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Let $X_i=\theta +E_i$ for $n\ge 2$ independent and identically distributed observations, with $E_i$ having an exponential distribution of unknown scale $\beta >0$. Test the null hypothesis $\theta =0$ within $\theta \ge 0$. For nondegenerate data, the likelihood function isMaximizing over the scale gives $\hat{\beta }(\theta )=\overline{X}-\theta$, and the resulting profile likelihood increases with $\theta$. Thus the unrestricted maximum-likelihood estimates are $\hat{\theta }=X_{(1)}$ and $\hat{\beta }=\overline{X}-X_{(1)}$, while the null estimate is $\tilde{\beta }=\overline{X}$. The likelihood ratio is consequently

Under the null, put $A=n{X}_{(1)}$ and $D={\sum }_i(X_i-X_{(1)})$. The memoryless property gives independent gamma distributions of shapes $1$ and $n-1$ and common scale $\beta$. For a direct verification, fix which observation is the minimum and write its value as $z$ and the other observations as $z+e_j$. Their joint density is proportional to $e^{-nz/\beta }{\prod }_j{e}^{-e_j/\beta }$, which factors into an exponential density for the minimum and independent exponential densities for the excesses. The minimum's index does not affect the resulting laws of $A,D$.

By beta sampling by gamma ratios, $D/(A+D)\sim \mathrm{Beta}(n-1,1)$. ThereforeAn exact test size $\alpha$ is obtained by rejecting for $R\le {\alpha }^{n/(n-1)}$. Equivalently,Although the alternative adds one endpoint parameter, a regular one-parameter Wilks theorem calibration is incorrect: the support depends on the endpoint, and the null lies on its boundary. In particular, $-2\log R$ tends to a chi-squared distribution with two statistical degrees of freedom, not one.

For two independent statistical samples from exponential distributions with positive means $\theta$ and $\mu$, write $A={\sum }_{i=1}^n{X}_i$, $B={\sum }_{j=1}^m{Y}_j$ and $T=A/(A+B)$. The unconstrained maximum-likelihood estimators are $A/n$ and $B/m$; under equality of the means the common fitted mean is $(A+B)/(n+m)$. Dividing the constrained maximum likelihood function by its unrestricted maximum gives the displayed likelihood ratio. Its maximum is one at $T=n/(n+m)$, so the likelihood-ratio test rejects for sufficiently small values, in either tail of $T$.

Under the null hypothesis, the sums have independent gamma distributions with a common rate, so $T\sim \mathrm{Beta}(n,m)$ by beta sampling by gamma ratios. This permits exact calibration. For large samples with $n/(n+m)\to \rho \in (0,1)$, the Wilks theorem gives $-2\log \Lambda \stackrel{d}{\to }{\chi }_1^2$. Its upper-tail probability is the asymptotic p-value. The one statistical degree of freedom is the difference between the unrestricted and constrained parameter dimensions.

For Beta sampling by gamma ratios, use either preceding chi-squared distribution algorithm to generate independent $S\sim {\chi }_{2\alpha }^2$ and $T\sim {\chi }_{2\beta }^2$, and returnWrite $W=S+T$, so $S=WR$ and $T=W(1-R)$. The Jacobian determinant is $W$. The joint probability density function after this change of variables factors asIntegrating out $w$ proves the Beta distribution for $R$.

A new method is Beta sampling by uniform order statistics. Generate $m=\alpha +\beta -1$ independent uniforms, sort them, and return the $\alpha$th order statistic. There are $\alpha -1$ uniforms below the selected value and $\beta -1$ above it. Selecting their labels and placing the selected observation in $dr$ gives densitywhich is the Beta distribution density. The integrality of the shape parameters is what makes this order-statistic construction possible.

Write $a=x_{(1)}={\min }_i{x}_i$, $T={\sum }_i{x}_i$ and $\overline{x}=T/n$. For nondegenerate data from the shifted exponential distribution, the likelihood function isFor fixed feasible $\theta$, differentiating its log-likelihood with respect to $\beta$ givesThe derivative changes from positive to negative there. Substituting gives the profile likelihood $e^{-n}(\overline{x}-\theta {)}^{-n}$, which increases with $\theta$. Therefore the unrestricted maximum-likelihood estimates arewhile under the null hypothesis $\theta =0$ the estimate is $\tilde{\beta }=\overline{x}$. Their likelihood ratio isSince the likelihood-ratio test rejects for small $R$, the requested form is obtained withThe formulas assume $\overline{x}>a$, which holds almost surely for a continuous sample with $n\ge 2$; coincident observations give an unbounded unrestricted likelihood and can be treated by its supremum.

An exact calibration is available and also explains why Wilks theorem should not be applied here. Under the null, the memoryless property makes $A=n{X}_{(1)}$ and $D={\sum }_i(X_i-X_{(1)})$ independent gamma distributions with common scale $\beta$ and shapes $1$ and $n-1$. By beta sampling by gamma ratios,Consequently, for $0\le k\le 1$,This yields an exact test size $\alpha$, independent of the unknown scale. The support endpoint and boundary null make this a likelihood-ratio test for an exponential support endpoint, with a different calibration from a regular one-parameter test.