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Saddlepoint approximation methods retain the nonquadratic shape of a distribution through its cumulant-generating function. Suppose $X_1,\dots ,X_n$ are independent and identically distributed, $K(t)=\log E(e^{t{X}_1})$ is finite on an open interval containing zero, and $K^{\prime \prime }>0$. For an interior target mean $x$, solve the saddlepoint equation $K^{\prime }(t)=x$. Introduce exponential tilting byUnder the tilted distribution the mean and variance are $K^{\prime }(t)=x$ and $K^{\prime \prime }(t)$. For the product sample the Radon-Nikodym derivative is $e^{t{\sum }_i{X}_i-nK(t)}$. Consequently the exact density relation for the sample mean isA local central limit approximation to the tilted density at its own mean gives $f_{\overline{X},t}(x)\simeq \sqrt{n/(2\pi K^{\prime \prime }(t))}$. HenceFor the sum $S_n$, the equivalent expression is $f_{S_n}(s)\simeq e^{nK(t)-ts}/\sqrt{2\pi n{K}^{\prime \prime }(t)}$, with $K^{\prime }(t)=s/n$. Under the usual smooth nonlattice density conditions this approximation has relative error of order $n^{-1}$ for a fixed interior target. In the tilted Edgeworth expansion the order $n^{-1/2}$ odd term vanishes at the mean. The exponent retains the Legendre-transform rate $I(x)=tx-K(t)$, rather than a quadratic Taylor approximation to that rate, explaining the useful tail and skewness behavior. Normalization is not automatic, and accuracy statements need their regularity and range restrictions.

For a normal distribution, $K(t)=\mu t+{\sigma }^2{t}^2/2$ and $t=(x-\mu )/{\sigma }^2$. Substitution recovers exactly the normal density of the sample mean. For an exponential distribution of rate $\lambda$, $K(t)=-\log (1-t/\lambda )$, $t=\lambda -1/x$ and $K^{\prime \prime }(t)=x^2$. The approximation becomesIt has exactly the gamma density shape. Replacing its leading constant by the exact normalizing constant makes it exact, illustrating why exactness of shape and exactness of the unnormalized approximation are separate questions.

For a vector sample mean, tilt by $e^{t^{\mathsf{T}}y-K(t)}$, solve $\nabla K(t)=x$, and use the local multivariate normal density at the tilted mean. This givesJoint and marginal saddlepoint densities can also be divided to approximate conditional densities, for example after conditioning on a nuisance sufficient statistic. The p* approximation uses likelihood and fitted-information coordinates to obtain a closely related approximation to a maximum-likelihood estimator's density, usually conditional on an ancillary statistic.

Tail probabilities need a corresponding integration approximation. The Lugannani-Rice saddlepoint tail approximation usesand givesHere $\Phi ,\varphi$ are the standard normal distribution functions. The upper-tail correction has the opposite sign: $1-\Phi (w)+\varphi (w)(1/u-1/w)$. Inverting a cumulative distribution function introduces a pole in addition to the density's saddlepoint; treating both supplies the correction term, which is why simply integrating a normal approximation at the untilted mean is less accurate. At $x=K^{\prime }(0)$ the apparent singularities cancel. With $v=K^{\prime \prime }(0)$ and $k=K^{\prime \prime \prime }(0)$, expansions give $w=t\sqrt{nv}(1+kt/(3v)+O(t^2))$ and $u=t\sqrt{nv}(1+kt/(2v)+O(t^2))$, so $1/w-1/u\to k/(6\sqrt{n}{v}^{3/2})$. For the normal distribution the correction is zero everywhere.

These methods require care for lattice distributions, where summation and lattice corrections replace the density inversion, and near boundaries or singular covariance matrices. A distribution with no moment-generating function in a neighborhood of zero does not satisfy the starting assumptions. A successful saddlepoint approximation is therefore a structured refinement of regular asymptotic inference, not a formula guaranteed uniformly in every tail or for every model.

Use the Probabilists' Hermite polynomials, rather than the differently scaled physicists' convention:Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The cumulant-generating function of an exponential variable of mean $\theta$ is $-\log (1-\theta t)$ near zero, so its $r$th cumulant is $(r-1)!{\theta }^r$. Hence the standardized third and fourth cumulants are ${\rho }_3=2$ and ${\rho }_4=6$; the fourth is a cumulant, not the fourth raw central moment.

