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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.

The bufferless queue output isPut $r_L=L\lambda +B{L}^{(1+\alpha )/2}$. Since $\alpha <\beta$, we have $r_L<u_L$ for all sufficiently large $L$. Consequently the two queue overflow events are exactly equal eventually:Apply the Poisson moderate deviation principle with exponent $\alpha$ to obtainThere is also an exponential equivalence proof, which describes the whole output law at this smaller scale. For each $\epsilon >0$,The right side has strictly negative logarithmic exponent at speed $L^{\beta }$. Dividing instead by $L^{\alpha }$ multiplies that negative exponent by $L^{\beta -\alpha }\to \infty$, giving $-\infty$. Thus the centered, scaled input and output are exponentially equivalent and have the same good rate function. This is the moderate deviations below a bufferless queue cap.

The input admits positive moderate deviations on every scale $L^{(1+\delta )/2}$ with $0<\delta <1$. The bufferless queue output retains exactly the same overflow events at thresholds below the cap, so clipping does not improve the exponents on the smaller $\alpha$ scale. It removes every positive burst at thresholds on the larger $\gamma$ scale.

At the cap's own scale $\beta$, apply the contraction principle for large deviations to the map $x\mapsto \min (x,C)$. The output has good rate functionAt $C$ the cost is ${\inf }_{x\ge C}{x}^2/(2\lambda )=C^2/(2\lambda )$. Thus the burstiness of queue input has been reduced in its sufficiently large positive tail, while its smaller-scale rare fluctuations persist. These are the moderate deviation scales of a clipped Poisson aggregate.

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.

The contraction principle for large deviations states that if random elements $X^L$ in a space $E$ satisfy a large deviation principle with speed $a_L\to \infty$ and good rate function $I$, and $f:E\to F$ is continuous, then $f(X^L)$ satisfies a large deviation principle at the same speed withFor example, take $E,F$ to be Polish spaces. The rate function $J$ is good: its finite sublevel sets are the continuous images $f(\{I\le r\})$ of compact sublevel sets. Compactness and lower semicontinuity ensure that a finite fibre infimum relevant to such a sublevel set is attained. The probability bounds follow by taking preimages of open and closed sets.

A standard form of Varadhan's integral theorem is as follows. Let ${\mu }_L$ be probability measures on a Polish space $E$ satisfying a large deviation principle with speed $L\to \infty$ and good rate function $I$. If $F:E\to \mathbb{R}$ is continuous and bounded above, thenThe usual bounded-continuous version follows immediately. If $F$ is unbounded above, the same conclusion holds under the additional exponential tail conditionThis is the exponential tail extension of Varadhan's lemma. A convenient sufficient hypothesis isAn integrability hypothesis is essential for general unbounded functions; the ordinary large deviation principle alone does not control arbitrarily large exponential weights.

For example, let $X^L=L$ with probability $e^{-L^2}$ and $X^L=0$ otherwise. It satisfies the large deviation principle at speed $L$ with rate zero at zero and infinite elsewhere. But for $F(x)=x^2$, $\mathbb{E}{e}^{LF(X^L)}\ge e^{L^3-L^2}$, whose logarithmic rate diverges, while $\sup (F-I)=0$. This shows why the tail hypothesis cannot simply be omitted.

For the lower bound, fix $x$ with $I(x)<\infty$ and $\epsilon >0$. By continuity choose an open set $G$ containing $x$ on which $F(y)\ge F(x)-\epsilon$. ThenApply the open-set lower bound of the large deviation principle:Let $\epsilon \downarrow 0$ and take the supremum over $x$ to obtain the required lower bound. This argument also applies when $F$ is not bounded above.

For the upper bound, suppose $F\le M$ and set $S={\sup }_E(F-I)$, which is finite. Choose a large $K$ and partition $[-K,M]$ into finitely many closed intervals $[a_j,a_j+\epsilon ]$ of length at most $\epsilon$. The preimagesare closed sets. The overlapping endpoints do not matter for an upper estimate. Split off the lower tail and bound each remaining exponential weight:For a finite sum of nonnegative terms the upper logarithmic rate is at most the largest of their upper logarithmic rates. Applying the closed-set upper bound givesFor $x\in C_j$, $a_j+\epsilon \le F(x)+\epsilon$, hence every inner term is at most $S+\epsilon$. Empty preimages contribute no term. Let $K\to \infty$ and then $\epsilon \downarrow 0$ to get the upper bound $S$. Together with the lower bound this proves the bounded-above theorem.

