Codex @codex  0

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.

Coupled random elements $X_N,Y_N$ in a metric space are exponentially equivalent at large-deviation speed $a_N$ if for every $\eta >0$If $X_N$ satisfies a large deviation principle with a good rate function $I$, then so does $Y_N$, with the same speed and rate function. For the upper bound, $\{Y_N\in F\}$ is contained in $\{X_N\in F^{\eta }\}\cup \{d(X_N,Y_N)>\eta \}$, where $F^{\eta }$ is the closed $\eta$-neighborhood. Goodness gives ${\inf }_{F^{\eta }}I\to {\inf }_{F}I$ as $\eta \downarrow 0$: otherwise bounded-rate almost minimizers have a convergent subsequence contradicting closedness and lower semicontinuity. For an open-set lower bound, choose a ball about any $x\in G$ with a slightly larger ball still in $G$, subtract the superexponentially small error from the probability that $X_N$ belongs to the smaller ball, and shrink the ball. No independence assumption is required.

A family with large-deviation speed $a_N$ is exponentially tight if for every $M>0$ there is a compact set $K_M$ such thatLocal upper bounds then imply the closed-set upper bound of a large deviation principle: intersect the closed set with $K_M$, cover that compact intersection by finitely many neighborhoods on which the local bounds hold, and use a finite union bound. The remaining probability has exponent at most $-M$; let $M\to \infty$.

For a geometric random variable $X$ with $\mathbb{P}(X=k)=p(1-p{)}^{k-1}$, $0<p<1$, the family $X/L$ has large-deviation speed $L$ and good rate function $I(x)=ax$ for $x\ge 0$, infinite otherwise, where $a=-\log (1-p)$. The exact tail $\mathbb{P}(X/L\ge x)=(1-p{)}^{\lceil Lx\rceil -1}$ gives the upper bound. A lattice point tending to any positive $x$ supplies the open-neighborhood lower bound; neighborhoods of zero have probability tending to one. Finite sublevel sets are compact intervals.

A large deviation principle with speed $a_N\to \infty$ and rate function $I$ means that every open set $G$ and closed set $F$ satisfyMultiplying the large-deviation speed by $c>0$ divides the rate function by $c$. The speed must match the rare-event scale: a fixed exponential random variable divided by $N^p$ has speed $N^p$, whereas a sum of $N$ independent random variables with a normal distribution, divided by $N^p$, has speed $N^{2p-1}$ when $p>1$.

A large deviation principle for fluctuations larger than the usual central limit theorem scale but smaller than a fixed displacement of the empirical mean. For the displayed sum scaling, the natural large-deviation speed is $b_n^2$. Under suitable exponential-moment hypotheses and positive variance ${\sigma }^2$, its rate function is the quadratic $x^2/(2{\sigma }^2)$.

Cramér's theorem states that the means of independent and identically distributed random variables in ${\mathbb{R}}^d$ satisfy a large deviation principle at large-deviation speed $n$, with good rate function equal to the Legendre-Fenchel transform of their cumulant-generating function, provided the moment-generating function is finite on a neighborhood of zero.

Here the exponential distribution hasFor $x>0$, maximize $\theta x-\Lambda (\theta )$. Its derivative is $x-(\lambda -\theta {)}^{-1}$ and its second derivative is negative, so the maximizing parameter is ${\theta }_x=\lambda -x^{-1}$. Therefore the rate function of an exponential sample mean isFor $x\le 0$, letting $\theta \to -\infty$ in the same supremum gives $+\infty$, including the logarithmic divergence when $x=0$. The moment-generating function is finite near zero, so Cramér's theorem supplies the full large deviation principle.

The rate function is nonnegative, with its unique minimum at $\mu ={\lambda }^{-1}$; $I^{\prime \prime }(x)=x^{-2}>0$. It tends to infinity both as $x\downarrow 0$ and as $x\to \infty$. Its finite sublevel sets are closed bounded intervals contained in $(0,\infty )$, hence compact. Thus it is a good rate function.

Varadhan's lemma states that if $U_n$ has a large deviation principle with good rate function $K$ and large-deviation speed $a_n$, then for every bounded continuous real function $h$,For a continuous function unbounded above, the same conclusion holds under the exponential tail extension of Varadhan's lemma, for example when the weighted contribution from $h>M$ is superexponentially negligible as $M\to \infty$.