Write $F_n$ for the cumulative distribution function of the pivot in part (i). Its Edgeworth expansion, valid here because the exponential law is smooth, nonlattice and has exponential moments, isat the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\varphi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the symmetric Edgeworth coverage cancellation givesIn particular its coverage probability error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.

A conditional likelihood is the conditional density of the observed data given a chosen statistic, viewed as a function of the parameter. Conditioning can eliminate a nuisance parameter if the statistic is sufficient for it within the relevant family. The conditioning statistic need not be ancillary; conditioning can also discard some information about the parameter of interest.

For the canonical family, let $T_1={\sum }_i{\tau }_1(Y_i)$ and $s=n^{-1}{\sum }_i{\tau }_2(Y_i)$. Normalization of the density gives the cumulant-generating function of one nuisance statistic:For the sample mean $S$, it is $K_S(t)=n[d(\psi ,\lambda +t/n)-d(\psi ,\lambda )]$. Differentiation gives $d_{\lambda }=\mathbb{E}{\tau }_2(Y_i)$ and $d_{\lambda \lambda }=\mathrm{Var}({\tau }_2(Y_i))$.

Assume an interior saddle with positive curvature and the usual smooth nonlattice density conditions. Define ${\hat{\lambda }}_{\psi }$ byThis is the constrained maximum-likelihood estimator of $\lambda$ with $\psi$ held fixed. The exponential tilt parameter for one summand is $\hat{t}={\hat{\lambda }}_{\psi }-\lambda$. Tilting puts the sample mean at $s$, where its tilted density has leading normal value $\sqrt{n/[2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })]}$. Undoing the tilt gives the saddlepoint density approximationThis is the density of the mean, not the sum; their factors differ by $n$.

The sample log-likelihood is $\ell (\psi ,\lambda )=\psi T_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(Y_i)$. Conditional on $S=s$, the factor involving $\lambda$ is constant on the conditioning surface, so the conditional distribution is free of $\lambda$. Subtracting the saddlepoint log-density from the joint log-likelihood now cancels all the original nuisance terms:The final term, and any conditioning-coordinate factor, are data-only constants. Thus the saddlepoint conditional likelihood adjustment isThe absolute-value notation agrees with the question; positive variance makes it unnecessary in this regular setting. The sign of this correction is positive, because a nuisance-statistic density with inverse-square-root curvature has been divided out. A lattice-valued statistic requires the corresponding mass approximation, and a boundary or zero-variance saddle is outside the stated regular density calculation.

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).

Consider a stationary nonnegative work-input process in slotted time, with cumulative work $A_n$ over the last $n$ slots and a work-conserving server of capacity $c$ per slot. Assume its past average tends almost surely to $m<c$. The Lindley recursion then defines the finite stationary queue workloadThis is work remaining, rather than the number of customers when customer service requirements vary. With unit work per customer the two notions coincide. Stationarity makes the distribution of a past cumulative window equal to that of a forward window of the same length.

For a parameter $\theta >0$ and a positive window length $n$, the effective bandwidth isIt measures a traffic rate through exponential moments, so it retains information about rare bursts that the mean alone discards. If the limiting scaled cumulant-generating function exists, putThe finite-window effective bandwidth matters when the relevant overflow window is short or when arrivals are correlated; one should not replace it by its long-window limit without justification.

The following proof gives an actual upper bound for the infinite-horizon queue workload. For $B>0$, if $Q>B$, at least one positive integer $n$ has $A_n-cn>B$. The union bound followed by the Chernoff bound givesAssume each finite-window exponential moment is finite and $\kappa (\theta )<c\theta$. Choose $\eta >0$ such that $\kappa (\theta )+\eta <c\theta$. For all sufficiently large $n$, the summand is at most $e^{-n(c\theta -\kappa (\theta )-\eta )}$. Its tail is a convergent geometric series, and its finite prefix is finite. Denote the resulting sum by $K_{\theta }$. The exponential workload bound from cumulative arrival moments therefore proves, for $b>0$,Optimizing over admissible parameters givesOnly parameters with the stated finite-window moments are included. If there are no such positive parameters, the argument yields only the trivial exponent zero. Infinite ${\theta }_{\ast }$ means that the upper rate is $-\infty$. For a non-strict threshold $Q\ge Lb$, first bound it by $Q>L(b-\epsilon )$ and then let $\epsilon \downarrow 0$. This proof permits dependence between slots. It proves the required upper bound; it does not assert a matching lower bound for every arrival process.

