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A defective renewal equation is $z(u)=g(u)+{\int }_{[0,u]}z(u-x)Q(dx)$ where the nonnegative kernel measure has total mass less than one. Its renewal representation is a convergent sum of convolutions. Exponential tilting can turn the kernel into a probability measure, enabling the key renewal theorem.

For an isonormal Gaussian process $X$ on a real Hilbert space $H$ and $k\in H$, the positive density $\exp (X(k)-\Vert k{\Vert }^2/2)$ has expectation one and defines an equivalent probability measure. Under this measure, $X(h)$ has mean $⟨h,k⟩$ and the same covariance as before. The shifted family $X(h)-⟨h,k⟩$ is again an isonormal Gaussian process. This follows by evaluating the joint Gaussian exponential formula for any finite collection of arguments. It is the Hilbert-space version of exponential tilting.

Let $X_N\in {\mathbb{R}}^d$ and $a_N\to \infty$. Define the scaled cumulant-generating functionSuppose its pointwise limit $\Lambda$ exists, is lower semicontinuous, has $0$ in the interior of its effective domain, and is essentially smooth: it is differentiable on that interior, which is nonempty, and its gradient norm diverges at every finite boundary point approached from the interior. Then $X_N$ satisfies a large deviation principle with good rate function given by the Legendre-Fenchel transformThe Chernoff bound supplies the upper estimate. The lower estimate uses exponential tilting at a parameter whose gradient selects the required mean; essential smoothness supplies the boundary approximation needed for the full lower bound. A finite boundary slope can leave an affine branch of ${\Lambda }^{\ast }$ outside this argument, so a separate lower-bound proof is then necessary.

Let the increments define an upward skip-free random walk, and suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X_1}$ is finite for all $\lambda \ge 0$. If ${\lambda }_0>0$ satisfies $\varphi ({\lambda }_0)=1$ and ${\varphi }^{\prime }({\lambda }_0)>0$, the exponential tilting at ${\lambda }_0$ gives integrable increments of positive mean. The strong law of large numbers makes every nonnegative integer level attainable under this tilted law. Upward jumps are at most one, so the value at a first level hit is exactly that level. The stopped likelihood ratio under exponential tilting therefore gives hitting probability $e^{-{\lambda }_{0}k}$ for integer $k\ge 0$. Subtracting consecutive tail probabilities yields the displayed geometric distribution on $j=0,1,\dots$. For real thresholds use $\lceil k\rceil$ in place of $k$.

The Hoeffding lemma controls the entire centered moment-generating function using only an interval width. For an essentially bounded real random variable $X$ with finite deterministic bounds $a\le X\le b$ almost surely, every real $\lambda$ satisfiesEquivalently, $\mathbb{E}{e}^{\lambda (X-\mathbb{E}X)}\le e^{{\lambda }^2(b-a{)}^2/8}$. This single-random variable statement needs no independence assumption; independence enters when applying it to the factors for a sum in the Hoeffding inequality. Both signs of $\lambda$ are allowed, and equality holds at $\lambda =0$ or when $a=b$.^[ref]

In the convention $\log \mathbb{E}{e}^{\lambda Y}\le {\lambda }^2{\sigma }^2/2$ for a centered sub-Gaussian random variable $Y$, the Hoeffding lemma supplies ${\sigma }^2=(b-a{)}^2/4$. This upper parameter need not equal the actual variance. For a Bernoulli random variable $B$ of success probability $p$,The actual variance is $p(1-p)$, whereas the supplied parameter is $1/4$ for every $p$. This explains part of the conservatism visible in Hoeffding sample size for Bernoulli frequencies.

A short proof exposes why the constant is $1/8$. Put $Y=X-\mathbb{E}X$ and define exponential tilting byThe tilted probability measure has the same interval of possible values, and differentiating the cumulant-generating function gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{{\mathbb{P}}_{\lambda }}(Y)$. Boundedness justifies these derivatives for every finite $\lambda$. Under any such tilted law, Popoviciu's inequality on variances gives ${\psi }^{\prime \prime }(\lambda )\le (b-a{)}^2/4$: the variance is no larger than the expected value of the squared distance from the interval midpoint, itself at most $(b-a{)}^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, the integral form of Taylor's theorem yieldsThis expression handles either sign of $\lambda$ without changing the argument.

