If $X_i$ follows $ U( \theta, \theta+1)$ and n is even. How do I find the probability distribution of median?

Question

I have used the following transformation to find the joint pdf: $u= X_\frac{n}{2}$ and $ v = \frac{X_\frac{n}{2}+X_{\frac{n}{2}+1}}{2}$

The joint pdf I have found is like below: $f_{(U,V)}(u,v)= \frac{2n!}{(\frac{n}{2}-1)!*(\frac{n}{2}-1)!}(u-\theta)^{\frac{n}{2}-1}(\theta+1-2v+u)^{\frac{n}{2}-1}$ and the support for u , v is $\theta<u<\theta+1$ and $u<v<\frac{u+\theta+1}{2}$

I am trying to integrate the joint pdf with respect to u to obtain the marginal pdf of v=Median. But things are getting complicated. How should I approach this problem. Please help me!

Answer 1

Comment: A simulation illustrates the approximate shape of the density function of the median--in the special case $n = 10, \theta = 5.$

set.seed(303)
h = replicate(10^5, median(runif(10, 5,6)))
hist(h, prob=T, ylim=c(0,3), col="skyblue2", 
     main="n=10: Dist'n of Medians of Uniform Data")
 curve(dnorm(x, mean(h), sd(h)), add=T, lwd=2, col="orange")

There is a 'Central Limit Theorem for Medians'. It requires that the population density be positive at the population median ($f(5.5)>0$ in my example).

In my example, $n = 10$ is yields sample medians that are 'nearly' normal, but the fit it not perfect. If you run the simulation with $n = 100,$ you'll get a better fit to normal.

Ref: The Irwin-Hall distribution may be relevant.