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In the completed natural Brownian filtration, suppose $d{S}_t/S_t={\mu }_{t}dt+{\sigma }_{t}d{W}_t$ with ${\sigma }_t\ne 0$, and let $d{\zeta }_t=-{\zeta }_t(r_{t}dt+{\kappa }_{t}d{W}_t)$, ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. For an affordable nonnegative terminal claim $X$, represent the conditional-expectation martingale $N_t=\mathbb{E}[{\zeta }_{T}X\mid {\mathcal{F}}_t]$ as $d{N}_t=h_{t}d{W}_t$. Then $w_t=N_t/{\zeta }_t$ and the displayed dollar amount in the stock replicate $X$. The Itô product rule gives $d(\zeta w)=\zeta (\sigma \theta -\kappa w)dW$, proving the formula; share holdings are ${\theta }_t/S_t$.

For increasing sigma-algebras ${\mathcal{G}}_n$, put $f_n=I_{\mu }(\xi \mid {\mathcal{G}}_n)$ and $F={\sup }_n{f}_n$. For each atom $A\in \xi$ and $t\ge 0$,Together with the bound by $\mu (A)$ and the tail integral formula for moments, this yields $\int Fd\mu \le H_{\mu }(\xi )+1$. The inequality follows by stopping the conditional-expectation martingale $\mathbb{E}[1_A\mid {\mathcal{G}}_n]$ the first time it falls below $e^{-t}$.

Let $P_{s,t}>0$ denote the time-$s$ price of a zero-coupon bond paying one at maturity $t$, with $P_{t,t}=1$. Information available at $s$ is the sigma-algebra ${\mathcal{F}}_s$ in the market filtration. For a curve differentiable in maturity, the instantaneous forward rate and its diagonal short rate areThe last identity follows by integrating the maturity derivative and using $P_{s,s}=1$. The continuous-time bank account reinvesting at the short rate is $B_s=\exp ({\int }_0^s{R}_{u}du)$, and its reciprocal is the discount factor. Modelling forward rates means specifying a random maturity-indexed curve with an evolution consistent with the same pricing measure for every bond. The expectation below is under that money-market risk-neutral measure, not an unspecified physical measure.

For fixed maturity $t$, if $P_{s,t}/B_s$ is a true martingale on $0\le s\le t$, its terminal value is $1/B_t$. ConsequentlyConversely, this representation gives $P_{s,t}/B_s={\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_s]$, a conditional-expectation martingale by the tower property. For example, for $r<s$, conditioning the right side on ${\mathcal{F}}_r$ returns ${\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_r]$. These claims assume ${\mathbb{E}}_Q{B}_t^{-1}<\infty$. A positive discounted local martingale alone need not equal the conditional expectation of its terminal value, so the word “martingale” cannot be weakened without an extra integrability argument.

In the Vasicek model, the Brownian motion $W$ and coefficients in $d{R}_s=\alpha (\beta -R_s)ds+\sigma d{W}_s$ must be understood under this pricing measure. Multiplying by $e^{\alpha s}$ and integrating gives, conditionally on ${\mathcal{F}}_s$,Assume first $\alpha \ne 0$, and put $\tau =t-s$ and $B(\tau )=(1-e^{-\alpha \tau })/\alpha$. Integrating the explicit solution over $s\le u\le t$ yieldsThe interchange of the stochastic integrals is justified by their deterministic square-integrable kernels on the finite triangular domain; equivalently approximate the kernels by step functions, interchange finite sums, and pass to the limit using the Itô isometry. The last integral conditionally has a centered normal distribution independent of ${\mathcal{F}}_s$, with varianceA deterministic Brownian integral is Gaussian because its step approximations are sums of independent normal increments; their variances converge by the Itô isometry. Completing the square in the normal density proves $\mathbb{E}{e}^{-Y}=e^{-m+v/2}$ for a normal variable of mean $m$ and variance $v$. Hence the zero-coupon bond price is exponential-affine:An equivalent familiar form isAt $t=s$, both coefficients are zero and the bond value is one. These formulas are valid for negative as well as positive nonzero $\alpha$ on finite horizons, though a genuinely mean-reverting Vasicek model has $\alpha >0$. For $\alpha =0$, direct integration gives $B(\tau )=\tau$, $J(\tau )={\tau }^3/3$ andAll the exponential moments used above are finite, so the prices defined by the conditional expectation do give true discounted martingales.

