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For the unperturbed Hamiltonian system, choose a conserved energy $H(x,y)$. Along the perturbed system,The energy balance method integrates this expression around each unperturbed periodic orbit. A zero of the resulting $O(\epsilon )$ energy change is a necessary leading-order condition for a nearby periodic orbit. A simple zero at which the drift changes from positive inside to negative outside predicts a stable orbit; the opposite change predicts an unstable one.

For a binary function class $\mathcal{F}$, the shattering coefficient isIts VC dimension is the largest $n$ for which $s(\mathcal{F},n)=2^n$, with value infinity if there is no largest such $n$. The Sauer-Shelah lemma states that, when $\mathrm{VC}(\mathcal{F})=d<\infty$,

At time $n$, the minimizing feedback from part (b) isStarting with $u_1^{\ast }=-X_0/(N+1)$ and repeatedly substituting
$X_j=X_{j-1}+u_j^{\ast }+{\xi }_j$ gives by inductionThereforeas required.

Set $V(n,x)=A_n+B_{n}x+C_n{x}^2$. The terminal values are $(A_N,B_N,C_N)=(0,0,1)$. Since the noise is centred, backward induction keeps $B_n=0$. Completing the square in the Bellman equation givesHence(with the empty sum equal to zero). This proves the asserted quadratic form of the value function.

Let $V(n,x)$ be the minimum expected remaining cost after time $n$, given $X_n=x$. The terminal condition is $V(N,x)=x^2$, and the Bellman equation is

Since $\rho ={\sigma }^2$ and the reparametrization is one-to-one, invariance of the maximum-likelihood estimator gives

The one-observation score function isIts variance, or equivalently minus the expected second derivative of the log-likelihood, is the Fisher information

For observations $X_1,\dots ,X_n$, the log-likelihood is, up to an additive constant,Its unique maximum isNow $\mathbb{E}[X_i^2]={\sigma }^2$ and, using $\mathbb{E}[Z^4]=3$, $\mathrm{Var}(X_i^2)=2{\sigma }^4$. The central limit theorem followed by the delta method for $g(x)=\sqrt{x}$ gives

Continuous differentiability gives the first-order expansionThe assumed convergence in distribution implies $\sqrt{n}({\hat{\theta }}_n-{\theta }_0)=O_p(1)$, so after multiplication by $\sqrt{n}$ the remainder is $o_p(1)$. The Slutsky theorem gives the multivariate delta method