How to derive the distribution of mean estimator based on ranked set sampling from normal distribution

17 Views Asked by At

I am studying about ranked set sampling from normal distribution.

From wolfe,2004 the joint pdf of RSS is $$f_{1,2,...,n:n}(x_1,x_2,...,x_n)=\prod_{i=1}^nf_{i:n}(x_i)$$

where $$f_{i:n}=\frac{n!}{(i-1)!(n-1)!}(F(x_i))^{i-1}(1-F(x_i))^{n-1}f(x_i)$$

My goal is to find the distribution of mean based on RSS with sample of size n

I began from n=2, and used the convolution then i got

$$\int_{-\infty}^{\infty} 2\Phi(x)\phi(x) 2\Phi(z-x)\phi(z-x) \mathrm{d}x,$$ where $\Phi(.)$ is the standard normal cumulative distribution function

$\phi(.)$ is the standard normal probability density function

I have no idea how to determine this integration.

I tried to combine the pdf term like $$\frac{1}{2\pi}e^{\frac{-x^2}{2}}e^{\frac{-(z-x)^2}{2}}=\frac{1}{2\pi}e^{\frac{-z^2}{4}}e^{\frac{-(x-z)^2}{2}},$$ and used integration by parts but I still cannot find the answer.

Thank for any advice.