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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.

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.

For the final distribution, both tails decay polynomially. Any positive exponential tilt makes the positive tail nonintegrable, and any negative tilt makes the negative tail nonintegrable. Its moment-generating function is therefore infinite away from zero, leaving no open interval on which to form the cumulant-generating function and solve a saddle equation. Here polynomial tails prevent exponential tilting, so the conventional saddlepoint density approximation is unavailable. A finite mean and variance, and even an ordinary central limit theorem, do not supply the missing exponential moments.

For every $s\ge 0$, the pointwise inequality $1_{\{Y\ge 0\}}\le e^{sY}$ holds: on the event the exponential is at least one, and off it the indicator is zero. Taking expected values and then the infimum proves the Chernoff boundAn infinite exponential moment merely gives a vacuous bound at that $s$; no existence assumption on moments is needed for this extended-value inequality.

For a positive $s$ at which the relevant moment-generating functions are finite, apply the Chernoff bound to $Y=Z-C$. Independence of the sources givesThusClasses with $n_j=0$ make no contribution, so no moment assumption on those inactive classes is necessary. The effective bandwidth ${\alpha }_j(s)$ is the exponential-moment capacity demand of one class-$j$ source over this observation interval. Independent sources have additive effective bandwidths; reserving the margin $\gamma /s$ below capacity certifies the desired overflow bound. If $X_{ji}$ describes traffic during one unit of time, this is a bandwidth in traffic per unit time; for a window of length $t$, divide $\log \mathbb{E}{e}^{sX(t)}$ by $st$ instead.

To establish monotonicity and bounds of effective bandwidth, write $K(s)=\log \mathbb{E}{e}^{sX}$ for a representative source. The Holder inequality gives, for $0<t<s$ in its moment domain,Equivalently, the cumulant-generating function is convex with $K(0)=0$, so its secant slope from zero is nondecreasing. Hence $\alpha (s)$ is nondecreasing, rather than necessarily strictly increasing: if $X$ is deterministic its effective bandwidth is constant.

Assuming the mean is defined and finite, the Jensen inequality gives $\mathbb{E}{e}^{sX}\ge e^{s\mathbb{E}X}$, hence $\alpha (s)\ge \mathbb{E}X$. LetFor finite $M$, $X\le M$ almost surely, so $\mathbb{E}{e}^{sX}\le e^{sM}$ and $\alpha (s)\le M$. Combining these provesThe upper bound is vacuous if the essential supremum is infinite. For nonnegative traffic these comparisons also make sense with extended values; for an arbitrary real-valued source with undefined mean, the printed mean comparison itself requires an integrability qualification. The positive parameter domain is essential: $\alpha (0)$ is defined only by a limit when the appropriate moment assumptions hold, and the stated monotonicity comparisons concern $s>0$.

Write $m_j=\mathbb{E}{Y}^j$ and $M=M_Y$. The derivatives of the cumulant-generating function areSince $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three cumulants areThe last equality follows by expanding the cube and taking expectations. We use the paper's unstandardized third central moment throughout. The usual skewness coefficient divides this quantity by $\mathrm{Var}(Y{)}^{3/2}$; its sign is the same whenever the variance is positive.

There is a minor domain qualification: finite ordinary moments do not guarantee a finite moment-generating function for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite moments, all formulas also follow from left derivatives at zero, because $Y^j{e}^{tY}\le Y^j$ for $t\le 0$ and dominated convergence applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.

For the random sum of independent claims, condition on $N=k$. Independence of the claims and independence from the count giveTaking expectations and then logarithms yields the random-sum transform identity and the cumulant composition for a random sum:Here $G_N$ is the probability generating function. These identities hold wherever the transforms are finite, in particular for $t\le 0$ with positive claims.

For reuse in the three cases, write $m_j=\mathbb{E}{X}_1^j$ and $a_j={\kappa }_{N,j}$. Applying the chain rule three times givesor equivalentlyThis expresses the requested third central moment in raw claim moments, while keeping the contribution of claim-count variability explicit.

