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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.