Contraction principle for large deviations ($J(y)={\inf }_{f(x)=y}I(x)$)

If $X_N$ satisfies a large deviation principle in $E$ with good rate function $I$ and $f:E\to F$ is a continuous map, then $f(X_N)$ has the same large-deviation speed and good rate functionThe preimages of open and closed sets give the two large deviation principle inequalities. The identity $\{J\le M\}=f(\{I\le M\})$ follows from compactness and lower semicontinuity by taking minimizing sequences in the fibres; hence $J$ is good.

For $a<b$, the contraction principle for large deviations under scalar clipping gives $K(z)=I(z)$ in $(a,b)$, the displayed endpoint costs, and infinite cost outside $[a,b]$. The endpoint costs can differ from $I(a)$ and $I(b)$ because clipping collects a whole input tail into a single output atom.

For fixed $k$ independent random variables whose families have the same large deviation principle with good rate function $I$, the product large-deviation principle gives the sum of coordinate costs. Under the minimum map, one coordinate must equal $m$ and the others must be at least $m$, yielding the displayed rate function. If $I$ is convex with unique zero $\mu$, then $J(m)=I(m)$ for $m\le \mu$ and $J(m)=kI(m)$ for $m\ge \mu$.

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 $z_{\ast }=(k+1)/(k\lambda )$ 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 $-\log \lambda$.