The constant $1/8$ cannot be reduced uniformly over all essentially bounded real random variables. Equal endpoint probabilities at $a$ and $b$ give $Y=\pm (b-a)/2$, a scaled Rademacher random variable, andThus the sharpness is already forced near zero by the variance; the bound need not be equality at nonzero parameters for a particular probability distribution.

Let the possible numerical outcomes be $-1,0,1$, and give the reference law $p$ uniform distribution on a finite set: $p=(1/3,1/3,1/3)$. We want a law $q$ with expected value $1/2$. Write the three probabilities in outcome order as $(q_{-},q_0,q_{+})$. The mean constraint is $q_{+}-q_{-}=1/2$, while normalization requires $q_{-}+q_0+q_{+}=1$.

These conditions leave a whole line of possible probability distributions:For example, $(0,1/2,1/2)$ and $(1/4,0,3/4)$ both have the required mean. The constraint alone does not choose between them. Information projection selects the point on this line with the smallest relative entropy from the reference.

Exponential tilting gives weights proportional to $e^{-\theta },1,e^{\theta }$. Set $t=e^{\theta }>0$ and multiply these weights by $t$ to obtain $q^{\ast }=(1,t,t^2)/(1+t+t^2)$. Enforcing the mean givesThe positive root is $t=(1+\sqrt{13})/2$, henceThe most favored outcome becomes $1$, but all three remain possible. This is an exact constructed example with a stipulated reference law and constraint, not a fitted distribution from measurements.

Suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X}<\infty$ for all $\lambda \ge 0$. Under the exponential tilting $d{\mu }^{\lambda }=e^{\lambda x}d\mu /\varphi (\lambda )$, $X$ is an integrable random variable whenever $\lambda >0$: its negative exponential tail is bounded, and its positive contribution is dominated by a larger positive exponential moment. Differentiation under the integral sign gives the displayed mean. At the endpoint $\lambda =0$, the positive part of $X$ is integrable, but the negative part need not be. The right derivative ${\varphi }_{+}^{\prime }(0)=\mathbb{E}X$ is valid as an extended value, possibly $-\infty$; a finite mean at zero requires an extra negative-tail integrability assumption.

Let $K(t)=\log \mathbb{E}{e}^{tY}$ be the one-observation cumulant-generating function, finite on an open interval containing zero. At a value $y$, find the real tilt $\hat{t}$ solvingIf the distribution is nondegenerate then $K^{\prime \prime }(t)>0$ in the interior, so the solution is unique whenever $y$ is in the range of $K^{\prime }$. The leading saddlepoint density approximation to the sample mean isTo derive it, use exponential tilting:Under this tilted distribution the mean is $K^{\prime }(t)$ and the variance is $K^{\prime \prime }(t)$, by differentiating its cumulant function $K(t+u)-K(t)$. For $S_n={\sum }_i{Y}_i$, changing measure gives the exact density identityAt $t=\hat{t}$, $ny$ is the tilted mean. A local normal approximation there gives $f_{S_n,\hat{t}}(ny)\simeq (2\pi n{K}^{\prime \prime }(\hat{t}){)}^{-1/2}$. The Jacobian determinant for $S_n=n\bar{Y}$ contributes the factor $n$, yielding the boxed formula.

This approximation retains the exponential large-deviation factor, while the curvature is evaluated at the tilt appropriate to each $y$. At the population mean, $\hat{t}=0$, so it reduces to the usual central normal approximation. Under suitable smooth nonlattice transform-inversion conditions, its relative error at fixed interior $y$ is $O(n^{-1})$; existence of a few moments alone does not establish this density accuracy. Boundary values and absent exponential moments need separate treatment. The leading density need not integrate exactly to one, so it can be divided by its integral when a normalized approximation is desired.

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.

Let $N=4700$ and let the daily distances be independent and identically distributed random variables with probabilities $p_1=p_3=0.4995$, $p_{100}=0.001$. The unconditioned mean daily distance is $2.098$, whereas the observed trip requires a daily average near $10$. Conditioning on that rare average changes the typical frequencies substantially. It does not determine an exact boost count.

