Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 32 (Advanced Financial Models)

The positive state-price density alternative characterizes one-period arbitrage. Interpret $A_0$ as the vector of current asset prices and $A_1$ as their terminal payments. A fixed vector of holdings produces initial cost $x^{\top }{A}_0$ and terminal payment $x^{\top }{A}_1$. The first alternative is a nonpositive-cost investment with a nonnegative payment and strict improvement either initially or on an event of positive probability. A strictly positive state-price density prices every such payment and excludes both forms of arbitrage, as proved below.

If there is a traded bank account costing one and paying a deterministic $R>0$, the density satisfies $1=R\mathbb{E}\nu$. Hence $dQ/dP=R\nu$ defines an equivalent martingale measure andConversely this pricing relation under an equivalent martingale measure gives $\nu =R^{-1}dQ/dP$. Without such an account, $\nu$ is a pricing density, not necessarily a probability density or even an integrable random variable by itself; the stated requirement is integrability of $\nu A_1$. This distinction keeps the abstract alternative valid without an unstated asset paying one.

Suppose a strictly positive pricing density exists. Its weighted payment for any fixed holdings is integrable, andIf the terminal payment is nonnegative and strictly positive on an event of positive probability, positivity of $\nu$ makes this expected value strictly positive. It cannot therefore have nonpositive initial cost. Thus the first form of arbitrage is incompatible with a positive pricing density. The existence argument in part (b) proves that this form occurs whenever separation within the payment span obstructs a density.

The same pricing relation gives $x^{\top }{A}_0\ge 0$ whenever the terminal payment is nonnegative. Therefore negative initial cost also contradicts a strictly positive state-price density, even if the terminal payment is identically zero.

This second form is needed for directions invisible in the terminal payments. Let $L$ be the linear span of the essential values of $A_1$, equivalentlyIf $A_0\notin L$, let $v$ be its nonzero orthogonal projection on $L^{\perp }$ and set $x=-v$. Then $x^{\top }{A}_1=0$ almost surely while $x^{\top }{A}_0=-\Vert v{\Vert }^2<0$. Hence the second form exists in this case. Part (b) handles all remaining cases.

We prove exhaustiveness, including existence of the positive density when neither form of arbitrage occurs. Use the subspace $L$ from part (a)(ii), and considerThis is a nonempty convex cone closed under positive scaling: ${\nu }_0=(1+\Vert A_1\Vert {)}^{-1}$ supplies one element and guarantees weighted integrability even when $A_1$ itself has no first moment. Crucially, $\mathcal{K}$ is relatively open in $L$; it need not be closed.

To prove openness at a point $z=\mathbb{E}[\nu A_1]$, put $Z=\nu A_1$. The vectors $\mathbb{E}[hZ]$, for bounded real measurable $h$, span $L$. Indeed, a vector orthogonal to all of them would satisfy $\mathbb{E}[h{u}^{\top }Z]=0$ for every such $h$, forcing $u^{\top }Z=0$ almost surely. Since $\nu >0$, this says $u\in L^{\perp }$. Choose finitely many bounded $h_j$ for which $\mathbb{E}[h_{j}Z]$ form a basis of $L$. For coefficients $t_j$ sufficiently small,This gives a relative neighbourhood of $z$. It also proves the openness step behind the strictly positive barycentre cone lemma without assuming a closed moment cone.

If $A_0\notin L$, the second arbitrage form was already constructed. If $A_0\in L\setminus \mathcal{K}$, the separation of a point and an open convex set, applied inside the finite-dimensional space $L$, gives a nonzero $x\in L$ with $x^{\top }{A}_0\le {\inf }_{z\in \mathcal{K}}{x}^{\top }z$. Positive scaling in the cone forcesWe claim $x^{\top }{A}_1\ge 0$ almost surely. Otherwise $E=\{x^{\top }{A}_1<0\}$ has positive probability, and the admissible positive weights ${\nu }_m={\nu }_0(1+m{1}_E)$ givecontradicting the separation inequality. Since $x\in L$ is nonzero, $x^{\top }{A}_1$ cannot vanish almost surely; thus it is strictly positive on an event of positive probability. This is the first arbitrage form.

