Ordered-rate exponential M-step (${\lambda }_1=N/U,{\lambda }_2=N/V$)

For $N$ observations from each of two labeled exponential distributions, an E-step may produce positive completed time totals $U,V$. Its expected complete-data log-likelihood is $Q=N\log {\lambda }_1+N\log {\lambda }_2-U{\lambda }_1-V{\lambda }_2$. This is strictly concave. On the closed ordering constraint ${\lambda }_1\ge {\lambda }_2>0$, its maximum is $(N/U,N/V)$ when $U<V$; otherwise it lies at ${\lambda }_1={\lambda }_2=2N/(U+V)$. To see the boundary case, first maximize along the equality line. At that maximum, moving into either feasible rate-separating direction has directional derivative $(V-U)/2\le 0$, so concavity proves global optimality. For the open constraint ${\lambda }_1>{\lambda }_2$, this boundary value is a supremum rather than an attained M-step.

// Target: probability-and-statistics.bigb