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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?

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The definition of "best" is ambiguous and researchers in statistics spent a good deal of effort to work out "best" estimators under different criteria.

For example, uniformly minimum variance unbiased estimators (UMVUE) are "best" in the sense of minimum variance among the class of unbiased estimators. Similarly, there are Bayesian estimators that will be biased, but perform better under mean squared error loss.

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.

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