Consequently absence of both forms forces $A_0\in \mathcal{K}$, which supplies the required $\nu$. Conversely either form excludes every such density by part (a). We have proved exactly one of arbitrage or a strictly positive pricing density holds. If $L=\{0\}$, the same argument reduces to the elementary cases $A_0=0$ and $A_0\ne 0$, so no nondegeneracy assumption has been hidden.

Specify the trading conventions first. Let the bank account have value $R_r>0$, with $R_0=1$ and its next gross return known at time $r$. Write $B_r=1/R_r$ for the discount factor. Let $S_r$ be the ex-dividend asset-price vector, and $d_r$ the vector of cash dividends paid per unit at time $r$; set $d_r=0$ for non-dividend-paying assets. Under an equivalent martingale measure $Q$, each discounted cumulative asset gainis a martingale. Equivalently, ${\mathbb{E}}_Q[B_{r+1}(S_{r+1}+d_{r+1})\mid {\mathcal{F}}_r]=B_r{S}_r$. This is the gain-process form of risk-neutral pricing, rather than a martingale assumption on raw ex-dividend prices.

Let $X_r$ be the ${\mathcal{F}}_r$-measurable holdings after trading and distributions at time $r$. The previous holdings $X_{r-1}$ earn the next price change and dividends. Define the strategy's net cash payment and its residual portfolio wealth byA negative payment is an external infusion; a self-financing portfolio with no consumption has $D_r^X=0$. The discounted budget equation isAssume these gains and payments are integrable under $Q$, for example by bounded predictable holdings and integrable discounted asset gains. Thenis a true martingale, by conditioning the displayed one-step gains. Merely assuming a local martingale without appropriate integrability would not justify the conditional equality.

Conditioning $M_n$ on ${\mathcal{F}}_r$ proves the general multi-period dividend pricing identityFor a strategy liquidated at time $n$, $X_n=0$ and all liquidation proceeds are included in $D_n^X$. The terminal value disappears, giving exactly the requested dividend-only relation. That liquidation convention is necessary: holding one unit of the bank account with no payments would otherwise have a positive left side and zero right side.

An attainable claim $H$ paid at $n$ can be replicated by a strategy with no earlier payments and final liquidation payment $D_n^X=H$. Its time-$r$ value is thereforeTwo admissible replicating strategies have the same value by this identity, and the value is the same under every equivalent martingale measure for which the replication gains are true martingales. For unattainable claims this expectation is a possible model price, but the replication argument does not establish a unique price. With density process $Z_r={\mathbb{E}}_P[dQ/dP\mid {\mathcal{F}}_r]$, the equivalent state-price density form is $V_r=(B_r{Z}_r{)}^{-1}{\mathbb{E}}_P[B_n{Z}_{n}H\mid {\mathcal{F}}_r]$.

For the Brownian first-passage time to a line, set $\lambda =\theta >0$ and choose $h=b+\sqrt{b^2+2\lambda }>0$, so $h^2/2-bh=\lambda$. The Exponential martingale for Brownian motion $M_s=e^{h{W}_s-h^{2}s/2}$, stopped at $T_{a,b}$, obeysThe optional stopping theorem at the bounded time $s\wedge T_{a,b}$ gives expectation one. The contribution from $\{T_{a,b}>s\}$ is at most $e^{ha-\lambda s}$, and tends to zero. Taking $e^{-\lambda \infty }=0$ therefore yieldsLetting $\theta \downarrow 0$ shows that the hitting probability is one for $b\le 0$ and $e^{-2ab}$ for $b>0$; in the latter case the hitting-time law has mass at infinity.

