Let $X_i$ be a random sample from a standard exponential distribution.
The goal is to find $p$ such that
$$n^p X_{(2)} \to c >0$$
in probability, where $X_{(2)}$ is the second order statistic (ie. second lowest value) of the sample.
I'm stumped. I've tried two different ways of attack:
Use the fact that $F(X_{(2)})=_d U_{(2)}$ where $U_{(2)}$ is the order statistic of a uniform random sample. Once I find the distribution of $U_{(2)}$ I can then transform back to $X_{(2)}$ since it's CDF is invertible.
Use Renyi representation to write $X_{(2)} =_d \frac{Z_1}{n} + \frac{Z_2}{n-1}$ where $Z_i$ are independent $\exp (1)$ random variables.
When I try either method, I just get lost in algebra with no clear way of linking it back to $n^p$.
For $t > 0$, $\mathbb P(X_{(2)} > t)$ is the probability that there is at most one member of the sample $\le t$, which is $$e^{-tn} + n (1-e^{-t})e^{-(n-1)t} $$ That is, the CDF for $X_{(2)}$ is $$F_{(2)}(t) = 1 - e^{-tn} - n (1-e^{-t})e^{-(n-1)t} $$ In particular, for each positive constant $c$, $$\eqalign{F_{(2)}(c/n) &= 1 - e^{-c} - n (1 - e^{-c/n}) e^{c(1-1/n)}\cr &= 1 - (1+c) e^{-c} + O(1/n)\ \text{as $n \to \infty$}} $$ That says $n X_{(2)}$ tends in distribution to a nontrivial limit (not a constant, though).