= High moments of a clipped minimum of exponential sample means
{title2=$\lim_n n^{-1}\log\mathbb E Z_n^n=(k+1)\log(1+1/k)-\log\lambda-1$}
Let $Z_n$ clip the minimum of $k$ independent exponential sample means into $[a,b]$, with $0<a<1/\lambda<b$. The <rate function under scalar clipping> is the minimum cost on this interval. Apply <Varadhan's lemma> to $\log z$. On the lower side its objective is increasing; on the upper side its derivative is $(k+1)/z-k\lambda$. Thus the optimum is $z_*=(k+1)/(k\lambda)$ once $k\ge(\lambda b-1)^{-1}$, giving the displayed limit. The <expectation> is of the high power, rather than a power of the mean; the latter has limit $-\log\lambda$.
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