 Solution

Put $A={\sum }_i{X}_i$, $B={\sum }_j{Y}_j$ and $N=n+m$. Independence gives the likelihood functionThe unrestricted maximum-likelihood estimators are $\hat{\theta }=A/n$ and $\hat{\mu }=B/m$. Under the null hypothesis, the common mean $\eta =\theta =\mu$ maximizes ${\eta }^{-N}{e}^{-(A+B)/\eta }$ at $\hat{\eta }=(A+B)/N$. These are maxima because each one-parameter score function has the sign of its observed sum minus its sample size times its proposed mean.

Define the generalized likelihood ratio to be the constrained maximum divided by the unrestricted maximum. The exponential factors cancel, givingSince $T=A/(A+B)$ and $1-T=B/(A+B)$,It has maximum one at $T=n/N$, when the two fitted means agree, and is small toward either endpoint. A likelihood-ratio test rejects for small $\Lambda$, or equivalently large $W=-2\log \Lambda$. This is the likelihood-ratio test for equality of exponential means.