Use the method of types: for a fixed finite alphabet, the probability of an empirical distribution $q$ iswhere $D(q\Vert p)={\sum }_i{q}_i\log (q_i/p_i)$ is relative entropy, using natural logarithms. This follows directly from Stirling's approximation. There are only polynomially many types, so the empirical distribution conditional on a mean constraint concentrates at its unique relative entropy minimizer, when admissible lattice types approximate that minimizer. Any fixed neighborhood of it has a strictly smaller minimum cost than its complement.

Minimize $D(q\Vert p)$ subject to ${\sum }_i{q}_i=1$ and $q_1+3q_3+100q_{100}=10$. The entropy minimization under a prescribed sample mean gives the exponential tiltingTo verify minimization without relying only on Lagrange multipliers, any feasible distribution $r$ satisfiesRelative entropy nonnegativity makes $q$ the unique minimizer. Its mean determines $\theta$ through $Z^{\prime }(\theta )/Z(\theta )=10$, orThe derivative of the tilted mean is the tilted variance, which is strictly positive. The mean rises from $2.098$ at $\theta =0$ to $100$ as $\theta \to \infty$, so this equation has one positive solution. Numerically,The predicted boost count is $N{q}_{100}=381.6597211$. Thus the conditional large-deviation estimate is about 382 boosts.

For exact arrival at total distance $47000$, let $k,m,l$ count days of distances $100,3,1$. The constraints giveConsequently $k$ can be any even integer from $340$ to $426$. In particular, a unique count cannot be inferred from the return time. The exact conditional weights are proportional toTheir most probable count is $k=382$, consistent with the exponential tilting calculation. This finite-sample inference is a conditional probability statement, not an assertion that all journeys with the stated duration have that count.

If returning home means first crossing the distance threshold on day $4700$, the final total instead lies between $47000$ and $47099$. The last increment must exceed the overshoot; for fixed counts this selects the corresponding fraction of possible last days. Near the tilted optimum that factor is bounded away from zero, so it does not alter the leading exponential type cost. The same approximate answer follows. Exact conditional enumeration of this first-arrival interpretation also has mode $382$.

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.

For the amount of work $A_j(0,t]$ offered by one flow in a time interval of length $t$, its effective bandwidth at parameter $\theta >0$ and timescale $t>0$ isIts units are work per unit time. The parameter $\theta$ measures the price placed on bursts: a stringent rare-overflow target generally requires a larger value of $\theta$ than a loose target. The timescale $t$ describes the duration of the congestion episode. When the limit exists, writeThis is the long-time effective bandwidth. It includes a burstiness cost, rather than just the mean offered rate. For a finite-time flow with the required moments, ${\lim }_{\theta \downarrow 0}{\alpha }_j(\theta ,t)=\mathbb{E}{A}_j(0,t]/t$, and effective bandwidth is increasing in $\theta$. If $A_j(0,t]/t$ is bounded, its limit as $\theta \to \infty$ is the essential supremum of that interval's input rate, the peak-rate limit. The analogous long-time mean limit requires the usual interchange or regularity assumptions.

For independent flows their cumulant-generating functions add, soThe Chernoff bound therefore gives, for any $\theta >0$ in the common domain,Optimizing over $\theta$ produces the relevant Legendre-Fenchel transform. The Cramér theorem states that sample means of independent and identically distributed inputs whose moment-generating function is finite near zero have a large deviation principle, with rate function equal to the Legendre-Fenchel transform of their cumulant-generating function. More generally, the Gärtner–Ellis theorem applies to a convergent scaled log moment-generating function when zero lies inside its domain and the limit is lower semicontinuous and essentially smooth. It gives the same transform as a good rate function. A sample-path version, together with the contraction principle for large deviations for the queue workload, translates input deviations into overflow deviations. The compact-time and infinite-horizon continuity/tail hypotheses must be checked for the chosen traffic model.