To fit an initial bond curve, let its observed prices $p_0(t)$ be positive, satisfy $p_0(0)=1$, and have a sufficiently smooth instantaneous forward rate $f_0(t)=-d\log p_0(t)/dt$. Set $R_0=f_0(0)$; this compatibility is forced by the infinitesimal-maturity bond price. Replace $\beta$ by a deterministic function $\theta (t)$, with $\alpha \ne 0$, so thatIts mean $m(t)={\mathbb{E}}_Q{R}_t$ satisfies $m^{\prime }=\alpha (\theta -m)$, $m(0)=R_0$. The variance of the integrated rate remains ${\sigma }^{2}J(t)$, since changing deterministic drift does not change the centered Gaussian process. ThereforeThe required mean is consequently $m(t)=f_0(t)+{\sigma }^{2}B(t{)}^2/2$. Solve its mean equation for $\theta$:Combining the last two terms gives the explicit Hull-White model calibration in the question's parameter convention:With $R_0=f_0(0)$, the specified mean solves the same linear differential equation and initial condition as the model mean, so it is the actual mean. The resulting model forward curve equals $f_0$; integrating and using $P_{0,0}=p_0(0)=1$ proves $P_{0,t}=p_0(t)$ for every maturity. If the drift is instead written $\tilde{\theta }(t)-\alpha R_t$, its calibration parameter is $\tilde{\theta }=\alpha \theta$, not $\theta$ itself.

The printed allowance of arbitrary constants needs a qualification for this final claim. If $\alpha =0$, replacing $\beta$ by $\theta (t)$ changes no coefficient at all: $R_t=R_0+\sigma W_t$ regardless of $\theta$. Its initial curve is fixed at $P_{0,t}=\exp (-R_{0}t+{\sigma }^2{t}^3/6)$. With $\sigma \ne 0$, the compatible smooth observed curve $p_0(t)=e^{-R_{0}t}$ is already a counterexample to the unconditional fitting claim. Thus arbitrary compatible smooth positive initial curves can be fitted by the stated replacement when $\alpha \ne 0$; it cannot fit arbitrary curves when $\alpha =0$. In that degenerate case one can instead introduce an independent deterministic drift $k(t)$ in $d{R}_t=k(t)dt+\sigma d{W}_t$, choosing $k(t)=f_0^{\prime }(t)+{\sigma }^{2}t$ to recover the desired initial curve. Likewise an externally fixed $R_0\ne f_0(0)$ is incompatible with exact fitting, and an arbitrary nonsmooth set of prices need not define the differentiable forward curve assumed in the model.

Put $Z_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$. It is a conditional-expectation martingale. We justify why its limit is $X_{\infty }$, rather than just quoting convergence to an unspecified random variable.

First the family is uniformly integrable. For $A=\{Z_n>K\}\in {\mathcal{F}}_n$, nonnegativity and the conditional expectation identity giveThus its tail expectations are at most twice the integrable tail of $X_{\infty }$, uniformly in $n$. The uniformly integrable martingale convergence theorem gives a limit $Z$ almost surely and in $L^1$. For any $A\in {\mathcal{F}}_m$ and $n\ge m$,Pass to the $L^1$ limit and extend from $\mathcal{A}$ to all Borel sets by the Monotone class theorem. The equality of these integrals forces $Z=X_{\infty }$ almost surely. Combining this with the preceding convergence of $X_n$ proves

Put $X_t=g_t(x)-W_t$, $Y_t=g_t(y)-W_t$ and $D_t=Y_t-X_t$. Before $T(x)$, order preservation gives $0<X_t<Y_t$ and $D_t>0$. The common noise cancels from their difference:Since $D$ has finite variation, applying the Itô formula to $Z=Y/D$ gives the SLE two-boundary-point ratio diffusionWith the strictly increasing clock $u(t)={\int }_0^t{D}_s^{-2}ds$, the Dambis-Dubins-Schwarz theorem yields a Brownian motion $B$ for which the time-changed process obeysThe sign of the new Brownian motion has absorbed the minus sign above.

The domain Markov property of a chordal Loewner chain says that, conditionally on the past before $T(x)$, the mapped future is a fresh SLE. Its boundary marked points are $X_t,Y_t$. HenceLocalize, for example, where $D_t\ge 1/n$, $Z_t\in [1+1/n,n]$ and $t\le n$. On each such stopped interval this is a bounded conditional-expectation martingale, and therefore a genuine martingale. No smoothness assumption on $f$ is needed for this argument. The exit calculation below identifies it explicitly.

Let ${\xi }_0^{n-1}={\bigvee }_{j=0}^{n-1}{T}^{-j}\xi$ denote the join of measurable partitions recording the first $n$ observations, and set $H_n=H_{\mu }({\xi }_0^{n-1})$, with $H_0=0$. The entropy of a countable measurable partition is $H_{\mu }(\xi )=-{\sum }_{A\in \xi }\mu (A)\log \mu (A)$, using natural logarithms and $0\log 0=0$. Null atoms can be discarded. Throughout, $\mu$ is a probability measure.

The stronger property needed for monotonicity of normalized block entropy is that its increments decrease. DefineSince conditioning reduces entropy, $c_{k+1}\le c_k$. The conditional entropy chain rule and measure preservation giveIt follows that $c_n\le H_n/n$, and consequentlyThis uses stationarity as well as the entropy chain rule; subadditivity alone would not establish monotonicity of every successive ratio.