To derive the finite-time linear-boundary Brownian first-passage distribution, put $Z_s=W_s-bs$. This is Brownian motion with drift $\mu =-b$, and the event is $\{{\sup }_{s\le t}{Z}_s\ge a\}$. For driftless Brownian motion, the Brownian reflection principle says that the joint endpoint density on this event equals the normal density $p_t(x)$ when $x\ge a$ and the reflected density $p_t(2a-x)$ when $x<a$, where $p_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$. The constant-drift change of measure weights these densities by $e^{\mu x-{\mu }^{2}t/2}$. Thus the crossing probability isCompleting the square gives respectively $\Phi ((\mu t-a)/\sqrt{t})$ and $e^{2\mu a}\Phi ((-\mu t-a)/\sqrt{t})$. Substituting $\mu =-b$ provesThe density weighting follows directly from the Exponential martingale for Brownian motion, or the Girsanov theorem, on the fixed finite horizon; it does not require eventual hitting.

For the barrier digital call, under the risk-neutral measure in the Black-Scholes model,Assume $\sigma >0$, put $a=\log (c/S_0)/\sigma >0$ and $b=-(\rho -{\sigma }^2/2)/\sigma$. The upper stock barrier is reached exactly when $W_s^Q$ reaches $a+bs$. The payment is at the fixed maturity $t_0$, not at the random hitting time. Hence its price isIf $\sigma =0$, use the deterministic path $S_s=S_0{e}^{\rho s}$ instead: the value is $e^{-\rho t_0}$ if that path reaches $c$ by $t_0$, and zero otherwise. The nondegenerate formula concerns the usual positive-volatility model.

Assume $\sigma >0$, $x>0$ and sufficient $C^{2,1}$ regularity to use the Itô formula. Under the physical measure write $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ and let the bank account be $R_t=e^{\rho t}$. If $V_t=f(S_t,t)$ is a self-financing portfolio with ${\Delta }_t$ stock units and ${\beta }_t$ bank-account units, thenThe Itô formula gives $df=(f_t+\mu S_t{f}_x+\frac{1}{2}{\sigma }^2{S}_t^2{f}_{xx})dt+\sigma S_t{f}_{x}d{W}_t$. Equating diffusion terms forces ${\Delta }_t=f_x(S_t,t)$; equating drifts and eliminating ${\beta }_t{R}_t=f-S_t{f}_x$ gives the Black-Scholes equationConversely, if this equation holds, take $\Delta =f_x$ and $\beta =(f-x{f}_x)/R$. The same comparison shows $df=\Delta dS+\beta dR$, so the constructed portfolio is self-financing. This is a local replication equivalence for smooth functions on the state space; global admissibility and a terminal condition specify a unique economically admissible claim price.

With continuous stock-dividend yield $\theta$, holding ${\Delta }_t$ stock units earns $\theta {\Delta }_t{S}_{t}dt$ in addition to their price changes. A self-financing portfolio reinvests that income, soThe same diffusion comparison gives $\Delta =f_x$, and the drift comparison yields the Black-Scholes equation with continuous stock dividendsUnder the risk-neutral measure, the ex-dividend stock consequently has drift $\rho -\theta$, while cash is discounted at $\rho$.

For the last claim, distinguish the terminal payoff function $F$ from a time-dependent price. Put $\tau =t_0-t$. The price with yield $\theta$ is $q(x,t)=e^{-\rho \tau }\mathbb{E}[F(x{e}^{(\rho -\theta -{\sigma }^2/2)\tau +\sigma \sqrt{\tau }Z})]$, with $Z$ a standard normal random variable. In a no-dividend model with interest rate $\rho -\theta$, the stock-transition law is identical but its discount factor is $e^{-(\rho -\theta )\tau }$. Thus, whenever the payoff expectation is finite, the dividend-yield discount shift givesEquivalently substitute this expression in the dividend equation: $q_t=e^{-\theta \tau }(p_t+\theta p)$, and the equation reduces exactly to the no-dividend Black-Scholes equation with rate $\rho -\theta$, with the same terminal data.

