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Work with the filtration relative to which the given Brownian motion or Markov jump process has its stated properties. All martingale assertions below concern every finite time interval; no infinite-horizon uniform integrability is claimed.

First put $C={\sup }_x|\sigma (x)|$. The solution has the representation $X_t={\int }_0^t\sigma (X_s)d{B}_s$, and the Itô isometry givesThus $X$ is a square-integrable martingale, not just a local martingale. Applying the Itô formula to its square cancels the finite-variation term and givesThe expected squared integrand integral is bounded byTherefore both $X$ and $Z$ are true square-integrable martingales. This is the bounded diffusion coefficient gives square martingales criterion. Only bounded measurability of the coefficient and existence of the stipulated solution have been used.

For the jump process, the Lévy kernel of a Markov jump process uses displacements: a mark $y$ sends the state $x$ to $x+y$. Let $\Lambda ={\sup }_x\lambda (x)$ and $V={\sup }_{x}v(x)$. The bounded total rate prevents explosion; the number of jumps on a finite interval is dominated by a rate-$\Lambda$ Poisson process. Let $N(ds,dy)$ record the jump displacements. Its predictable compensator of a jump measure isThe Cauchy-Schwarz inequality gives $\int |y|K(x,dy)\le \sqrt{\lambda (x)v(x)}$, so the zero first moment is absolutely meaningful. Since that moment vanishes,The compensated integral is a square-integrable martingale: its isometry follows first for bounded marks from conditional compensation and then by truncation in the squared-integrand norm.

The jump version of the square Itô formula is $X_t^2=2{\int }_0^t{X}_{s-}d{X}_s+{\sum }_{s\le t}(\Delta X_s{)}^2$. HenceThe first term is square integrable becauseThe second is an integrable martingale: the expected absolute uncompensated jump-square sum is at most $Vt$, and the compensator identity, multiplied by any bounded event known at an earlier time, makes its increments conditionally centered. Values $X_s$ and $X_{s-}$ agree for Lebesgue-almost every time, so the displayed compensation is exactly the one defining $Z$. Thus $X$ and $Z$ are true martingales in the jump case as well. A fourth moment of the jumps is unnecessary; in particular we have not asserted that this $Z$ must be square integrable. This is centered finite-rate jump kernel gives square martingales.

A simple predictable process can be writtenwhere each ${\xi }_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. Define its integral to start at zero:Refining the partition leaves this sum unchanged. Every summand is a martingale, by conditioning its future increment on the information at $t_j$. Bounded coefficients and the $L^2$ bound on $M$ make the integral an L2-bounded continuous martingale. Its terminal value is the same sum with $M_{\infty }$ substituted at any infinite endpoint; $L^2$ martingale convergence supplies this limit.

We need the conditional square identity, rather than assuming independent increments. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_t|M_t{|}^2<\infty$. Localize the square identity for $M^2-[M]$ at level and bracket stopping times. The stopped identity and this maximal bound give $\mathbb{E}[M{]}_{\infty }<\infty$ by monotone convergence. Nowis integrable, so localization passes to conditional expectations and $M^2-[M]$ is a uniformly integrable martingale. Expanding a squared increment and using $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ provesThe infinite endpoint follows from $L^2$ convergence and integrability of the bracket. The initial value $M_0$ need not be zero: the integral itself is zero-starting.

In the square of the terminal integral, all cross terms vanish. For $j<k$, the $j$th increment and both coefficients in its cross term are measurable at $t_k$, whereas the conditional mean of the $k$th increment there is zero. For the diagonal terms, the preceding conditional identity can be multiplied by the bounded ${\xi }_j^2$. ThereforeThis proves the Itô isometry for these simple integrands. Ordered stopping-time intervals also work, replacing the conditional identity by optional stopping theorem, but deterministic intervals suffice for the approximation requested here.

