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At a simple saddle point, ${\varphi }^{\prime }(z_0)=0$ and ${\varphi }^{\prime \prime }(z_0)\ne 0$. On a method of steepest descent ray leaving the saddle, choose the oriented local coordinatewith the square-root branch determined by that ray. Then $\varphi (z)=\varphi (z_0)-s^2/2+O(s^3)$ and $dz=ids/\sqrt{{\varphi }^{\prime \prime }(z_0)}$ to leading order. For large positive $x$, the effective range is $s=O(x^{-1/2})$. The half-Gaussian integral givesAs usual this saddle contribution presumes the remaining contour is lower in real phase and integrable. Reversing the contour orientation reverses the sign; the square-root convention encodes the chosen outgoing ray.

For the second integral set $t=\sqrt{n}z$. Using the specified logarithm branch,The saddles are $z=\pm i$, and only $z_0=i$ is in the upper half-plane. HereThe contour can be deformed to the horizontal line through $i$, oriented from left to right, without crossing the pole at zero. Along $z=s+i$,which has a unique strict maximum at $s=0$ and decays at infinity. This explicitly verifies saddle dominance. Locally $\varphi (i+s)=1-i\pi -2s^2+O(s^3)$. Both sides of the saddle contribute a full Gaussian, soFor integer $n$, $e^{-i\pi n}=(-1{)}^n$, and henceThere is no extra factor of two from another saddle: the full Gaussian already accounts for the incoming and outgoing parts at the upper saddle.

The stage cost is $(x+u{)}^2+6u^2\ge 0$. The finite-horizon Bellman equation isInductively write $C_{s-1}(x)={\Pi }_{s-1}{x}^2$. Completing the square in $u$ gives the unique optimal first controlandFor $0\le q<2$, the recurrence map $F(q)$ satisfies $F(q)<2$ andThus ${\Pi }_s$ increases and is bounded above by two. Its limit solves $L=6(1+L)/(7+L)$, or $(L-2)(L+3)=0$; nonnegativity forces $L=2$. In particular $C_{\infty }(x_0)\ge 2x_0^2$, and indeed this recurrence already identifies the limit.

For the proposed policy, induction on the state equation gives $x_t=(2/3{)}^t{x}_0$ and $u_t=-x_t/3$. The stage cost is $(10/9)x_t^2$, so the infinite geometric series givesThis supplies an admissible policy attaining the lower bound:The last formula is the requested closed-loop control: it uses the observed current state rather than an initial-state schedule.

Given the complete stopped sample $(x_1,\dots ,x_n)$, the likelihood function isThe stopping factor is positive and independent of $\theta$. The Fisher-Neyman factorization theorem makes $(N,S)$ sufficient. For two possible stopped samples of lengths $n,m$ and sums $S,S^{\prime }$, their likelihood ratio is a parameter-free factor times ${\theta }^{n-m}{e}^{-\theta (S-S^{\prime })}$. This is independent of $\theta$ exactly when $n=m$ and $S=S^{\prime }$: differentiating its logarithm gives $(n-m)/\theta -(S-S^{\prime })=0$ for all $\theta >0$. By the likelihood-ratio criterion,The likelihood principle says proportional likelihoods give the same evidence about the parameter. Thus with the full sample observed, this parameter-free stopping rule adds no likelihood evidence beyond the realized values and length; it gives the same likelihood-based inference as a fixed-length sample with those values. This statement does not assert that frequentist sampling distributions are unchanged.

Let $q(\theta )={\int }_0^{\infty }\alpha (x)\theta e^{-\theta x}dx\in (0,1)$ be the chance of stopping at a given stage. Independence of successive stages givesTherefore $a(y)=\log \alpha (y)$ and $k(\theta )=\log {\int }_0^{\infty }{e}^{a(y)-\theta y}dy$ give the requested exponential family form.

Writing $p=p_Y$, differentiation in $y$ gives $p^{\prime }=(a^{\prime }-\theta )p$. Integrating and using $p(0)=p(\infty )=0$ shows ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$. LikewiseThe stipulated vanishing of $p^{\prime }$ at both boundaries givesThese integrations require the displayed moments and integrals to exist, the usual regularity interpretation of the question; when establishing a finite variance this assumption should be explicit.

Differentiating the normalization in $\theta$ gives $k^{\prime }(\theta )=-{\mathbb{E}}_{\theta }Y$ and $k^{\prime \prime }(\theta )={\mathrm{Var}}_{\theta }Y$. The score function is $-k^{\prime }(\theta )-Y$, so the Fisher information is $I(\theta )=k^{\prime \prime }(\theta )$. The Cramér-Rao bound states that a regular unbiased estimator of $\theta$ has variance at least $1/I(\theta )$. Apply it to $a^{\prime }(Y)$:Equivalently differentiate ${\mathbb{E}}_{\theta }{a}^{\prime }(Y)=\theta$ and use the Cauchy-Schwarz inequality to get $1\le \sqrt{\mathrm{Var}(a^{\prime }(Y))\mathrm{Var}(Y)}$, which gives the same conclusion.