Use the physical dynamics of the Black-Scholes model, $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with bank rate $\rho$ and $\sigma >0$. Write $\lambda =(\mu -\rho )/\sigma$ for the market price of risk. If $w_t$ is portfolio wealth, ${\pi }_t$ the dollar amount in stock, and $c_t\ge 0$ the consumption rate, the budget dynamics areAdmissible controls are progressively measurable, locally integrable in the required drift and diffusion norms, and keep $w_t\ge 0$. Such a solvency restriction excludes doubling-type strategies. In a complete Brownian market, the investment-consumption problem is to maximizewhere the utility functions are increasing and concave. A terminal-only problem omits the integral. Strict concavity gives uniqueness of optimal consumption and terminal wealth when an optimum exists.

The state-price density isThe Itô formula givesThus $H_t{w}_t+{\int }_0^t{H}_s{c}_{s}ds$ is a nonnegative local martingale, hence a supermartingale, yielding the state-price budget constraintConversely, a nonnegative consumption plan and terminal payoff $\xi$ satisfying budget equality and the necessary integrability can be financed in the Brownian complete market. DefineThe Brownian martingale representation theorem for the total future budget supplies a stochastic-integral coefficient; comparing it with $H_t(\sigma {\pi }_t-\lambda w_t)$ defines ${\pi }_t$. This reconstructs the wealth equation with $w_t\ge 0$. Localization gives the representation for integrable positive budgets when an $L^2$ assumption is not available.

For smooth strictly concave utilities with the usual marginal-utility range, let $I_u=(u^{\prime }{)}^{-1}$ and $I_U=(U^{\prime }{)}^{-1}$ be the inverse marginal utility functions. The Lagrange-multiplier solution iswhere $y>0$ makes the budget an equality. This is a global argument, not merely formal differentiation: the concave supporting-tangent inequality giveswith the analogous terminal inequality. Taking expectations, integrating, and applying the budget bound makes the difference in objectives nonpositive. Replication of the candidate budget proves attainability and optimality whenever these expectations are well-defined and finite. If the marginal-utility ranges or integrability conditions fail, existence must be established separately rather than inferred from the first-order equations.

A detailed example is constant relative risk aversion utility $u(c)=c^{1-\gamma }/(1-\gamma )$ and $U(w)=\eta w^{1-\gamma }/(1-\gamma )$, with $\gamma >0$, $\gamma \ne 1$, $\eta >0$, and $\delta \ge 0$. PutThe optimal plans from the marginal-utility equations are $c_t^{\ast }=(y{e}^{\delta t}{H}_t{)}^{-1/\gamma }$ and ${\xi }^{\ast }={\eta }^{1/\gamma }(y{e}^{\delta T}{H}_T{)}^{-1/\gamma }$. The normal distribution of $W_t$ gives, for $a=1-1/\gamma$,The budget therefore becomes $y^{-1/\gamma }b(T)=w_0$, determining the multiplier explicitly. Independent Brownian increments in the conditional budget formula give $w_t^{\ast }=c_t^{\ast }b(T-t)$. The diffusion coefficient of $c_t^{\ast }$ is $(\lambda /\gamma )c_t^{\ast }$, so matching the wealth diffusion determines the dollar stock holding. Consequently the finite-horizon power-utility investment and consumption optimum isAll candidate moments are finite on a finite horizon, since powers of a lognormal state-price density have finite moments. The budget and tangent argument above therefore verify optimality in this example, including admissibility and positive wealth.

