= Ordered-rate exponential M-step
{title2=$\lambda_1=N/U,\quad\lambda_2=N/V$}
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 $\lambda_1\geq\lambda_2>0$, 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> $(V-U)/2\leq0$, 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
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