The entropy rate of a measurable partition and the Kolmogorov-Sinai entropy are, respectively,One may equivalently take the supremum over countable finite-entropy measurable partitions.

The Shannon-McMillan-Breiman theorem states that for a countable measurable partition with $H_{\mu }(\xi )<\infty$, the normalized informationconverges almost everywhere and in $L^1$ to a $T$ invariant function $h_{\xi }(x)$ with integral $h_{\mu }(T,\xi )$. Here ${\xi }_0^{n-1}(x)$ is the atom containing $x$. If $T$ is an ergodic transformation, then $h_{\xi }(x)=h_{\mu }(T,\xi )$ almost everywhere. We prove the general form, including an explicit formula for its limit.

Let ${\mathcal{F}}_k$ be the sigma-algebra generated by ${\bigvee }_{j=1}^k{T}^{-j}\xi$, let ${\mathcal{F}}_0$ be trivial, and put ${\mathcal{F}}_{\infty }={\bigvee }_{j=1}^{\infty }{T}^{-j}\sigma (\xi )$. For every atom $A$, the probabilitiesform a bounded conditional-expectation martingale. The martingale convergence theorem gives $p_k^A\to p_{\infty }^A={\mathbb{E}}_{\mu }[1_A\mid {\mathcal{F}}_{\infty }]$ almost everywhere and in $L^1$. Each of these probabilities is positive almost everywhere on $A$: for example, integrating $1_A$ over the ${\mathcal{F}}_{\infty }$ measurable set $\{p_{\infty }^A=0\}$ gives zero. Countability of $\xi$ lets us choose a common full-measure set for every atom.

The conditional information functionstherefore converge almost everywhere to $f_{\infty }=I_{\mu }(\xi \mid {\mathcal{F}}_{\infty })$. We need an integrable bound on this sequence; convergence of the probabilities alone would not supply one after taking logarithms. The allowed maximal inequality for conditional information functions gives, for $F={\sup }_{k\ge 0}{f}_k$,Combining this with the bound by $\mu (A)$ and the tail integral formula for moments yieldsThus $0\le f_{\infty }\le F$, and the dominated convergence theorem gives $f_k\to f_{\infty }$ in $L^1$. In particular,because $H_n/n$ is the average of the decreasing sequence $(c_k)$.

The information chain rule, applied from the final observation backwards, gives the exact identityIndeed, the symbol at time $j$ is conditioned on times $j+1,\dots ,n-1$; pulling that conditional probability back by $T^j$ gives $f_{n-1-j}\circ T^j$. Measure preservation is sufficient for this pullback identity.

We now prove the required triangular ergodic averaging lemma in this instance. Put $g_M={\sup }_{k\ge M}|f_k-f_{\infty }|$. Then $g_M\downarrow 0$ almost everywhere, $g_M\le 2F$, and $\Vert g_M{\Vert }_1\to 0$. Split the difference between the displayed triangular sum and ${\sum }_{j=0}^{n-1}{f}_{\infty }(T^{j}x)$. Terms whose index is at least $M$ are bounded by ${\sum }_{j=0}^{n-1}{g}_M(T^{j}x)$. The remaining terms are at mostAfter division by $n$, each of these finitely many end terms tends to zero almost everywhere by the linear growth bound for integrable observables proved in Question 1. Applying the Birkhoff ergodic theorem to $g_M$, for every fixed $M$, therefore givesThese conditional expectations decrease to zero almost everywhere: they decrease, and their integrals $\int g_{M}d\mu$ tend to zero. Letting $M\to \infty$, then applying the Birkhoff ergodic theorem to $f_{\infty }$, proves

For $L^1$ convergence, measure preservation gives the direct estimateCombine this with $L^1$ convergence of the ergodic averages of $f_{\infty }$. Finally, $\int h_{\xi }d\mu =\int f_{\infty }d\mu =h_{\mu }(T,\xi )$. In the ergodic case the invariant sigma-algebra is trivial, completing the familiar form of the theorem:

Put $M_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$. Each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, with $|M_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]\le 1$. Since a filtration is increasing, the tower property of conditional expectation givesThus $(M_n)$ is the conditional-expectation martingale associated with $Z$.

An $L^1$-bounded supermartingale has an almost sure convergence limit $X_{\infty }$ in $L^1$. The conditional-expectation martingale $M_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ has uniform integrability and converges to $X_{\infty }$ both almost surely and in $L^1$. To identify its limit, test against every event in every finite-time sigma-algebra and use uniqueness on their generated sigma-algebra. Then $Y=X-M$ is a supermartingale tending to zero almost surely. There is no general nonnegativity conclusion for $Y$, nor general convergence in L1 to zero: the negative of the fair-coin doubling martingale is a counterexample to both stronger assertions.