized.

To make sure we are all on the same page. $\mu = 1/ \lambda$, $\sigma = 1 / \lambda$ and $\tilde{x} = ln(2) / \lambda$ for the pdf. Clearly I can calculate $\mu$, $\sigma$, $\sigma_\mu = \sigma / \sqrt{N}$ and $\tilde{x}$ from my sample but the $\sigma_\tilde{x}$ is more complicated.

As explained in Central limit theorem for sample medians, $\sigma_\tilde{x} = \frac{1}{2 \sqrt{N} f(\tilde{x})}$, where $f(\tilde{x})$ is the value of the pdf at the median. Substituting $\tilde{x} = ln(2) / \lambda$ into the pdf $f(\tilde{x}) = \lambda e^{- \lambda \tilde{x} } = \lambda / 2$ so the formula above gives $\sigma_\tilde{x} = \frac{1}{\lambda\sqrt{N}}$

The problem at hand is that I do not know how to calculate $\lambda$. $\lambda=1/\mu$ and $\lambda=1/ \sigma$ so I could just take the calculation of $\mu$ or $\sigma$ from my sample. Would they give the same exact value on the same set? Furthermore, if $\lambda=1/ \sigma$ is taken then $\sigma_\tilde{x} = \sigma_\mu$ which I have a hard time believing is a coincidence.

Alternatively I could fit the sample to the assumed form of the pdf to get $\lambda$. I would think that an appropriate choice of fitting algorithm would return the same value.

So what is the "best" way to obtain $\lambda$ from $X_i$ and why?

However, to answer the OP's question, you can estimate $\hat{\lambda} = 1/\bar{x}$, where $\bar{x}$ denotes the sample mean, as a reasonable estimate for a first pass. Generally speaking your estimates will not differ by much with a reasonably large sized set of data.