A particularly useful rigorous queue bound holds for inputs with stationary increments and independent increments, such that $\mathbb{E}{e}^{\theta A(t)}=e^{t\Lambda (\theta )}$. If $\Lambda (\theta )\le C\theta$, thenis a nonnegative supermartingale, since its conditional increment factor is $e^{(t-s)(\Lambda (\theta )-C\theta )}\le 1$. The maximal inequality for a nonnegative supermartingale on $[0,T]$, followed by $T\to \infty$, gives the continuous-time exponential workload boundHere $W$ is the stationary queue workload of the stable infinite-buffer queue, represented by input from the past. This directly bounds the supremum over time; applying a one-time Chernoff bound alone would not establish it.

For a nonzero Compound Poisson process with bounded positive packet sizes and mean input rate less than $C$, a positive root ${\theta }_{\ast }$ of $\Lambda (\theta )=C\theta$ exists and is unique. Here the root theorem can be made explicit:The upper bound was just proved. For the lower bound, use exponential tilting with density $e^{{\theta }_{\ast }(A(t)-Ct)}$ at time $t$. Under the tilted law the net input has positive mean drift ${\Lambda }^{\prime }({\theta }_{\ast })-C>0$, by strict convexity and the positive-root equation. The strong law of large numbers then makes the first crossing time ${\tau }_b$ finite almost surely under that law. Its overshoot is at most $s_{\max }$. Changing measure at ${\tau }_b$, first on $\{{\tau }_b\le T\}$ and then letting $T\to \infty$, givesDividing the logarithmic bounds by $b$ proves ${\lim }_{b\to \infty }{b}^{-1}\log \mathbb{P}(W\ge b)=-{\theta }_{\ast }$.

Thus the familiar admission rule compares ${\sum }_j{a}_j(\theta )$ with $C$, choosing $\theta$ from the desired buffer-tail exponent. For more general inputs, a logarithmic root theorem requires its own light-tail and renewal or sample-path hypotheses; the displayed supermartingale upper bound needs only the moment and increment assumptions stated above.

The profile likelihood for the parameter of interest replaces the nuisance parameter by its constrained best fit: $L_p(\psi )={\sup }_{\lambda }L(\psi ,\lambda )$, with ${\ell }_p(\psi )=\ell (\psi ,{\hat{\lambda }}_{\psi })$. It is not in general a normalized sampling density. A conditional likelihood instead uses the sampling law conditional on a statistic carrying the nuisance dependence. In the canonical exponential family here, conditioning on $S$ removes $\lambda$ because the factor involving $\lambda$ is constant over the conditioning fibre.

Let ${\bar{T}}_1=n^{-1}{\sum }_i{\tau }_1(y_i)$ and let the observed $S$ equal $s$. For the cumulant-generating function of ${\tau }_2(Y)$ at fixed $\psi$,The saddlepoint solves $K^{\prime }(\hat{t})=s$, equivalently $d_{\lambda }(\psi ,{\hat{\lambda }}_{\psi })=s$ with ${\hat{\lambda }}_{\psi }=\lambda +\hat{t}$. This is also the constrained likelihood equation, since ${\ell }_{\lambda }=n(s-d_{\lambda })$. Assume an interior saddle, $d_{\lambda \lambda }>0$, and the usual smooth nonlattice conditions under which $S$ has a density. Exponential tilting to ${\hat{\lambda }}_{\psi }$ makes $s$ the tilted mean. A local normal approximation gives tilted density $\{n/(2\pi d_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })){\}}^{1/2}$ at that mean; undoing the tilt gives the saddlepoint density approximationThe factor $n^{1/2}$ is for the sample mean, rather than the sum. A transform-based derivation gives the same expression by expanding $K$ to second order at its stationary point.

The joint log-likelihood is $n\psi {\bar{T}}_1+n\lambda s-nd(\psi ,\lambda )-{\sum }_{i}Q(y_i)$. Subtracting $\log {\hat{f}}_S(s;\psi ,\lambda )$ cancels every appearance of the original $\lambda$, leavingAny change from the ambient density to a conditional density on $S=s$ contributes only a data-dependent Jacobian determinant. Thus, up to a constant independent of $\psi$, the saddlepoint conditional likelihood adjustment isEquivalently $B=\frac{1}{2}\log j_{\lambda \lambda }(\psi ,{\hat{\lambda }}_{\psi })$ up to the constant $-\frac{1}{2}\log n$, where $j_{\lambda \lambda }=n{d}_{\lambda \lambda }$ is the nuisance observed information. The adjustment has a positive sign. The printed exponential-family form alone does not guarantee a density for $S$: for example a continuous $Y$ can have ${\tau }_2(Y)=1_{\{Y>0\}}$. In a lattice case one must approximate the probability mass of the sum instead; the displayed density calculation uses the regular continuous interpretation.