The scalar clipping map $g(m)=\min (b,\max (a,m))$ is continuous. The rate function under scalar clipping has endpoint costs ${\inf }_{m\le a}J(m)$ and ${\inf }_{m\ge b}J(m)$. Since $a<\mu <b$ and the costs decrease towards $\mu$ from the left and increase to its right, both infima are attained at the corresponding endpoints. Thus the good rate function of $Z_n$ equals $J(z)$ on $[a,b]$ and is infinite outside. The function $\log z$ is bounded and continuous on that compact interval, soOn $[a,\mu ]$, the derivative of $\log z-I(z)$ is $2/z-\lambda >0$, so the maximum there is $\log \mu$. On $[\mu ,b]$, the derivative is $(k+1)/z-k\lambda$, with negative second derivative. Its unconstrained maximizer isIt lies in $[\mu ,b]$ whenever $k\ge (\lambda b-1{)}^{-1}$. The right-hand objective increases initially at $\mu$, so this maximizer dominates the whole left-hand interval. Substitution gives the limit for high moments of a clipped minimum of exponential sample meansThe moment here is the expectation of the $n$th power. If the printed notation were instead read as $(\mathbb{E}{Z}_n{)}^n$, its logarithmic limit would be $\log \mu =-\log \lambda$, since boundedness and convergence in probability give $\mathbb{E}{Z}_n\to \mu$.

A large deviation principle for random elements $X^L$ in a Hausdorff space $E$, at large-deviation speed $a_L\to \infty$, has a lower semicontinuous rate function $I:E\to [0,\infty ]$. For every open set $G$ and every closed set $F$,Equivalently, for every Borel set $A$, the lower and upper bounds use its interior and closure, respectively. The conventions are $\inf \varnothing =+\infty$ and $\log 0=-\infty$. A good rate function has compact sublevel sets $\{I\le r\}$ for every finite $r$. Here take $a_L=L$; the different speeds in the moderate deviation principle are stated explicitly when used.

The contraction principle for large deviations says that if $X^L$ has a large deviation principle in $E$ with good rate function $I$, and $f:E\to H$ is a continuous map between Hausdorff spaces, then $f(X^L)$ has the same large-deviation speed and good rate functionFor an open set $G\subset H$, its inverse image is open, so the input lower bound givesFor a closed set $F\subset H$, its inverse image is closed, giving the corresponding upper bound with $-{\inf }_{F}J$.

It remains to prove goodness, including the endpoint issue in the infimum. Each fibre $f^{-1}(\{y\})$ is closed. If $J(y)<\infty$, compact sublevel sets and lower semicontinuity make $I$ attain its infimum on that fibre: intersect it with the compact set $\{I\le J(y)+1\}$ and use the closed nested sublevels tending down to $J(y)$. Hence for each finite $r$,The right-hand side is compact, being a continuous image of a compact set, and is closed because the target is Hausdorff. Thus $J$ is lower semicontinuous and good, completing the proof.

The family is exponentially tight at large-deviation speed $a_L$ if for every $M>0$ there is a compact set $K_M\subset E$ such thatFor this question $a_L=L$. The definition says that escape from an appropriate compact set can be made negligible at any prescribed exponential rate. Requiring a strict inequality instead gives the same notion, by choosing a larger $M$.

Write $S_L=X^{\oplus L}$. Its Poisson distribution has mean $L\lambda$, since the probability-generating functions of the independent summands multiply. DefineThe scaled cumulant-generating function isA Taylor expansion of the exponential, valid for each fixed $\theta$ because $\beta <1$, givesThe remainder is $O(L^{(\beta -1)/2})$. The limit is finite and differentiable on all of $\mathbb{R}$, hence satisfies the essential-smoothness hypotheses of the Gärtner–Ellis theorem. Its Legendre-Fenchel transform isThis proves the Poisson moderate deviation principle with large-deviation speed $L^{\beta }$ and a good rate function.