For that approximation, put $\mu (A)=\mathbb{E}{\int }_0^1{1}_A(\omega ,s)ds$ on the predictable sigma-algebra. This is a finite measure. That sigma-algebra is generated, apart from the zero-time slice of measure zero, by rectangles $A\times (s,t]$ with $A\in {\mathcal{F}}_s$. The indicator of every such rectangle is a bounded simple predictable process.

Here is a density proof. Let $V$ be the closed linear span of these indicators in $L^2(\mu )$, and let $\mathcal{D}$ contain the predictable sets whose indicators belong to $V$. The whole space belongs to $\mathcal{D}$; complements stay in $\mathcal{D}$; and disjoint countable unions do also, because their finite partial indicator sums converge in $L^2$ under the finite measure. Thus $\mathcal{D}$ is a Dynkin system containing the generating rectangle pi-system. The Pi-lambda theorem makes it the entire predictable sigma-algebra. Bounded predictable simple functions therefore belong to $V$. Truncating any $L^2(\mu )$ function and then approximating its truncated range by finitely many levels shows that every such function belongs to $V$. A finite sum of rectangle indicators can be rewritten on one common deterministic time partition, with the coefficient on each interval measurable at its left endpoint; it is exactly an allowed simple process.

Consequently choose $H^n$ of this form with $\Vert H^n-H{\Vert }_{L^2(\mu )}<1/n$. This provesIt is density of simple predictable processes for finite measures, with a proof of the actual approximation step.

Finally apply the preceding isometry to $M_t=B_{t\wedge 1}$, whose bracket is $t\wedge 1$. The elementary integrals $I_n={\int }_0^1{H}_s^{n}d{B}_s$ satisfyDefine the Itô integral as the $L^2$ limit of these elementary integrals. The same identity shows independence of the approximating sequence, linearity, mean zero, and $\mathbb{E}({\int }_0^1{H}_{s}d{B}_s{)}^2=\mathbb{E}{\int }_0^1{H}_s^{2}ds$. Applying the Doob inequality to the difference processes also gives uniform-in-time $L^2$ convergence on $[0,1]$, with a continuous martingale version of the integral.

Subtract $M_0$, which does not affect either the integral or its quadratic variation. Let $N=M-M_0$. Choose a localizing sequence ${\tau }_n$ for $N$. By local boundedness of the previsible process $H$, there are increasing stopping times ${\rho }_n\uparrow \infty$ and finite deterministic constants $K_n$ with $|H_s|\le K_n$ on $0<s\le {\rho }_n$. PutContinuity and finiteness on compact intervals imply ${\sigma }_n\uparrow \infty$ almost surely. The stopped process $N^{{\sigma }_n}$ is a bounded square-integrable martingale, its bracket is bounded by $n$, and its stopped integrand is bounded by $K_n$. The indicator $1_{\{s\le {\sigma }_n\}}$ is predictable, so the stopped stochastic integral identity is valid.

Fix $n$ and abbreviateThe Itô isometry gives $\mathbb{E}{J}_t^2=\mathbb{E}{Q}_t\le K_n^{2}n$, so $J$ is a square-integrable martingale. For $F\in {\mathcal{F}}_s$, apply the same isometry to the predictable integrand $1_F{1}_{(s,t]}H$, stopped at ${\sigma }_n$. This yieldsAlso $\mathbb{E}[J_t-J_s\mid {\mathcal{F}}_s]=0$. Expanding $J_t^2-J_s^2$ therefore proves that $J^2-Q$ is a true martingale. The process $Q$ is continuous adapted increasing and starts at zero. By the characterization of continuous martingale quadratic variation as its increasing square compensator, and uniqueness of an increasing square compensator, we have $[J]=Q$.