For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

First take $k$ to be a nonnegative integer. The exponential tilting constructed from the independent increments restricts to ${\mathbb{P}}_n^{\lambda }$ on each ${\mathcal{F}}_n$. Since its Radon-Nikodym derivative is strictly positive, finite-time change of measure givesUnder the tilted law, $L_n=(M_n^{\lambda }{)}^{-1}$ is a martingale, becauseUse the stopped likelihood ratio under exponential tilting at this finite horizon. In detail, $\{{\tau }_k=j\}\in {\mathcal{F}}_j$, soHere the last expression is defined only on the displayed event, so no value of $M_{\infty }^{\lambda }$ is being assumed. The upward skip-free random walk has integer increments bounded above by one, which imply $S_{{\tau }_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. ConsequentlyAt $\lambda ={\lambda }_0$, the mean under one-sided exponential tilting is finite and positive. The strong law of large numbers gives $S_n/n\to {\varphi }^{\prime }({\lambda }_0)>0$ under ${\mathbb{P}}^{{\lambda }_0}$, so every integer level is hit almost surely. Since $\varphi ({\lambda }_0)=1$, passing to $n\to \infty$ givesLet $H={\sup }_{n\ge 0}{S}_n$. The event $\{H\ge k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb{P}(H=\infty )={\lim }_k{e}^{-{\lambda }_{0}k}=0$. Subtraction of consecutive tail probabilities gives the geometric maximum of an upward skip-free random walk:This is the geometric distribution counting failures before the first success, with success parameter $1-e^{-{\lambda }_0}$.

The PDF writes only $k\ge 0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then ${\tau }_k={\tau }_{\lceil k\rceil }$ and the displayed factor must be $e^{-\lambda \lceil k\rceil }$; the hitting probability is likewise $e^{-{\lambda }_0\lceil k\rceil }$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have ${\lambda }_0=\log 2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.

For independent observations in a natural exponential family, the log likelihood isIn a minimal exponential family the Observed Fisher information matrix is positive definite, so the log likelihood is strictly concave. Any finite solution of the likelihood equation $\nabla \kappa (\hat{\theta })=\bar{Y}$ is therefore the unique global maximum likelihood estimator. Existence is a separate issue: the observed sufficient-statistic mean must belong to the image of the mean parameter map $\nabla \kappa$.

For a full regular minimal natural exponential family this image is the interior of the convex support of an exponential family, $C=\overline{\mathrm{conv}(\mathrm{supp}(h\nu ))}$. The necessity follows from a supporting hyperplane argument. If a finite parameter had its mean on a boundary face, some nonzero $v$ would satisfy $v^{T}Y\le b$ almost surely and $\mathbb{E}(v^{T}Y)=b$. Equality of the expectation forces $v^{T}Y=b$ almost surely. Exponential tilting preserves the base-measure null sets, so this contradicts minimality.

For sufficiency, let $\bar{Y}$ be an interior point of $C$. Choose finitely many points $y_1,\dots ,y_m$ of the support whose convex hull contains a ball about $\bar{Y}$, say of radius $2\epsilon$. Such points exist because $\bar{Y}$ is interior to the closed convex hull; one may approximate the vertices of a small simplex surrounding it by finite convex combinations of support points. Take support neighbourhoods of radius at most $\epsilon$ with positive finite base-measure masses $a_j$. These masses can be taken finite by restricting to bounded neighbourhoods and using finiteness of the normalizing integral at any fixed interior parameter. For every vector $\theta$, at least one $j$ has ${\theta }^T(y_j-\bar{Y})\ge 2\epsilon \Vert \theta \Vert$. Integrating over that neighbourhood yieldsThus the negative log likelihood grows to infinity as $\Vert \theta \Vert \to \infty$. At a finite boundary point of the full open natural parameter space, the normalizing integral is infinite, since otherwise that point would itself belong to the full domain. Fatou's lemma makes $\kappa (\theta )$ diverge on approach to such a boundary. A minimizing sequence for $\kappa (\theta )-{\theta }^T\bar{Y}$ therefore stays in a bounded subset away from the boundary; it has an interior limit where the minimum is attained. Its gradient vanishes, giving the required likelihood equation. Strict convexity gives uniqueness.