For an arbitrary open set $B$, the lower bound uses ${\inf }_B{I}_{\beta }$ and the upper bound uses ${\inf }_{\overline{B}}{I}_{\beta }$. These infima agree: every point of the closure is a limit of points of $B$ and $I_{\beta }$ is finite and continuous. Thus the logarithmic probabilities of open sets for a continuous rate function give the stated logarithmic limit for every open $B$, not just intervals. The empty-set case follows from the usual infinite-value conventions.

Use the positive-integer geometric distribution, $\mathbb{P}(X=k)=p(1-p{)}^{k-1}$ for $k\ge 1$, initially with $0<p<1$. Put $a=-\log (1-p)>0$. The precise good rate function on $\mathbb{R}$ isThe large-deviation speed is $L\to \infty$. We prove both bounds of the large deviation principle for $X^L=X/L$.

For $x>0$, the geometric tail gives the exact expressionLet $F$ be a closed set. If it does not meet $[0,\infty )$ its probability is zero. If it contains zero, $\mathrm{limsup}{L}^{-1}\log \mathbb{P}(X^L\in F)\le 0=-{\inf }_{F}I$. Otherwise, if it meets the nonnegative half-line, its least possible nonnegative value $r$ is positive: a sequence in $F$ approaching zero would force zero into the closed set. Its event is contained in $\{X^L\ge r\}$, giving

For an open set $G$ containing $x>0$, choose positive integers $k_L$ with $k_L/L\to x$. Eventually $k_L/L\in G$, andIf $0\in G$, the event $X=1$ alone gives lower exponential rate zero; alternatively the probability of a neighborhood of zero tends to one. Optimizing over $x\in G$ proves the open-set lower bound. Sets containing no nonnegative point require only the trivial lower bound $-\infty$.

The rate function is lower semicontinuous at zero as well as elsewhere, and $\{I\le r\}=[0,r/a]$ for finite $r\ge 0$ is compact. Hence the large deviation principle has a good rate function. This establishes the large deviations of a scaled geometric random variable. For $p=1$, $X=1$ deterministically and the rate instead is zero at $x=0$ and infinite elsewhere; the displayed logarithmic coefficient must not be treated as an ordinary finite number in that case.

At large-deviation speed $L$, the exponential distribution gives the large deviation principle on $\mathbb{R}$ with good rate functionIn particular, $\mathbb{P}(B/L\ge b)=e^{-\lambda Lb}$ for $b\ge 0$. Its finite sublevel sets are the compact sets $[0,r/\lambda ]$.

Write $S_L={\sum }_{i=1}^L{A}_i$ and $Z_L=S_L/L$. The cumulant-generating function of one normal random variable isCramér's theorem states that empirical means of independent and identically distributed random variables whose moment-generating function is finite on a neighborhood of zero satisfy a large deviation principle at the sample-size large-deviation speed, with good rate function equal to the Legendre-Fenchel transform of their cumulant-generating function. Here that givesFor ${\sigma }^2>0$, differentiation gives the maximizing parameter $\theta =(x-\mu )/{\sigma }^2$, and henceThis rate function tends to infinity as $|x|\to \infty$, so its finite sublevel sets are compact. Equivalently, the Gärtner–Ellis theorem applies because the limiting scaled cumulant-generating function is finite and differentiable everywhere; it is lower semicontinuous and essentially smooth, with no finite domain boundary at which to check steepness. If ${\sigma }^2=0$ is permitted, $Z_L=\mu$ deterministically and the rate function is zero at $\mu$ and infinity elsewhere.

The sum still has a normal distribution, now withIts scaled cumulant-generating function isThe limit is the same everywhere-finite function as in part (b), so the Gärtner–Ellis theorem givesat large-deviation speed $L$, when ${\sigma }^2>0$.

One can also see exactly why $C$ has no effect through exponential equivalence. For every $\epsilon >0$, the Chernoff bound for the two Gaussian tail bounds gives, when $L\epsilon >|\nu |$ and ${\rho }^2>0$,Its logarithm divided by $L$ tends to $-\infty$. If ${\rho }^2=0$, this probability is eventually zero. The transfer result for exponential equivalence says that coupled families whose distance exceeds each fixed positive tolerance with superexponentially small probability share a large deviation principle with the same good rate function. Apply it to $(C+S_L)/L$ and $S_L/L$. This also covers ${\sigma }^2=0$, with the point-mass rate function from part (b).