This also gives a closed-set large-deviation upper bound for $Q/L$: set $I_{+}(q)={\theta }_{\ast }q$ for $q>0$, $I_{+}(0)=0$, and $I_{+}(q)=+\infty$ for $q<0$. If a closed set $F$ contains zero, its desired upper bound is the trivial bound zero. If $F$ contains positive points but not zero, its positive part has a smallest point $b>0$, and $\mathbb{P}(Q/L\in F)\le \mathbb{P}(Q\ge Lb)$ gives the bound $-{\inf }_F{I}_{+}$. If it contains no nonnegative points, its probability is zero. This verifies the upper half of a large deviation principle; the actual rate function can be larger when a matching lower bound is unavailable.

For independent and identically distributed slot arrivals $a_t$, let $\kappa (\theta )=\log \mathbb{E}{e}^{\theta a_0}$. Now $\log \mathbb{E}{e}^{\theta A_n}=n\kappa (\theta )$, so effective bandwidth is independent of window length. For $\kappa (\theta )<c\theta$, the geometric-series prefactor is explicitlyThus the familiar workload Chernoff bound for independent increments is a special case, and the largest admissible parameter is often the positive root of $\kappa (\theta )=c\theta$.

For these independent increments there is a sharper argument that works at the root itself when its exponential moment is finite. Along the past read in reverse order, put $S_n=A_n-cn$ and $M_n=e^{\theta S_n}$. If $\mathbb{E}{e}^{\theta (a_0-c)}\le 1$, then $M_n$ is a nonnegative supermartingale starting at one. For the first crossing time ${\tau }_B=\inf \{n\ge 0:S_n\ge B\}$, bounded optional stopping givesLet $N\to \infty$. Strict negative drift makes $S_n\to -\infty$ almost surely, so its supremum is attained and the crossing event agrees with $\{Q\ge B\}$. ConsequentlyThis strengthens the prefactor but uses temporal independence, which was unnecessary for the previous proof.

Several examples show what the effective bandwidth constraint means. Deterministic traffic of $r$ work units per slot has $\kappa (\theta )=\theta r$ and effective bandwidth $r$ at every parameter; if $c\ge r$ there is no stationary backlog. For independent on-off slots carrying $r$ units with probability $p$ and zero otherwise, the Bernoulli effective bandwidth isFor $0<p<1$, it rises from the mean $pr$ to the peak $r$. When $pr<c<r$, strict convexity of $\kappa (\theta )-c\theta$, its negative derivative at zero, and its divergence to infinity give a unique positive root. When $c\ge r$, no slot can increase the queue workload. In the specific case $r=2,c=1,p<1/2$, the workload increases by one with probability $p$ and decreases by one, reflected at zero, with probability $1-p$. Its stationary probabilities satisfyThe service-root equation has $e^{{\theta }_{\ast }}=(1-p)/p$, and thus $\mathbb{P}(Q\ge k)=q^k=e^{-{\theta }_{\ast }k}$. This example realizes the upper exponent exactly.

For independent Poisson slot counts of mean $\nu$, with unit work per arrival, the Poisson effective bandwidth isStability requires $c>\nu$. There is a unique positive root of $\nu (e^{\theta }-1)=c\theta$, again by strict convexity, negative initial derivative after subtracting $c\theta$, and growth to infinity. For independent exponentially distributed work with rate $\beta$, the corresponding expressions areIf $c>1/\beta$, the positive service root lies strictly below $\beta$. The finite moment-generating function domain cannot be ignored.

