UMVUE for population median of exponential distribution

Question

Suppose $ X_1 , \cdots , X_n$ is a random sample from $\text{Exp}(\lambda)$ . Then , is it true that the sample median is a uniformly minimum variance unbiased estimator (UMVUE) of the population median ?

Answer 1

Comment: Following @user10354138's suggestion.

Consider random samples of size $n = 5$ from $\mathsf{Exp}(rate=1/10).$ Thus, $E(X_i) = \mu = 1/\lambda = 10$ and $MED(X_i) = \log(2) \mu = 6.9315.$ From R statistical software:

qexp(.5, lam);  10*log(2)
[1] 6.931472
[1] 6.931472

Perhaps see Wikipedia on 'Exponential distribution' if your textbook or notes do not have everything you need to follow theoretical parts of this.

Look at $m = 10^6$ such random samples of size $n = 5$ and see what happens. By the WLLN simulated quantities should be 'reasonably' close to corresponding parameter values. From a simulation with a million samples, one can expect 2 or 3 place accuracy.

set.seed(1234)           # for reproducibility
m = 10^6;  n = 5;  lam = .1
x = rexp(m*n, lam)
MAT = matrix(x, nrow=m)  # m by n matrix each row a sample
a = rowMeans(MAT)        # vector of m mean
h = apply(MAT,1,median)  # vector of m medians

We see that $E(\bar X) = 10$ is closely approximated and that $SD(\bar X) \approx 4.479.$ You can get easily derive the exact value of $Var(\bar x).$

mean(a); sd(a)
[1] 10.00473
[1] 4.479718

We know from above that the population median is $\eta = 6.931472,$ so the the sample median is not unbiased for $\eta.$

mean(h); sd(h)
[1] 7.833585
[1] 4.625242

A better estimate of $\eta$ is $\hat \eta = \log(2)\bar X,$ which is based on the sufficient statistic. The MLE is unbiased and has a smaller SD (hence variance) than does the sample median. How would you prove that it is UMVUE?

est.h = log(2)*a
mean(est.h); sd(est.h)
[1] 6.934753           # aprx mean of MLE = 6.9315
[1] 3.105104           # SD(MLE) < 4.625242
2*sd(est.h)/sqrt(m)
[1] 0.006204349        # aprx marg of sim err for 6.934

Below is a histogram of the simulated distribution of $\hat \eta$ with $E(\hat \eta)$ shown as a vertical dotted line. Using moment generating functions, it is not difficult to find the density function of the exact distribution of $\hat \eta$ (solid curve), which is related to the gamma distribution of $\bar X.$