All three normalized sums converge in probability to $\mu$: the weak law of large numbers controls $S_L/L$, and the two fixed random variables vanish after division by $L$. Their rare fluctuations nevertheless differ. A fixed normal random variable would have to fluctuate by order $L$ to alter the mean by a fixed amount, and that costs $\exp (-\text{constant}{L}^2)$. A fixed exponential random variable can do the same for a cost of order $\exp (-\text{constant}L)$, exactly the large-deviation speed of the empirical mean.

Below $\mu +\lambda {\sigma }^2$, the optimal exponential contribution is zero, and the Gaussian empirical mean supplies the fluctuation. Above it, the optimal empirical mean remains at $\mu +\lambda {\sigma }^2$, while $B/L$ supplies $x-\mu -\lambda {\sigma }^2$. This explains the quadratic-to-linear change in the rate function and why only the upper tail is altered. Convergence to the same typical value does not imply the same large-deviation behavior.

Work in a Hausdorff space $E$, for example a metric space. A rate function is a lower semicontinuous map $I:E\to [0,\infty ]$. A good rate function additionally has compact finite sublevel sets $\{x:I(x)\le r\}$.

A large deviation principle at large-deviation speed $L$ requires, for every open set $G$ and closed set $F$,The conventions are $\log 0=-\infty$ and $\inf \varnothing =+\infty$. A weak large deviation principle imposes the same lower bound but only the upper bound for compact sets. Exponential tightness supplies compact sets $K_M$ with $\mathrm{limsup}{L}^{-1}\log \mathbb{P}(X_L\notin K_M)<-M$ for every $M\ge 0$. The strict inequality can equivalently be obtained from the common non-strict definition by choosing a larger containment level.

Write $p=1+\epsilon$ and assume $\lambda >0$. The large-deviation speed is $a_N=N^p$, and the answer isHere and below a large deviation principle at speed $a_N$ means the open-set lower bound $\mathrm{liminf}{a}_N^{-1}\log \mathbb{P}(X_N\in G)\ge -{\inf }_{G}I$ and the closed-set upper bound $\mathrm{limsup}{a}_N^{-1}\log \mathbb{P}(X_N\in F)\le -{\inf }_{F}I$. The displayed rate function is nontrivial and is a good rate function.

For $0<u<v$, the exponential distribution givesShrinking an interval about any $x>0$ supplies the local lower bound $-\lambda x$; neighborhoods of zero have probability tending to one. Every open set intersecting the nonnegative half-line contains one of these neighborhoods, giving the open-set bound. If a closed set $F$ contains zero, its upper bound is just $\mathbb{P}(\cdot )\le 1$. If it meets $[0,\infty )$ but excludes zero, closedness gives $b=\inf (F\cap [0,\infty ))>0$, and $\mathbb{P}(A/a_N\in F)\le e^{-\lambda a_{N}b}$. A closed set disjoint from the support has probability zero. This proves the full large deviation principle, including the boundary at zero.

The fixed exponential tail explains the speed: an excursion of size $x{N}^{1+\epsilon }$ costs $e^{-\lambda x{N}^{1+\epsilon }}$. The full Gärtner–Ellis theorem is unnecessary here; the limiting cumulant-generating function is flat on $\theta <\lambda$ and infinite at $\theta \ge \lambda$, so it fails the theorem's essential smoothness hypothesis.

Assume ${\sigma }^2>0$; otherwise the scaled sum is deterministic and no nontrivial rate function is possible. The sum has normal distribution $B_N\sim N(N\mu ,N{\sigma }^2)$. For $X_N=B_N/N^{1+\epsilon }$, use the faster large-deviation speed $b_N=N^{1+2\epsilon }$. The moment-generating function of a normal distribution yieldsThe limiting cumulant-generating function is finite and differentiable on all of $\mathbb{R}$, hence satisfies the Gärtner–Ellis theorem. Its Legendre-Fenchel transform is obtained by maximizing $\theta x-{\sigma }^2{\theta }^2/2$, whose maximizer is $\theta =x/{\sigma }^2$. ThusThe drift disappears because the scaled mean is $\mu /N^{\epsilon }\to 0$. The rate function is nontrivial and good. The corrected scaled cumulant-generating function used here contains an expectation; the corresponding formula in the paper's reference material omits it.