Stochastic integration and quadratic variation commute with stopping; for the latter this follows either from squared-increment sums or from the same square-compensator characterization. Hence the desired identity holds up to every ${\sigma }_n$. Since these times increase to infinity, they exhaust every finite time interval almost surely, and the continuous identities patch to giveThis is the localized isometry proof of stochastic-integral quadratic variation. No unstopped second-moment or bracket-integrability assumption has been made.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

Define the stochastic integralIt is adapted to ${\mathcal{G}}_t$, starts at zero and has continuous paths. Integrals over disjoint clock intervals are centered Gaussian and independent, and an increment from $s$ to $t$ is independent of ${\mathcal{G}}_s$. The Itô isometry gives its varianceThese are precisely the defining properties of a ${\mathcal{G}}_t$-Brownian motion; in particular its future increments are independent of ${\mathcal{G}}_0={\mathcal{F}}_1$.

Deterministic substitution in the stochastic integral, first for step integrands and then by the Itô isometry, gives $d{X}_s=\sqrt{2\lambda }{e}^{\lambda s}d{W}_s$. HenceThe initial term is essential: a stochastic integral from time zero alone cannot equal $X$, since $X_0=B_1$. This is the unscaled part of the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process.

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.

For times $t_1,\dots ,t_n<1$ and real coefficients $a_i$, the linear combination of coordinates isThe Gaussianity of deterministic Brownian stochastic integrals makes this a centered normal random variable. Indeed step-function approximations give linear combinations of independent normal increments, and the Itô isometry gives convergence in $L^2$; their characteristic functions converge to that of a normal variable with the limiting variance. Since every linear combination is normal, the coordinate vector is a Gaussian random vector. Appending the deterministic coordinate $X_1=0$ preserves that property, so $X$ is a centered Gaussian process on $[0,1]$.

For $0\le s\le t<1$, the covariance version of the Itô isometry givesAt $t=1$ the covariance is zero because $X_1=0$, agreeing with the formula. Thus the Brownian bridge covariance kernel isIts finite-dimensional distributions identify it as a Brownian bridge; continuity at the terminal endpoint is established next.

The polarization identity for quadratic covariation, together with $\mathbb{E}[M\pm N{]}_s=\mathbb{E}(M_s\pm N_s{)}^2$, yieldsThe bracket identity for each zero-starting square-integrable martingale is the Itô isometry for integration of the constant integrand against that martingale. Polarizing both identities gives the displayed covariance identity for continuous square-integrable martingales. The bracket is integrable: its total variation is at most $\sqrt{[M{]}_s[N{]}_s}$ by the Kunita-Watanabe inequality, whose expectation is finite by the Cauchy-Schwarz inequality. Therefore condition (iii) is equivalent to (ii).

For the integrable terminal bracket criterion, assume $M_0=0$, or more generally $M_0\in L^2$. This initial-value assumption is necessary if it is not already part of the notation: an integrable but non-square-integrable random constant process has zero quadratic variation and still fails square-integrability.

For the zero-starting case, choose a localizing sequence ${\tau }_n$ bounding both $|M|$ and its quadratic variation. The stopped processes are true square-integrable martingales, and the Itô isometry givesFor each fixed time these variables are uniformly integrable by their bounded second moments. Almost sure convergence as $n\to \infty$ is therefore convergence in $L^1$. Passing to the limit in the stopped conditional-expectation identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$: the local martingale is a true martingale. The Fatou lemma also givesso $M$ belongs to the global ${\mathcal{M}}_c^2$ space. For general square-integrable $M_0$, apply this argument to $M-M_0$ and use $|M_t{|}^2\le 2|M_0{|}^2+2|M_t-M_0{|}^2$.

For $t>1$, $B_{t^2}$ is not measurable with respect to the original Brownian filtration ${\mathcal{F}}_t$: its conditional variance is $t^2-t>0$. Thus it is not even adapted on the full half-line. It also fails the martingale condition before time one. With $s=1/4$, $t=3/4$,The appropriate filtration isBrownian independent increments make $M_t=B_{t^2}$ a continuous ${\mathcal{G}}_t$-martingale, with quadratic variation $[M{]}_t=t^2$. Define the stochastic integralThe value of the integrand at zero is immaterial; it is square-integrable against the bracket because ${\int }_0^t(2s{)}^{-1}d(s^2)=t$. Consequently $[W{]}_t=t$, and the Lévy characterization of Brownian motion makes $W$ a ${\mathcal{G}}_t$-Brownian motion. Associativity of stochastic integrals gives the Brownian motion under a quadratic time change representationThe function $f$ is continuous at zero. One can first verify the inverse-integrand identity on $[\epsilon ,t]$ and then let $\epsilon \downarrow 0$ using the Itô isometry, since $\mathbb{E}{M}_{\epsilon }^2={\epsilon }^2$.

