The distribution of $X_i-X_{(1)}$

Question

If $X_1,…,X_n$ be a random sample from $Exp(1)$ and $X_{(1)}$ is the smallest exponential order statistic, how can I get the distribution of $X_i-X_{(1)}$?

Answer 1

Assuming the $X_i$ are i.i.d.,

$$\begin{aligned} P(X_i-X_{(1)} \leq t) &= P(X_i-t\leq X_{(1)})\\ &= P\left(\bigcap_k X_i-t\leq X_k\right)\\ &= E\left(1_{X_i-t\leq X_1} \ldots 1_{X_i-t\leq X_n}\right)\\ &= \int 1_{x_i-t\leq x_1} \ldots 1_{x_i-t\leq x_n} e^{-x_1}1_{x_1\geq 0}\ldots e^{-x_n}1_{x_n\geq 0} dx_1\ldots dx_n \\ &= \int 1_{-t\leq 0} \left(\int_{\max(0,x_i-t)}^\infty e^{-x_1}dx_1 \right)^{n-1} e^{-x_i} dx_i\\ &= 1_{t\geq 0}\left[\int_0^t e^{-x_i}dx_i + \int_t^\infty e^{-(x_i-t)(n-1) }e^{-x_i}dx_i\right]\\ &= 1_{t\geq 0}\left[1-e^{-t}\left(1-\frac 1n \right)\right] \end{aligned} $$

This yields the cdf of $X_i-X_{(1)}$ (which is independent of $i$ as expected). The distribution is neither discrete nor continuous.