To see statistical aggregation explicitly, let the entire arrival processes of $k$ sources be independent, though each source may itself have dependence over time. Their cumulative totals satisfyThus the shared queue has an exponential overflow bound when ${\sum }_j{\alpha }_j(\theta )<c$, with the finite-window conditions proved above. One must use a common parameter $\theta$; adding individual service-root exponents is not the rule. For example, an independent Poisson source of mean $\nu$ and an independent on-off source of size $r$ and probability $p$ have aggregate cumulant-generating functionFor $\nu >0$ and $c>\nu +pr$, their positive service root exists and determines the optimized upper exponent. Several independent Poisson sources similarly have the same formula as one source with intensity ${\sum }_j{\nu }_j$.

The connection with large deviation principles also appears in a many-source scaling. For $L$ independent copies of a source, aggregate service $Lc$ and buffer $Lb$, fix a window $n$. Let $X_i(n)$ be each copy's cumulative input in that window. Their total exceeds $L(b+cn)$ only if the empirical mean exceeds $b+cn$. Its Chernoff bound has exponentIf exponential moments exist near zero, Cramér's theorem gives the corresponding finite-window large deviation principle. The threshold exceeds the mean because $m<c$ and $b>0$. A finite union of windows has upper exponential rate at most $-{\min }_n{V}_n(b)$ over those windows. Passing to infinitely many windows requires uniform control of long-window probabilities, not simply an interchange of a limit with an infinite union. A good sample-path large deviation principle in a topology where the queue workload is continuous, as in Question 3, instead permits the contraction principle for large deviations to derive a workload rate function and its closed-set overflow upper bound.

Finally, the mean and peak limits of effective bandwidth quantify the safety margin. With exponential moments near zero, independent slot work of mean $m$ and variance $v$ hasFor bounded work, its large-parameter limit is the essential peak. Stronger tail requirements select larger $\theta$ and hence more capacity; finite-window effective bandwidths also reflect temporal burst correlations. A stable heavy-tailed input can lack every positive exponential moment, in which case this exponential-bound method supplies no positive decay rate. Stability and exponentially small overflow are distinct requirements.

Let U be the total payload arriving in one slot. Conditional on its Poisson packet count, use the exponential moment of each independent payload to get, for $\theta <1$,For net work $Z=U-C\delta$, its cumulant-generating function isThe mean net work is $\delta (\lambda -C)<0$. Under the arrivals-before-service slot convention the Lindley recursion gives stationary queue workload $R_{\delta }\stackrel{d}{=}{\sup }_{m\ge 0}{\sum }_{j=1}^m{Z}_j$. The positive root of the exponential-moment equation is ${\theta }_{\ast }=1-\rho$. Thus $\exp ({\theta }_{\ast }{\sum }_{j=1}^m{Z}_j)$ is a nonnegative martingale. Stop it at the first crossing of r or a finite horizon; its value on the crossing event is at least $e^{{\theta }_{\ast }r}$. Increasing the horizon provesThis is an infinite-horizon bound, not a one-slot Chernoff estimate.

A useful refinement identifies both the exponential rate and the small-slot prefactor. Couple the random walk to the continuous compound Poisson input U(t), and write $W={\sup }_{t\ge 0}(U(t)-Ct)$ and $W_{\delta }={\sup }_{k\ge 0}(U(k\delta )-Ck\delta )$. Their laws are the continuous and slot-boundary stationary queue workloads. Monotonicity of U implies the grid-sampled compound Poisson workload bound: $W_{\delta }\le W\le W_{\delta }+C\delta$, by rounding each t up to the next grid point.

The continuous queue workload law can be obtained directly from the stationary packet count. Conditional on Q=k, the k remaining payloads are independent unit-mean exponentials: unserved payloads have not affected earlier service completions, and the in-service residual uses the memoryless property. HenceThus W has mass $1-\rho$ at zero, and its positive part is exponential with rate $1-\rho$. Combining this stationary workload of an exponential-payload queue with the grid bound gives the explicit estimateIn particular ${\lim }_{r\to \infty }{r}^{-1}\log \mathbb{P}(R_{\delta }\ge r)=-(1-\rho )$, and for small delta the estimate is $\rho e^{-(1-\rho )r}$. The exact continuous-time tail is that latter expression. Another slot convention may shift queue workload by the within-slot input/service timing; its convention must be specified before asserting a prefactor, while this exponential tail rate is unchanged.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

Let the summands have mean mu, variance ${\sigma }^2>0$, and cumulants ${\kappa }_j$. For the standardized sum $Z_n=(S_n-n\mu )/(\sigma \sqrt{n})$, expansion of its log-characteristic function gives a normal leading term and successively smaller cumulant terms. Fourier inversion turns these terms into Probabilists' Hermite polynomials. Under sufficient moments and a smooth nonlattice condition, the Edgeworth expansion beginsfor bounded z. Higher orders include the fourth cumulant and products of lower cumulants. This is a central expansion in powers of $n^{-1/2}$; it need not remain positive, normalized, or accurate in far tails after truncation.