For a discrete-time martingale relative to the filtration $({\mathcal{F}}_n)$, each $M_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|M_n|<\infty$, and $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$ for $n\ge 1$. The tower property then gives the corresponding identity conditional on any earlier time.

A bounded previsible process has ${\varphi }_i$ measurable at time $i-1$ and $|{\varphi }_i|\le C$. Its martingale transform $V$ is adapted and integrable, since each increment has absolute expectation at most $C(\mathbb{E}|M_i|+\mathbb{E}|M_{i-1}|)$. Moreover,Thus $V$ is a martingale, with $V_0=0$.

For the Itô integral, begin with a bounded adapted step integrand $H_s=H_i$ on $(t_i,t_{i+1}]$. Its integral is $I={\sum }_i{H}_i(W_{t_{i+1}}-W_{t_i})$. Each summand has zero conditional mean. For $i<j$, conditioning the cross product on ${\mathcal{F}}_{t_j}$ makes its expectation zero. The conditional second moment of a Brownian increment is $t_{i+1}-t_i$, henceApproximate the predictable integrand $g(s,W_s)$ in $L^2(dsd\mathbb{P})$ by such step processes. The same identity makes their integrals Cauchy in $L^2$ and defines the limit; both its first and second moments converge. This gives the requested mean-zero result and the Itô isometry:

For the Ornstein-Uhlenbeck process, the deterministic integrating factor has no quadratic covariation with $X$. The Itô formula therefore gives $d(e^{\kappa t}{X}_t)=\kappa \theta e^{\kappa t}dt+\sigma e^{\kappa t}d{W}_t$. Integrating and dividing by $e^{\kappa t}$ yieldsThe mean-zero integral and Itô isometry giveAt $\kappa =0$ these formulas are read by continuity: $X_t=x+\sigma W_t$, with mean $x$ and variance ${\sigma }^{2}t$.

For the CIR model, let $C_0=c_0\ge 0$ and use the usual positive parameters. Before a possible visit to zero, the derivatives of $\sqrt{c}$ are $1/(2\sqrt{c})$ and $-1/(4c^{3/2})$. Applying the Itô formula gives the Square root of a CIR diffusion:The PDF's claim under $2ab={\sigma }^2$ is false: that choice leaves the nonzero drift ${\sigma }^2/(8Z_t)-a{Z}_t/2$, which is not affine in $Z_t$. The actual interior cancellation condition is $4ab={\sigma }^2$. Even then the principal square root is nonnegative and cannot be an unrestricted Gaussian OU process.

To give the intended corrected construction, let a signed process solve $d{Y}_t=-a{Y}_{t}dt/2+(\sigma /2)d{B}_t$, with $Y_0=\sqrt{c_0}$. The Itô formula givesWriting ${\tilde{W}}_t={\int }_0^t\mathrm{sgn}(Y_s)d{B}_s$, with either sign chosen at zero, produces a Brownian motion; the diffusion term is $\sigma |Y_t|d{\tilde{W}}_t$. Thus $Y^2$ is a CIR model with $ab={\sigma }^2/4$, while $Z=|Y|$ is reflected at zero. The reflection contributes local time of a semimartingale, so the interior SDE alone is not a global unrestricted OU equation. Its second moment is $c_0{e}^{-at}+{\sigma }^2(1-e^{-at})/(4a)$, agreeing with the CIR mean for that corrected parameter choice.

