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For an independent statistical sample from an Inverse Gaussian distribution with mean $\mu >0$ and known shape $\lambda >0$, the maximum-likelihood estimator of $\mu$ is the sample mean. To prove this, expand the log-likelihood function to get $\ell (\mu )=C-\lambda {\sum }_i{y}_i/(2{\mu }^2)+n\lambda /\mu$. Its score function is $\lambda ({\sum }_i{y}_i-n\mu )/{\mu }^3$, positive below $\bar{y}$ and negative above it. Taking the negative expected second derivative gives one-observation Fisher information $I_1(\mu )=\lambda /{\mu }^3$. Thus the regular asymptotic normality of a maximum likelihood estimator yields $\sqrt{n}(\bar{Y}-\mu )\stackrel{d}{\to }N(0,{\mu }^3/\lambda )$. The same conclusion follows from the central limit theorem and the distribution's variance ${\mu }^3/\lambda$.

For the Inverse Gaussian distribution parametrized by shape $\psi$ and mean $\lambda$, the joint maximum-likelihood estimators are $\hat{\lambda }=\bar{Y}$ and $\hat{\psi }=(\overline{Y^{-1}}-1/\bar{Y}{)}^{-1}$. With known $\lambda$, the shape estimator is instead $(\overline{Y^{-1}}-2/\lambda +\bar{Y}/{\lambda }^2{)}^{-1}$. Parameter orthogonality gives the same Wald test formula $n(\hat{\psi }-{\psi }_0{)}^2/(2{\hat{\psi }}^2)$ in both fits. The two estimators differ by $O_p(n^{-1})$ under regular model sampling, so the test statistics agree to first order under the null; they need not be numerically equal in finite samples.

In the parametrization with density proportional to $y^{-3/2}\exp [-(\lambda /y+\varphi y)/2]$, the Inverse Gaussian distribution has cumulant-generating function $K(t)=\sqrt{\lambda \varphi }-\sqrt{\lambda (\varphi -2t)}$. For $n$ independent copies, the average has this same family with parameters $(n\varphi ,n\lambda )$, while the sum has parameters $(\varphi ,n^2\lambda )$. These assertions follow by multiplying moment-generating functions and scaling the density. The more usual mean parameter is $\mu =\sqrt{\lambda /\varphi }$, so the sum has mean $n\mu$ and shape $n^2\lambda$.

For independent Inverse Gaussian distributions with shape $\psi$ and mean $\lambda$, the constrained and full nuisance maximum-likelihood estimates both equal $\bar{Y}$, giving a unit sample-coordinate Jacobian determinant. The nuisance observed information there is $n\psi /{\bar{Y}}^3$. Hence subtracting half its logarithm from the profile log-likelihood replaces the coefficient $n/2$ of $\log \psi$ by $(n-1)/2$. For a nonconstant sample the modified fit is $(n-1)/({\sum }_i{Y}_i^{-1}-n/\bar{Y})$.

Partition the statistical parameter as $\theta =(\psi ,\lambda )$, with $\psi$ of interest. For each fixed $\psi$, maximize over the nuisance parameter to obtain ${\hat{\lambda }}_{\psi }$. The profile likelihood and profile log-likelihood areTheir maximizer is the interest component of the joint maximum-likelihood estimator. At a regular nuisance maximum, ${\ell }_p^{\prime }(\psi )={\ell }_{\psi }(\psi ,{\hat{\lambda }}_{\psi })$ and its negative Hessian matrix is the Schur complementThe profile likelihood ratio is unchanged by a one-to-one nuisance reparametrization, and transforms naturally under a one-to-one interest reparametrization. Under the regular conditions of Wilks theorem, $2\{{\ell }_p(\hat{\psi })-{\ell }_p({\psi }_0)\}$ has a limiting chi-squared distribution with dimension of $\psi$ degrees of freedom. Profiling is maximization, rather than integration or conditioning; $L_p$ need not be a normalized data density, and nuisance fitting can cause appreciable higher-order distortion.

The modified profile likelihood corrects for nuisance curvature and sample-coordinate distortion. In suitable coordinates $(\hat{\psi },\hat{\lambda },a)$, with $a$ an ancillary statistic, setholding $\hat{\psi },a$ fixed when differentiating the data coordinates. Up to an interest-independent factor, the Barndorff-Nielsen modified profile likelihood isEquivalently, implicit differentiation of the nuisance score equation gives $J_{\psi }=|j_{\lambda \lambda }{|}^{-1}|{\ell }_{\lambda ;\hat{\lambda }}|$, andThe semicolon denotes sample-coordinate differentiation, not a second parameter derivative. In orthogonal coordinates, the Cox-Reid adjusted profile likelihood uses the information adjustment alone; it agrees with this modification when $J_{\psi }=1$, as happens below.

For the Inverse Gaussian distribution, write $s={\sum }_i{y}_i$, $r={\sum }_i{y}_i^{-1}$ and $\bar{y}=s/n$. Terms independent of both parameters can be absorbed into $c(y)$, givingThe nuisance score function isIt is positive below $\bar{y}$ and negative above it. Thus its global maximum isindependent of $\psi$. Put $D=r-n/\bar{y}$. ThenThe nuisance observed information at this fit isSince the constrained nuisance estimator is literally the full nuisance estimator as a function of the data, $J_{\psi }=1$ in the specified sample coordinates. Hence the modified profile likelihood for inverse Gaussian shape isFor $n\ge 2$ and a nonconstant positive sample, the Cauchy-Schwarz inequality gives $D>0$. The corresponding fits are ${\hat{\psi }}_p=n/D$ and ${\hat{\psi }}_m=(n-1)/D$. A constant sample has $D=0$ and no finite profile maximum; under continuous independent sampling this event has probability zero when $n\ge 2$. In particular, the modified coefficient accounts for the fitted mean nuisance parameter.