Hence, under these full, regular and minimal conventions,This is maximum likelihood existence in a full regular natural exponential family. Regularity alone does not make every sample admissible: in a Bernoulli distribution an all-zero or all-one sample has its sample mean on the boundary of $C=[0,1]$, and the canonical maximum is approached only as $\theta \to -\infty$ or $+\infty$. The usual probability parameter then has a boundary maximum in an enlarged model, but no finite canonical maximizer. A regular restriction of the natural parameter space requires membership in its actual mean-map image, rather than merely in $\mathrm{int}C$. Without minimality, affine redundancies may prevent uniqueness of the canonical parameter even when the fitted probability distribution is unique.

Write $K(t)=\log \mathbb{E}[e^{tY}]$ for the cumulant-generating function in the question's $(\varphi ,\lambda )$ parametrization. Multiplying the density by $e^{ty}$ replaces $\varphi$ in the decaying part of its exponent by $\varphi -2t$. For $t<\varphi /2$, the remaining integral is the normalizing integral for an Inverse Gaussian distribution with parameters $(\varphi -2t,\lambda )$. Comparing the two normalizing constants givesIn particular $K^{\prime }(0)=\sqrt{\lambda /\varphi }$, so the mean parameter used in the more usual Inverse Gaussian distribution notation is $\mu =\sqrt{\lambda /\varphi }$; it must not be confused with the present parameter $\varphi$.

First compute the cumulant-generating function of the sample average, as requested. Independence givesThis is exactly the cumulant-generating function of an Inverse Gaussian distribution with parameters $(n\varphi ,n\lambda )$. The uniqueness theorem for moment-generating functions therefore identifies that law. Scaling $S_n=n{\overline{Y}}_n$ and including the density Jacobian $1/n$ givesEquivalently, the inverse Gaussian sum closure is $S_n\sim IG(\varphi ,n^2\lambda )$ in the parametrization of this question. Its mean is $n\sqrt{\lambda /\varphi }$ and its variance is $n\sqrt{\lambda }/{\varphi }^{3/2}$, in agreement with addition of independent means and variances.

For a general continuous iid sample with cumulant-generating function $K$, finite near zero, set $x=s/n$ and choose a real saddle $\hat{t}$ in the interior of the finite moment domain satisfying $K^{\prime }(\hat{t})=x$. With $K^{\prime \prime }(\hat{t})>0$, the saddlepoint density approximation for the sum isUnder the usual smooth nonlattice conditions, at a fixed interior $x$ the true density has this leading expression times $1+O(n^{-1})$. This is a relative density approximation; the leading expression is not automatically an exactly normalized probability density in every model.

One way to understand the statistical saddlepoint approximation is through exponential tilting. The tilted one-observation density is $f_t(y)=e^{ty-K(t)}f(y)$, with mean $K^{\prime }(t)$ and variance $K^{\prime \prime }(t)$. Undoing this tilt for the sum multiplies its tilted density at $s$ by $e^{nK(t)-ts}$. At $t=\hat{t}$, $s$ is exactly the tilted mean of the sum. A local version of the central limit theorem gives tilted density there approximately $[2\pi n{K}^{\prime \prime }(\hat{t}){]}^{-1/2}$. Multiplying these two factors gives the displayed saddlepoint density approximation. Higher local expansion terms produce the stated relative error order.

For the Inverse Gaussian distribution, differentiation givesThus for every $x>0$ the saddle exists, withAt this saddle,Substitution into the saddlepoint density approximation gives precisely the exact density already found for $S_n$. ThereforeThis exact saddlepoint density of an inverse Gaussian sum is stronger than a relative $O(n^{-1})$ claim: the leading formula is exact and needs no renormalization.