For the unbounded extension, put $F_M=\min (F,M)$. This is continuous and bounded above, so the proved result applies, withThe full integral is at most the integral with $F_M$ plus the integral on $\{F>M\}$. The upper logarithmic rate is therefore at most the larger of $S_M$ and the tail rate. The tail hypothesis and then $M\to \infty$ give the upper bound $S$; the earlier local argument gives the lower bound. If $S=+\infty$, that lower argument already yields the asserted infinite limit.

Finally, the sufficient exponential-moment condition really implies the required tail estimate: on $F>M$,After integration the upper logarithmic rate is bounded by a finite constant minus $(\gamma -1)M$, which tends to $-\infty$. This completes the statement and proof of Varadhan's lemma, including the hypotheses needed for the unbounded version.

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 product large-deviation principle states that independent families satisfying large deviation principles at the same speed with good rate functions $J$ and $I$ have joint good rate function $J(b)+I(z)$. The contraction principle for large deviations states that a continuous map $f$ sends such a principle to one with good rate function ${\inf }_{f(u)=x}I(u)$. Apply these results to $(B/L,S_L/L)$ and addition. The resulting infimal convolution isFor ${\sigma }^2>0$, the unconstrained stationary point is $b=x-\mu -\lambda {\sigma }^2$. Convexity gives the constrained minimum at $b_{\ast }=(x-\mu -\lambda {\sigma }^2{)}^{+}$, soThe two branches agree in value and first derivative at their junction. Both tails tend to infinity, giving a good rate function. This is the fixed exponential perturbation of a Gaussian empirical mean.

There is a reason to use the product large-deviation principle rather than invoke the supplied Gärtner–Ellis theorem without checking its hypotheses. The limiting scaled cumulant-generating function here isAt $\lambda$ it is neither lower semicontinuous nor steep from the left: its left derivative tends to the finite value $\mu +\lambda {\sigma }^2$. Its Legendre-Fenchel transform still equals $K$, but the stated Gärtner–Ellis theorem does not establish the full lower bound on the linear branch. The product large-deviation principle and contraction principle for large deviations do establish it. In the degenerate case ${\sigma }^2=0$, the answer is $K(x)=\lambda (x-\mu )$ for $x\ge \mu$, and infinity otherwise.

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).

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.

The indexing defect also affects the downstream claim. For the two inputs in part (b), now choose $\lambda <d<c$. The constant input has zero downstream queue workload, while the input with $a_0=c+1$ hasTheir printed distance is zero, so the literal post-slot downstream continuity assertion is false as well.

In the corrected weighted cumulative-input topology for a slotted queue, apply the finite-window proof of part (b) separately with services $d$ and $c$. Both have strictly negative long-window drift because $\lambda <d<c$. There are a neighborhood of $a$ and finite constants $C_d,C_c$ such thatThis proves continuity, and indeed local Lipschitz continuity, for the intended downstream function.

Suppose a family of random arrival paths $a^{(L)}$ has a large deviation principle in this corrected input space, at speed $L$ with good rate function $\mathcal{I}$. The contraction principle for large deviations then gives the downstream good rate functionIt is infinity for $r<0$. The path large deviation principle, its topology, and its subcritical domain must actually be established for the chosen stochastic model: a collection of finite-dimensional large deviation principles alone does not control infinitely long workload windows. Nor should upstream departures be assumed independent; the path contraction avoids that unjustified assumption.

Write $a_n=n^{2(1-H)}$ and $b_n=n^{2(1-G)}$. Then $a_n/b_n=n^{2(G-H)}\to \infty$. The large deviation principle on the whole space forces $\inf I=0$; compact sublevel sets and lower semicontinuity make that infimum attained. The unique possible zero is therefore mu, and $I(\mu )=0$.

If a closed set C contains mu, its new infimum rate is zero and the claimed upper bound follows from probabilities being at most one. If it excludes mu, then $\eta ={\inf }_{C}I>0$, possibly infinite: otherwise a sequence of rates tending to zero would have a convergent subsequence in a compact sublevel set, with limit in C and zero rate, contradicting uniqueness. This is the fact that a good rate function is separated from zero away from its minimizer.

Choose any $0<c<\eta$. The original upper bound gives $\log \mathbb{P}(X_n\in C)\le -c{a}_n$ for all sufficiently large n. ThusTogether these cases prove the closed-set upper bound at the slower speed, with rate zero only at mu. Its finite sublevels are the compact singleton $\{\mu \}$, so it is a good rate function.

Use the stated continuity of the infinite-buffer and finite-buffer queue workload maps on the input path space. The contraction principle for large deviations gives, at the same speed as the input sequence,Empty fibers have infimum infinity. Both rates are good because they are continuous contractions of the good rate function I. In particular, at speed L the infinite-buffer queue workload satisfies, for every Borel set E,and the identical bounds with bars give the finite-buffer queue workload LDP. This states the entire principles, not just an informal tail approximation.

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.