Put $n=T-t$ and define the finite geometric seriesStarting the autoregressive process of order one at time $t$ gives $r_{t+k}={\beta }^k{r}_t+{\sum }_{j=1}^k{\beta }^{k-j}{\xi }_{t+j}$. ConsequentlyThe future innovations are independent and identically distributed random variables, independent of ${\mathcal{F}}_t$. Factorizing their exponential conditional expectation and using the cumulant-generating function $K$ yieldsThus the exponential-affine bond pricing coefficients areThey vanish at maturity, giving $P(T,T)=1$. Everywhere finiteness of $K$ ensures that each zero-coupon bond price is positive and finite, including when $\beta$ is zero, negative, or one.

An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let $X\ge 0$ be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write $\Lambda (s)=\log \mathbb{E}[e^{sX}]$. The Chernoff bound follows immediately from Markov's inequality applied to $e^{sX}$:Optimizing gives $\boxed{\mathbb{P}(X\ge b)\le \exp [-\underset{s>0}{\sup }\{sb-\Lambda (s)\}]}$, where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.

For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small $s$, $\boxed{\alpha (s)\to \mathbb{E}X}$, with the variance giving the first burstiness correction. convexity of $\Lambda$ and $\Lambda (0)=0$ show that its secant slope $\alpha (s)$ is nondecreasing; Jensen's inequality also gives $\alpha (s)\ge \mathbb{E}X$.

If $X$ is bounded with essential supremum $M$, then $\alpha (s)\le M$. For any $a<M$, the probability $p_a=\mathbb{P}(X>a)$ is positive andLet $s\to \infty$ and then $a\uparrow M$. For large $s$, $\boxed{\alpha (s)\to \mathrm{esssup}X}$, so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives $\alpha (s)\to \infty$. If exponential moments cease to exist beyond a finite parameter, $\alpha$ is infinite there; one cannot assign a finite large-$s$ peak interpretation to that traffic.

For example, deterministic traffic $X=d$ has $\alpha (s)=d$. A burst of size $b$ with Bernoulli distribution probability $p$ giveswhich moves from $pb$ to $b$. For Poisson packet count of mean $\lambda$ and packet size $b$, $\alpha (s)=\lambda (e^{sb}-1)/s$, showing a mean limit $\lambda b$ but no finite peak limit.

For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals $A_r(t)$ in a window of length $t$, use the rate-valued definition ${\alpha }_r(s,t)=\log \mathbb{E}[e^{s{A}_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfiesThus a sufficient admission condition for a target probability $\epsilon$ is ${\sum }_r{\alpha }_r(s,t)\le C-\log (1/\epsilon )/(st)$ for some $s>0$. Independence is needed for this additive form; correlated sources require the joint exponential moment.

A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments $X_i$, constant service $C>\mathbb{E}X$, and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever $a(s)=e^{\Lambda (s)-sC}<1$,The requirement is precisely $\alpha (s)<C$. This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.

Let $\mu =\mathbb{E}{X}_1$ and let the cumulant-generating function and its Legendre transform beUnder the permitted assumption that $\psi$ is finite everywhere, the right-tail form of the Cramér theorem is, for $a\ge \mu$,Use $\log 0=-\infty$. The restriction to nonnegative parameters is valid on this side of the mean: convexity gives $\psi (\theta )\ge \mu \theta$, so negative parameters cannot improve the value zero available at $\theta =0$. For $a<\mu$, the right-tail probability tends to one by the strong law of large numbers, so its logarithmic rate is zero; it is not generally $-I(a)$.

The full large deviation principle bounds probabilities of closed sets from above and open sets from below with rate $I$. For the displayed tail equality, a possible finite upper endpoint also needs attention. If $b=\mathrm{esssup}{X}_1<\infty$, then at $a=b$ the event requires every summand to equal $b$; its probability is $\mathbb{P}(X_1=b{)}^n$, and $I(b)=-\log \mathbb{P}(X_1=b)$. The identity follows by taking $\theta \to \infty$ in $\mathbb{E}{e}^{\theta (X_1-b)}$. Above $b$, both sides have infinite rate. The theorem describes exponential decay, including endpoints and impossible events.

