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Put $S_1={\sum }_i{Y}_i$, $S_2={\sum }_i{Y}_i^2$ and $R=\sqrt{S_2}$. Apart from a data-only constant, the log-likelihood isThe natural parameters in the ambient two-parameter exponential family of normal distributions are ${\eta }_1=1/\mu$ and ${\eta }_2=-1/(2{\mu }^2)$, restricted to the nonaffine curve ${\eta }_2=-{\eta }_1^2/2$, ${\eta }_1>0$. Its parametrization has rank one. Thus this is a one-dimensional curved exponential family inside a two-dimensional full family.

The Fisher-Neyman factorization theorem makes $(S_1,S_2)$ sufficient. For two samples $y,z$, the parameter-dependent part of their log density ratio is $\Delta S_1/\mu -\Delta S_2/(2{\mu }^2)$. A polynomial in $1/\mu$ is constant on $(0,\infty )$ only if both nonconstant coefficients vanish. The likelihood-ratio criterion for minimal sufficiency therefore proves $(S_1,S_2)$ is minimal sufficient.

Write $Y_i=\mu Z_i$ with independent $Z_i\sim N(1,1)$. Since $\mu >0$, $a=\sqrt{n}\sqrt{{\sum }_i{Z}_i^2}/{\sum }_i{Z}_i$ has a probability distribution independent of $\mu$: it is an ancillary statistic. The denominator is nonzero almost surely. The Cauchy-Schwarz inequality gives $|a|\ge 1$; on the requested positive branch, $S_1>0$ and $a\ge 1$.

The likelihood equation is $n{\mu }^2+S_1\mu -S_2=0$. Its positive root isThere is exactly one positive root. The sign of the score function changes from positive to negative there, and $\ell \to -\infty$ at both ends of $(0,\infty )$ when $S_2>0$, so it is the unique maximum-likelihood estimator. The useful identities areSubstitution yields, dropping the constant $-n/2$,Differentiating twice and using $q/a=q^2-1$ at the fitted value gives the observed informationThe scalar P-star approximation uses $\sqrt{j(v)}\exp \{\ell (\mu ;v,a)-\ell (v;v,a)\}$. Consequently, with its normalizing factor explicitly retained,Here $C_n(a)$ is chosen to make the density integrate to one, and is independent of $\mu$. To see this without assuming an approximate normalizer, defineThe integral is finite and positive, and normalized $p^{\ast }$ is exactlyIndeed $C_n(a)=\{\sqrt{n(1+q^2)/(2\pi )}\exp [n{q}^2/2-nq/a]J_n(a){\}}^{-1}$.

There is a direct check of this conditional scale density in a curved Gaussian family. In polar coordinates $y=Ru$, the sample density times volume element is proportional to ${\mu }^{-n}{R}^{n-1}\exp [-R^2/(2{\mu }^2)+R{\sum }_i{u}_i/\mu ]dRd\omega (u)$. Conditioning on $a$ fixes ${\sum }_i{u}_i=\sqrt{n}/a$. The remaining angular factor is independent of $R,\mu$, while $R=q\sqrt{n}v$. Thus the exact conditional radial density has the same normalized form. For $n=1$ the positive ancillary branch fixes the positive sign and the same one-dimensional radial calculation applies. In this scale model the normalized $p^{\ast }$ expression equals the exact conditional density, not merely its leading approximation.

For $t>0$, compute its conditional distribution by one-dimensional quadrature:It is zero for $t\le 0$; with the conditional-density calculation just given, this normalized formula is exact for almost every admissible $a$. Numerically, subtract the maximum log integrand before exponentiating to avoid overflow. The sampling probability still depends on the stipulated true $\mu$; replacing it by a fitted value would be a separate plug-in inference step.