Suppose, $X_1, X_2, \ldots X_n \sim \text{iid Exponential}(\theta)$. The median is given by $\log(2) \theta$. The MLE of the median is given by:

$$ \hat{M} = \log(2) \sum_{i=1}^n X_i / n $$

And the variance of the sample median is (according to Brookmeyer, Crowley 1982):

$$ \text{var}(\hat{M}) = \hat{M} / \sum_{i=1}^n \left(1-\exp(-T_i \log (2) / \hat{M}) \right) $$

However I can't seem to confirm these results with simulation. For 1000 simulated datasets, I get inconsistent values for the variance of the sample median versus the average of the median variance estimates.

l <- replicate(1000, {
x <- rexp(100, 10)
mx <- log(2) * length(x) / sum(x)
vx <- mx / sum(1- exp(- x * log(2) / mx))
c(mx, vx)

mean(l[2, ])


> var(l[1,])
[1] 0.4873672
> mean(l[2, ])
[1] 7.476596

Is the Bartholomew expression in fact correct?


Brookmeyer and Crowley seem to have a misprint in their version of Bartholomew's expression. It should read (translating Equation 5 of Bartholomew into the notation of this question):

$$ \text{var}(\hat{M}) = \hat{M}^2 / \sum_{i=1}^n \left(1-\exp(-T_i \log (2) / \hat{M}) \right) $$

In retrospect it's obvious: the scale of the variance should be the square of the scale of the median. See if that fixes the problem.

Also, Bartholemew and Crowley make the following point, which I haven't yet thought through (page 33):

In order to calculate the Bartholomew confidence interval ... the censoring times {$T_i$} of all patients must be known. This is often not the case; in particular, the censoring times are usually not known for those patients who actually died. (Emphasis added.)

I'm not quite sure how to interpret that.

  • $\begingroup$ +1, excellent find! I am especially perplexed by the issue of non-observed censoring times. In my example, I have simulated everyone as an event. To that end, if I put as their censoring times $\infty$, then the $\sum_{i=1}^n (1-\exp(-\infty )) = n$ actually gives $\hat{M}^2/n$ which is the variance of the MLE!!! $\endgroup$
    – AdamO
    Jun 9 '21 at 4:42

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