Write $b=\mathrm{esssup}{X}_1$. The hypothesis $\mathbb{P}(X_1>a)>0$ implies $a<b$ and rules out a constant random variable. In particular $I(a)<\infty$. The cumulant-generating function is strictly convex, since its second derivative is the positive variance under exponential tilting, and ${\psi }^{\prime }(0)=\mu$, ${\psi }^{\prime }(\theta )\to b$ as $\theta \to \infty$.

If $a>\mu$, there is a unique ${\theta }_a>0$ with ${\psi }^{\prime }({\theta }_a)=a$, and the Legendre transform gives $I(a)={\theta }_{a}a-\psi ({\theta }_a)$. ThusIf $a=\mu$, use differentiability at zero: $\psi (\theta )=\mu \theta +o(\theta )$, so for some sufficiently small $\theta >0$,In either case choose a parameter $\theta \ge 0$ with $J=\theta (a+\epsilon )-\psi (\theta )>I(a)$, and choose $0<\eta <J-I(a)$. The Cramér theorem gives, for all sufficiently large $n$,The Chernoff bound at the chosen parameter gives $\mathbb{P}(S_n/n\ge a+\epsilon )\le e^{-nJ}$. The conditioning event has positive probability for every $n$, since all summands can simultaneously exceed $a$. ThereforeTaking the complement within the conditioning event proves the conditional concentration at a large-deviation threshold:The farther tail has a strictly larger exponential cost. The argument also works when $a+\epsilon$ reaches or exceeds the upper support endpoint, without requiring an interior maximizer there.

An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.

Let $S={\sum }_i{Y}_i$ and $V_H={\sum }_i(b_i-a_i{)}^2$. To prove the Hoeffding lemma, take a centered random variable $Y\in [a,b]$ and its cumulant-generating function $\psi (t)=\log \mathbb{E}{e}^{tY}$. Under exponential tilting, ${\psi }^{\prime \prime }(t)$ is the variance of $Y$. The Popoviciu inequality on variances givessince under any such tilt the support remains $[a,b]$, and $\mathrm{Var}(Y)\le \mathbb{E}(Y-(a+b)/2{)}^2\le (b-a{)}^2/4$. Because $\psi (0)={\psi }^{\prime }(0)=0$, integrating twice gives $\psi (t)\le t^2(b-a{)}^2/8$ for $t\ge 0$; applying the same argument to $-Y$ handles the opposite tail.

The independence of the $Y_i$ gives $\mathbb{E}{e}^{tS}\le \exp (t^2{V}_H/8)$. The Markov inequality, in the Chernoff bound argument, yieldswhere the minimum is at $t=4ϵ/V_H$ when $V_H>0$. Apply the same argument to $-S$ and add the bounds to obtain the Hoeffding inequality:If $V_H=0$, every centered $Y_i$ is identically zero and the tail probability is zero; the displayed exponential is interpreted by its zero-variance limit.

For the Bennett inequality, write $v=n{\sigma }^2$, and first assume $M>0$ and $v>0$. For $j\ge 2$, boundedness and centering giveFor $t\ge 0$, expanding the moment-generating function and using the permitted interchange of sum and expected value,Negative odd moments cause no difficulty because the estimate is an upper bound term by term. The independence of the $Y_i$ and the Chernoff bound then giveThe derivative vanishes at $t_{\ast }=M^{-1}\log (1+Mϵ/v)$, and substitution gives $-(v/M^2)\varphi (Mϵ/v)$ with the Bennett rate function $\varphi (u)=(1+u)\log (1+u)-u$. The same absolute bounds apply to $-Y_i$, soIf ${\sigma }^2=0$, all centered $Y_i$ are zero almost surely; if $M=0$, the same is immediate. Handle those cases directly rather than divide by zero. Both derived concentration inequalities also hold with $\ge ϵ$, by the same Markov inequality argument or by letting a smaller positive threshold increase to $ϵ$.