For a saddlepoint density approximation, require the summand cumulant-generating function K to be finite on an interval about zero. At a sum value $s=ny$, find the interior saddle $\hat{t}$ satisfying $K^{\prime }(\hat{t})=y$. Exponential tilting makes the tilted mean y, so the tilted sum is approximated locally by its normal probability density at its mean. Undoing the tilt givesEquivalently this follows by deforming the transform-inversion contour through its stationary point. Suitable analytic and nonlattice regularity gives relative error $O(n^{-1})$ at fixed interior y. This approximation incorporates a nonquadratic tail exponent and is positive, while an Edgeworth truncation uses a normal probability density times a polynomial. It uses more information from the distribution and stronger exponential-moment assumptions. Its integral may differ from one by a higher-order amount, so positivity does not imply exact normalization.

For the centered Laplace distribution, $K(t)=t^2+t^4/2+\cdots$, hence ${\kappa }_2=2$ and ${\kappa }_3=0$. Its skewness vanishes, so the order-$n^{-1/2}$ Edgeworth correction is zero. With $z=s/\sqrt{2n}$, the requested expansion, omitting terms of order $n^{-1}$ in the standardized correction, isFor bounded z the exact probability density function is this expression times $1+O(n^{-1})$. The prefactor is $n^{-1/2}$ because this is the probability density function of the unstandardized sum; that change of units should not be confused with the orders inside the Edgeworth expansion.

For the saddlepoint, $K^{\prime }(t)=2t/(1-t^2)$. Write $y=s/n$ and $d=\sqrt{1+y^2}$. The root in $(-1,1)$ isThese formulas include $s=0$ without a division by y. Substitution yieldsThis is the leading saddlepoint approximation, without its order-$n^{-1}$ relative correction. Expanding d near one recovers the same central normal approximation, while retaining d captures the Laplace-sum large-deviation exponent away from the center.

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.

Interpret the additional copies in $Z_N$ as independent of the initial $T_1,\dots ,T_k$, as well as of one another. Set $r=k\lambda$. The Cramér theorem for rate-$r$ exponential random variables gives the rate functionIndeed their cumulant-generating function is $\log (r/(r-\theta ))$ for $\theta <r$, and the maximizer in its Legendre-Fenchel transform is $\theta =r-1/z$. For $Z_N/N$, replacing $N$ summands by $N-k$ does not change either the speed $N$ or this rate function. One can also see this directly from $N^{-1}\log \mathbb{E}{e}^{\theta Z_N}\to \log (r/(r-\theta ))$ and the Gärtner–Ellis theorem.

The product large-deviation principle, followed by the contraction principle for large deviations under addition, now givesFor $x>0$ this is the minimum over $0<z\le x$ ofIf $k>1$, its derivative with respect to $z$ is $r-\lambda -1/z$. Hence the optimum is $z=\min (x,1/[\lambda (k-1)])$, andThe two expressions agree at the transition and have the same derivative $\lambda$. The minimum rate zero occurs at the typical value $x=1/(k\lambda )$. If $k=1$, the minimand is $\lambda x-1-\log (\lambda z)$, decreasing in $z$, so insteadThese are good rate functions. The affine branch for $k>1$ is the slow exponential plus an exponential sample mean mechanism: the bulk ceases to carry additional excess, which instead comes from $T_1$. In particular, blindly applying the full Gärtner–Ellis theorem to the combined sum would be unjustified. Its limiting cumulant-generating function is $\log (r/(r-\theta ))$ only for $\theta <\lambda$, with a finite boundary slope when $k>1$; essential smoothness fails. The product and contraction argument proves the missing lower bounds.