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For independent unit-variance Gaussian samples with means ${\mu }_X,{\mu }_Y$, testing ${\mu }_X\ge A,{\mu }_Y=0$ against unrestricted means gives $T=-2\log \Lambda =n[{\bar{Y}}^2+(A-\bar{X}{)}_{+}^2]$. The unrestricted maximum-likelihood estimators are the two sample means, while the null estimators are $\max (A,\bar{X})$ and zero. Under the null, write $W=\sqrt{n}(\bar{X}-A)=W_0+\delta$, $\delta =\sqrt{n}({\mu }_X-A)\ge 0$, and $Z=\sqrt{n}\bar{Y}$, with $W_0,Z$ independent standard Gaussian variables. Then $T=Z^2+(-W_0-\delta {)}_{+}^2$ decreases pointwise with $\delta$, so the largest rejection probability occurs at the boundary ${\mu }_X=A$. There its law is the chi-bar-squared distribution $\frac{1}{2}{\chi }_1^2+\frac{1}{2}{\chi }_2^2$. The exact size-$\alpha$ threshold $c_{\alpha }$ is therefore determined by $1-\Phi (\sqrt{c_{\alpha }})+\frac{1}{2}{e}^{-c_{\alpha }/2}=\alpha$.

The unrestricted maximum-likelihood estimators remain $(\bar{X},\bar{Y})$. Subject to the null constraint, minimizing $({\mu }_X-\bar{X}{)}^2$ on $[A,\infty )$ gives ${\hat{\mu }}_X=\max (A,\bar{X})$, while ${\hat{\mu }}_Y=0$. ConsequentlyWe must calibrate this likelihood-ratio test over the entire composite null. Under that null let $W=\sqrt{n}(\bar{X}-A)$ and $Z=\sqrt{n}\bar{Y}$. Then $W\sim N(\delta ,1)$ with $\delta =\sqrt{n}({\mu }_X-A)\ge 0$, while $Z\sim N(0,1)$ independently, and $T=Z^2+(-W{)}_{+}^2$. Couple all null distributions by $W=W_0+\delta$, with a fixed standard Gaussian $W_0$. Increasing $\delta$ decreases $(-W_0-\delta {)}_{+}^2$ pointwise, so the largest rejection probability occurs at $\delta =0$, namely ${\mu }_X=A$.

At this boundary the sign of $W$ is independent of $W^2$, with each sign having probability one half. On $W\ge 0$, $T=Z^2\sim {\chi }_1^2$; on $W<0$, $T=Z^2+W^2\sim {\chi }_2^2$. Thus the least favourable null hypothesis law is the exact chi-bar-squared distribution $\frac{1}{2}{\chi }_1^2+\frac{1}{2}{\chi }_2^2$. Choose $c_{\alpha }>0$ byThe tail is a continuous function and strictly decreasing from one to zero, so this threshold is unique for $0<\alpha <1$. The null rejection probability is at most $\alpha$ and equals $\alpha$ at ${\mu }_X=A$. This proves the required size, rather than using an interior chi-squared approximation.

To compare, put $c_1=z_{1-\alpha /2}^2$, the threshold in (i). Since ${\chi }_2^2$ is stochastically larger than ${\chi }_1^2$ (add an independent squared standard Gaussian), the mixture threshold obeys $c_{\alpha }>c_1$. When $\bar{X}\ge A$, test (ii) therefore requires stronger evidence from $\bar{Y}$ than test (i). When $\bar{X}<A$, test (ii) also responds to violation of the mean constraint and can reject even with $\bar{Y}=0$. Neither critical region contains the other. In the coordinates $(W,Z)$, the acceptance region in (i) is the strip $|Z|\le \sqrt{c_1}$; in (ii) it is the right half-strip $W\ge 0$, $|Z|\le \sqrt{c_{\alpha }}$ together with the left half-disk $W<0$, $W^2+Z^2\le c_{\alpha }$. The extra constraint changes both the statistic and its correct null calibration.