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Use the maximizing form of the Pontryagin maximum principle. For a cost-minimization problem there are a nonnegative multiplier ${\lambda }_0$ and a costate $\psi$, not both identically zero, with Hamiltonian $\mathcal{H}=\psi b-{\lambda }_{0}c$. The optimal control maximizes $\mathcal{H}$, the costate obeys $\dot{\psi }=-{\mathcal{H}}_x$, and free terminal time at the fixed terminal state zero gives $\mathcal{H}(\tau )={\lambda }_0{C}^{\prime }(\tau )$. There is no fixed terminal costate value since the terminal state is fixed. In the normal case scale ${\lambda }_0=1$.

For the energy problem $b=-u$, $c=up(t)$, $C=-R$, so $\psi$ is constant and the normal Hamiltonian is $u(p_{\ast }-p(t))$, where $p_{\ast }=-\psi$. Maximization gives full power for $p(t)<p_{\ast }$ and zero power for $p(t)>p_{\ast }$; either value, or a fractional value, is allowed on a price tie. On the final active arc, the free-time condition givesThis is the threshold energy scheduling with a completion reward rule. In this example the normal case is justified: an abnormal constant costate would require either no processing or processing at all times, and free-time transversality would then force that nonzero costate to vanish.

For $p_{\ast }=1/4$, processing starts at $t=1/2$. Energy $1/2$ is completed at $\tau =1$, before the price rises again. Its cost and reward areAny threshold carrying completion into day two has $\tau \ge 2$ and reward at most $1/3$. Since energy cost is nonnegative, $J\ge -1/3>-11/24$, so it is suboptimal.

For a global check, even any completion time $\tau \ge 5/4$ has reward at most $4/9<11/24$ and cannot beat the displayed candidate. Also $\tau \ge 1/2$ since power is bounded by one. At a fixed $\tau \in [1/2,5/4]$, the cheapest half-unit of processing consists of the last half-unit interval $[\tau -1/2,\tau ]$: every earlier point has a price at least as large as every selected point. Exchanging any earlier powered interval with a later unpowered cheaper one proves this rearrangement claim, also for fractional controls.

Write $h=3/2-\tau \in [1/4,1]$. This minimal cost minus reward iswithIt has its unique minimum at $h=1/2$, where $J^{\prime }=0$. The selected interval is then $[1/2,1]$ and its threshold is $h^2$. ConsequentlyThe terminal condition also checks: $1/4+R^{\prime }(1)=1/4-1/4=p(1)=0$.

A sufficient statistic $T$ has a conditional distribution of the data given $T$ that does not depend on the parameter. In a dominated family, the factorization criterion for sufficiency states that this is equivalent to $p_{\mu }(x)=h(x)g_{\mu }(T(x))$ for a nonnegative parameter-free $h$. Bayes' formula then gives a posterior distribution proportional to $\pi (\mu )g_{\mu }(T(x))$, because $h(x)$ cancels from numerator and normalizing denominator. Thus the posterior depends on the data only through $T$.

For the normal sample,The first term is parameter-free, so the factorization proves sufficiency of $\bar{X}$. Combining its likelihood with the normal prior and completing the square givesThe prior predictive distribution of the sample mean is obtained from $\bar{X}=M+\epsilon$, $\epsilon \sim N(0,1/n)$ independent of $M$:

Define the Bayes factor $B_{01}$ as the marginal likelihood under $H_0$ divided by that under $H_1$. The common residual factor cancels, so it is the ratio of the densities of $\bar{X}$ under these two hypotheses:The reverse Bayes factor is its reciprocal. At $|\bar{x}|=1.96/\sqrt{n}$,for a fixed positive $\tau$. Thus the frequentist test is at its rejection threshold while fixed prior odds give increasingly strong posterior support for the point null. This is the Jeffreys-Lindley paradox for a Gaussian point null: the standardized statistic stays fixed, but the observed effect tends to zero and the fixed-scale alternative spreads predictive mass much more widely. There is no logical contradiction, since the procedures compare hypotheses using different calibrations. For the degenerate $\tau =0$ the hypotheses coincide and $B_{01}=1$.