The requested mean under the parameters actually printed is still available without the false OU assertion. Taking expectation in the original CIR equation gives $m^{\prime }=a(b-m)$; the stochastic integral has zero mean after the usual moment/localization justification. Consequently, for the printed choice as well as general admissible parameters,

An interest rate model must describe a whole term structure of interest rates, not just one future scalar rate. Let $P(t,T)$ be the time-$t$ price of a unit zero-coupon bond maturing at $T$. For a sufficiently regular maturity curve, define the instantaneous forward rate and short rate bySince $P(t,t)=1$, integration givesHere $B$ is the continuous-time bank account. Thus modelling the two-parameter field $f(t,T)$ determines both the bonds and the discounting account. A Gaussian random field means that every finite collection of its values is jointly Gaussian; merely making each individual forward rate normally distributed is insufficient.

A useful arbitrage-free construction works under a money-market risk-neutral measure $Q$. Take independent Brownian motions $W^j$, deterministic coefficient sequences ${\sigma }_j(t,T)$, and a deterministic initial forward curve $f(0,T)$. With the coefficients viewed as vectors in a separable Hilbert space $H$, writeFor infinitely many factors, the last term means the $L^2$ limit of ${\sum }_j{\sigma }_j(t,T)d{W}_t^j$, not an assumption that the infinite vector of Brownian motions is itself $H$-valued. Require ${\int }_0^t\Vert \sigma (s,T){\Vert }_H^{2}ds<\infty$ and the maturity/time regularity needed for stochastic integration and exchange of integrals. Deterministic drift and volatility make this a Gaussian forward-rate field.

The field covariance is determined byFor every finite set of maturities, $(c_t(T_i,T_j))$ must be positive semidefinite. A one-factor model has rank-one instantaneous covariance, while additional or infinitely many factors permit richer correlations along the yield curve. Equivalently one can drive the curve with a maturity-correlated Brownian field $Z_t(T)$ satisfying $d⟨Z(T),Z(U){⟩}_t=c(T,U)dt$. A Brownian sheet gives the exampleMultiplication by deterministic maturity-dependent amplitudes, or deterministic increasing changes of the two sheet coordinates, gives further Gaussian covariance structures. A covariance specification alone is not sufficient: the forward drift must also obey absence of arbitrage.

To derive that drift restriction, putDifferentiate the moving lower endpoint in $\log P(t,T)=-{\int }_t^{T}f(t,u)du$ to obtainThe Itô formula therefore givesDiscounted bond prices must be martingales under $Q$. Their drift vanishes precisely when ${\int }_t^T\alpha (t,u)du=\frac{1}{2}\Vert \Sigma (t,T){\Vert }_H^2$. Differentiating in maturity gives the Heath-Jarrow-Morton model restrictionFor scalar amplitudes $v(t,T)$ multiplying a field with covariance kernel $c(T,U)$, this becomes $\alpha (t,T)=v(t,T){\int }_t^{T}v(t,u)c(T,u)du$. The extra maturity correlation factor cannot in general be dropped.

The complete Gaussian specification is consequentlyIts mean and covariance follow immediately from deterministic integration and the Itô isometry. The initial yield curve is an input, so fitting initial bond prices requires choosing $f(0,T)$ rather than changing a few scalar parameters. With the displayed drift, the discounted bond obeysOn finite horizons, deterministic square-integrable bond volatility gives an expectation-one stochastic exponential, hence a true discounted bond price martingale. This supplies an arbitrage-free pricing model for admissible portfolios in any finite collection of these bonds. Under a physical measure, a market price of risk vector changes the drift: if $d{W}_t^Q=d{W}_t^P+{\lambda }_{t}dt$, then the physical forward drift is $\alpha (t,T)+⟨\sigma (t,T),{\lambda }_t{⟩}_H$. Historical mean changes must therefore not be inserted in place of the risk-neutral drift restriction. Gaussianity under both measures additionally requires suitable deterministic drift changes; state-dependent risk premiums need not preserve Gaussianity under the physical measure.