Write $u(Y_i;\theta )={\partial }_{\theta }\log f(Y_i;\theta )$ and $U_n(\theta )={\sum }_{i}u(Y_i;\theta )$ for the individual and total score functions. Under the usual regularity assumptions, ${\mathbb{E}}_{{\theta }_0}u=0$ and ${\mathrm{Cov}}_{{\theta }_0}(u)=I({\theta }_0)$, the nonsingular per-observation Fisher information matrix. The central limit theorem and weak law of large numbers giveRearranging the stipulated score expansion, including its remainder, givesSlutsky theorem therefore proves the asymptotic normality of a maximum likelihood estimator:A consistent plug-in Fisher information matrix yields the Wald statistic $W_n=n({\hat{\theta }}_n-{\theta }_0{)}^{T}I({\hat{\theta }}_n)({\hat{\theta }}_n-{\theta }_0)$. Under the null hypothesis, $W_n\stackrel{d}{\to }{\chi }_d^2$, so a test of asymptotic size $\alpha$ rejects when $W_n$ exceeds the $(1-\alpha )$ quantile of that chi-squared distribution. A consistent observed Fisher information matrix gives the same limiting test.

With an $r$-dimensional parameter of interest $\psi$ and nuisance parameter $\lambda$, partition $I$ accordingly. The marginal asymptotic covariance matrix of $\hat{\psi }$ is the $\psi \psi$ block of $I^{-1}$, whose inverse is the efficient informationThus the nuisance-adjusted Wald statistic is $W_{\psi ,n}=n(\hat{\psi }-{\psi }_0{)}^T{I}_{\mathrm{eff}}(\hat{\theta })(\hat{\psi }-{\psi }_0)$, with limiting null distribution ${\chi }_r^2$. Using $I_{\psi \psi }$ alone is generally incorrect when the nuisance parameter is estimated; it is valid when the cross-information vanishes.

For the Inverse Gaussian distribution in this problem, expand the exponent to obtain the log likelihood, up to terms independent of both parameters,Its score functions areFor each positive $\psi$, the first score changes sign from positive to negative at $\lambda =\bar{Y}$, so this is the global maximizing mean parameter of an exponential family. Substituting it in the second score givesThe reciprocal function is strictly convex, so the denominator is positive for a nonconstant sample. The profile log likelihood is strictly concave in $\psi$ and tends to minus infinity at both endpoints, proving this is a maximum. Under continuous model sampling, a sample of size at least two is nonconstant almost surely. For a constant sample, including a one-observation sample, no finite joint maximum-likelihood estimator exists: the likelihood increases without bound as $\psi \to \infty$ at $\lambda =\bar{Y}$.

For one observation, differentiating again and using $\mathbb{E}Y=\lambda$ givesThis parameter orthogonality proves that estimating $\lambda$ does not reduce the efficient information for $\psi$. The Wald statistic has the same algebraic form in the two cases:Using information evaluated at ${\psi }_0$ instead is another first-order equivalent convention.

The printed claim of coincidence needs a first-order interpretation. Known and unknown $\lambda$ need not give numerically identical Wald statistics, because their maximum-likelihood estimators differ. With $\lambda$ known, direct maximization givesFor example, observations $1,2$ and known $\lambda =1$ give ${\hat{\psi }}_K=4$ and ${\hat{\psi }}_U=12$. At ${\psi }_0=1$, the displayed Wald convention gives $9/16$ and $121/144$, respectively. Thus exact equality is false. Under the null hypothesis, however, $\bar{Y}-\lambda =O_p(n^{-1/2})$, hence ${\hat{\psi }}_K-{\hat{\psi }}_U=O_p(n^{-1})$. Their centered, standardized estimates differ by $o_p(1)$, so their Wald statistics differ by $o_p(1)$. This is the valid asymptotic coincidence supplied by inverse Gaussian shape estimation and nuisance orthogonality.

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 $x>\mu$, the corresponding smaller root is ${\mu }^2/x$. The conditional probability of selecting the larger root isIt is the same weight as a function of the actual selected value $x$. HenceCombining the two open intervals proves the output has the Inverse Gaussian distribution $IG(\mu ,\lambda )$. The event $Y=0$ has probability zero, so no atom occurs at $\mu$; the limiting density there is finite. The algorithm always selects one of its positive roots, so these branch densities exhaust its probability distribution.

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, $-A(\eta )=\sqrt{-2\eta }=1/\mu$. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

Here the natural statistic is $T(X)=X$. The exponential-family derivative identities giveandThus

Expanding the exponent of the Inverse Gaussian distribution givesThus the exponential dispersion family representationhasandThe standard cumulant identities yieldHence the variance function is $V(\mu )={\mu }^3$, and the canonical link function is $g(\mu )=\theta (\mu )=-1/(2{\mu }^2)$.

Expanding the exponent, the log-likelihood function iswhere $C$ is independent of $\theta$. Its score function and second derivative areThe first is positive below $\bar{y}$ and negative above it, so the unique maximum-likelihood estimator isUsing $\mathbb{E}{\sum }_i{Y}_i=n\theta$, the one-observation Fisher information is $I(\theta )=-n^{-1}\mathbb{E}{\ell }^{\prime \prime }(\theta )={\theta }^{-3}$. The support does not depend on the interior parameter $\theta >0$, and the regular asymptotic normality of a maximum likelihood estimator givesThis is Inverse Gaussian mean estimation with known shape one. Equivalently, the Inverse Gaussian distribution has variance ${\theta }^3$, so the same result follows directly from the central limit theorem applied to its sample mean.