For another verification of the formulas, the Hamilton-Jacobi-Bellman equation for a value function with discount measured from the current time isSubstitute $V(w,t)=b(T-t{)}^{\gamma }{w}^{1-\gamma }/(1-\gamma )$. The two concave maximizations give the displayed controls, and the remaining equation is $b^{\prime }(\tau )=1-kb(\tau )$ with $b(0)={\eta }^{1/\gamma }$, exactly solved by the formula above. ThusFor $k=0$, interpret the integral as $\tau$, so $b(\tau )=\tau +{\eta }^{1/\gamma }$. Unlike the infinite-horizon Merton consumption-investment problem, finite horizons do not require $k>0$. Consumption vanishes in the terminal-only formulation; then the integral term in $b$ is absent and the optimal stock fraction remains $(\mu -\rho )/(\gamma {\sigma }^2)$. Logarithmic utility is handled directly with inverse marginal utility $1/z$ and the same wealth-to-consumption formula at $\gamma =1$, with $k=\delta$.

A Gaussian one-factor short-rate model is most naturally specified under a chosen risk-neutral measure $Q$, not by using a physical drift unaltered in a pricing formula. The Vasicek model hasIts explicit solution is $r_s=\bar{r}+(r_t-\bar{r})e^{-\kappa (s-t)}+\eta {\int }_t^s{e}^{-\kappa (s-u)}d{W}_u^Q$. It is a Markov process, and its conditional mean and variance are $\bar{r}+(r_t-\bar{r})e^{-\kappa (s-t)}$ and ${\eta }^2(1-e^{-2\kappa (s-t)})/(2\kappa )$. Hence all finite-dimensional distributions are Gaussian. The model permits negative rates; Gaussianity does not enforce their positivity.

The unit zero-coupon bond price is $P(t,T)={\mathbb{E}}_Q[e^{-{\int }_t^T{r}_{s}ds}\mid {\mathcal{F}}_t]$. Write $\tau =T-t$ and $B(\tau )=(1-e^{-\kappa \tau })/\kappa$. Stochastic integration of the explicit solution givesThis Gaussian variable has variance ${\eta }^2{\int }_0^{\tau }B(v{)}^{2}dv$. Evaluating its exponential moment gives exponential-affine bond pricingFor example, ${\int }_0^{\tau }B(v{)}^{2}dv={\kappa }^{-2}[\tau -2B(\tau )+(1-e^{-2\kappa \tau })/(2\kappa )]$, which gives the displayed expression by direct simplification. The term structure of interest rates is therefore analytically tractable, and bond-price changes are driven by the current short rate alone.

The Hull-White model replaces the drift by $\alpha (t)-\kappa r_t$ with deterministic $\alpha$. The same calculation givesThis retains the one-factor Gaussian and Markov structure while allowing an exact initial-curve fit through $\alpha$. If $f_0(t)$ is the initial instantaneous forward rate and $r_0=f_0(0)$, differentiating the bond formula shows thatThus calibration flexibility is separated from stochastic mean reversion. Time-dependent calibration generally destroys time stationarity even though the state remains Markov. In the time-homogeneous Vasicek case, a stationary initialization is $r_0\sim N(\bar{r},{\eta }^2/(2\kappa ))$ independent of future increments; its covariance is ${\eta }^2{e}^{-\kappa |t-s|}/(2\kappa )$. A fixed initial short rate does not give a stationary process.

A Gaussian forward-rate field models the whole instantaneous forward rate curve. Let $H$ be a real separable Hilbert space, let $W$ be a cylindrical Brownian motion, and let deterministic $\sigma (t,T)\in H$ satisfy the maturity-integrability and square-integrability conditions needed below. For a deterministic initial curve, specify under $Q$Using an orthonormal basis, the noise is ${\sum }_j{\sigma }_j(t,T)d{W}_t^j$; an infinite-dimensional cylindrical noise need not exist as an $H$-valued random vector. Its covariation kernel across maturities is the positive semidefinite kernel $C_t(T,U)=⟨\sigma (t,T),\sigma (t,U)⟩$. Thus arbitrary suitable maturity-correlation kernels can be represented by Hilbert-space factors, while finitely many factors give finite-rank kernels.