For the binomial distribution, write $X={\sum }_{i=1}^n{Z}_i$ with independent and identically distributed random variables $Z_i$ having the Bernoulli distribution with parameter $p=p_n$. Then $Y_i=Z_i-p$ has range $[-p,1-p]$, width one, and variance $p(1-p)$. For large $n$, $p<1/2$, so $M=1-p$ and $v=np(1-p)$. Set $ϵ=C\sqrt{v}$.

The Hoeffding inequality givesAfter truncation at one, this is asymptotically the trivial probability bound. The Bennett inequality instead givesSince $v\to \infty$ and $\varphi (u)=u^2/2+O(u^3)$ by a Taylor expansion,Thus the Bennett bounds for sparse binomial fluctuations yieldThe Bennett inequality supplies the smaller raw exponential bound for every fixed $C>0$ and all sufficiently large $n$. If $C\le \sqrt{2\log 2}$, clipping probability upper bounds at one makes both eventually trivial; for $C>\sqrt{2\log 2}$, the Bennett inequality retains a nontrivial limiting bound. This distinction avoids claiming a probability improvement after clipping when both raw bounds exceed one.

Because the moment-generating function is finite in a neighborhood to the right of zero and $\mathbb{E}X=0$, its cumulant-generating function satisfiesDivide the defining Sub-Gamma random variable in the right tail inequality by ${\lambda }^2/2$ and let $\lambda \downarrow 0$. The right side converges to $\nu$, proving $\mathrm{Var}(X)\le \nu$.

For an integrable real random variable $X$, its cumulant-generating function is the extended-real convex functionwhere $\psi (\theta )=+\infty$ if the exponential moment diverges. Its Legendre transform of a cumulant-generating function is

For independent Poisson arrival counts of mean $\nu$ per slot, with one unit of work per arrival, the cumulant-generating function is $\nu (e^{\theta }-1)$. Its effective bandwidth is the displayed expression. If $c>\nu$, the equation $\nu (e^{\theta }-1)=c\theta$ has a unique positive root: strict convexity, negative initial derivative after subtracting $c\theta$, and divergence to infinity prove existence and uniqueness. Independent Poisson streams combine by replacing $\nu$ with the sum of their intensities.

For a Poisson random variable $S_L$ of mean $L\lambda >0$ and $0<\beta <1$, the scaled cumulant-generating function is $\lambda L^{1-\beta }(e^{\theta L^{(\beta -1)/2}}-1-\theta L^{(\beta -1)/2})$. Its Taylor expansion tends to $\lambda {\theta }^2/2$ for every real $\theta$. The Gärtner–Ellis theorem applies because the limit is finite and differentiable everywhere. Its Legendre-Fenchel transform is the displayed quadratic good rate function.

For sample means of Poisson random variables of mean $\lambda >0$, the cumulant-generating function is $\lambda (e^{\theta }-1)$. Its Legendre-Fenchel transform is the displayed good rate function, with $0\log 0=0$, and is infinite for $x<0$. Near its zero, $I_P(\lambda +h)=h^2/(2\lambda )+O(h^3)$. This local quadratic explains the Poisson moderate deviation principle.

For independent and identically distributed exponential random variables of rate $\lambda >0$, the sample mean has large-deviation speed $n$ and the displayed good rate function, infinite for $x\le 0$. The cumulant-generating function is $\log [\lambda /(\lambda -\theta )]$ for $\theta <\lambda$. Its Legendre-Fenchel transform has maximizing parameter $\theta =\lambda -1/x$. For $x\le 0$, sending $\theta \to -\infty$ gives infinite cost. The cost has its unique zero at $1/\lambda$ and diverges at both ends of its positive domain.