Likelihood-ratio test for equality of exponential means ($\Lambda =\frac{(n+m{)}^{n+m}}{n^n{m}^m}{T}^n(1-T{)}^m$)

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.