Subtracting the minimum from Independent Normal Distributions

Question

Say we pick $n$ independent identically distributee variables $x_i \sim N(0,\sigma^2)$.

Say we generated a new series of random variable $X_i = x_i-\min_jx_j$

What is the probability distribution of the $X_i$ NOT equal to zero (i.e. the $X_i$ corresponding to $x_i\neq\min_jx_j$)?

Trying to figure this out has been breaking my brain: the $x_i$ are independent, except we know that one of them is the minimum, so I'm trying to wrap around whether independence is maintained, or how to express the dependence. My gut tells me that the resulting $X_i$ should be gamma distributed, but that's only because all the results are necessarily nonnegative. Just a regular difference of normal variables doesn't cut it: the resulting distribution is normal as well, whereas the $X_i \geq 0$.

Answer 1

Let $\phi$ ($\Phi$) denote the $N(0,\,1)$ PDF (CDF). Without loss of generaliy the minimum is $x_1$, and when $x_1=m$ the $X_i$ with $i>1$are IIDs with conditional CDF $\frac{\Phi\left(\frac{X-m}{\sigma}\right)-\Phi\left(\frac{-m}{\sigma}\right)}{1-\Phi\left(\frac{-m}{\sigma}\right)}$ on $X\ge0$. So the CDF of a nonzero $X_i$ is$$\int_{\Bbb R}f(m)\frac{\Phi\left(\frac{X-m}{\sigma}\right)-\Phi\left(\frac{-m}{\sigma}\right)}{1-\Phi\left(\frac{-m}{\sigma}\right)}dm,$$where $f$ is the PDF of $\min_jx_j$. But this minimum has CDF $1-\Phi^n\left(\frac{-m}{\sigma}\right)$ and PDF $\frac{n}{\sigma}\Phi^{n-1}\left(\frac{-m}{\sigma}\right)\phi\left(\frac{-m}{\sigma}\right)$, so a nonzero $X_i$ has CDF$$\int_{\Bbb R}\frac{n}{\sigma}\Phi^{n-1}\left(\frac{-m}{\sigma}\right)\phi\left(\frac{-m}{\sigma}\right)\frac{\Phi\left(\frac{X-m}{\sigma}\right)-\Phi\left(\frac{-m}{\sigma}\right)}{1-\Phi\left(\frac{-m}{\sigma}\right)}dm.$$