No arbitrage requires a drift restriction. Define the zero-coupon bond price and integrated volatility byDifferentiation of the moving lower endpoint and the Itô formula giveThe discounted bond must have zero drift. Differentiating ${\int }_t^T\alpha (t,u)du=\frac{1}{2}\Vert \Sigma (t,T){\Vert }^2$ with respect to maturity yields the Heath-Jarrow-Morton model restrictionThis is why forward-rate drift and covariance cannot be selected independently in an arbitrage-free random-field model. With deterministic volatilities square-integrable on the finite horizon, the discounted bond stochastic exponential is a true martingale, not only a local martingale.

The solution is explicitly Gaussian:with mean given by the first two terms and covarianceThe entire Gaussian random field can therefore be calibrated through its initial curve and maturity covariance structure. Bond prices are exponential functionals of these Gaussian rates; conditional bond ratios have lognormal laws under appropriate forward measures. Negative rates are possible. A large or infinite volatility space provides rich maturity correlations, but Gaussianity alone supplies neither a low-dimensional Markov state nor positivity of rates.

An interest-rate cap is the sum of caplets over successive reset and payment dates. Consider one accrual period $[T,U]$ with year fraction $\delta >0$ and strike $K\ge 0$. The reset rate is $L(T;T,U)=[P(T,U{)}^{-1}-1]/\delta$; the unit-notional payment at $U$ is $\delta (L-K{)}^{+}$. At reset its value isHence the caplet is equivalent before reset to $c=1+\delta K$ puts expiring at $T$ on the maturity-$U$ zero-coupon bond, with strike $K_b=1/c$. This algebra accounts for payment at $U$ rather than incorrectly treating the payment date as $T$.

Under the T-forward measure, use the bond-ratio process $F_s=P(s,U)/P(s,T)$, which is a martingale. From the Gaussian HJM dynamics its relative volatility is $-[\Sigma (s,U)-\Sigma (s,T)]$. With deterministic volatilities its logarithm at reset is normal with integrated varianceFor $v>0$, write $d_1=\log (F_t/K_b)/v+v/2$ and $d_2=d_1-v$. A normal integral gives ${\mathbb{Q}}^T(F_T<K_b\mid {\mathcal{F}}_t)=\Phi (-d_2)$ and ${\mathbb{E}}_{Q^T}[F_T{1}_{\{F_T<K_b\}}\mid {\mathcal{F}}_t]=F_t\Phi (-d_1)$. Therefore the Gaussian caplet bond-put formula isFor $v=0$, the value is $(P(t,T)-(1+\delta K)P(t,U){)}^{+}$. This is also obtained from the Gaussian bond-option formula by put-call parity. A cap price is the sum of its caplet prices, multiplied by the notional principal.

In the Vasicek model, the instantaneous forward-rate volatility is $\sigma (s,u)=\eta e^{-\kappa (u-s)}$. ThusIndeed, $\Sigma (s,U)-\Sigma (s,T)=\eta e^{-\kappa (T-s)}(1-e^{-\kappa (U-T)})/\kappa$, whose squared integral gives this expression without ambiguity. Gaussian short and instantaneous forward rates do not mean the reset floating rate $L$ is Gaussian: the latter is an inverse bond price minus one. The lognormal bond-ratio calculation, under the proper forward measure, is the reason for the normal-CDF pricing formula.

The correct state in a general Gaussian forward-rate field is the whole current curve. In time-to-maturity coordinates $g_t(x)=f(t,t+x)$, assume deterministic time-homogeneous volatility $\sigma (x)\in H$. Set $A(x)=⟨\sigma (x),{\int }_0^x\sigma (u)du⟩$. The Musiela forward-curve equation isUnder the appropriate maturity differentiability, or in its mild interpretation on a suitable function space, its solution isFor any future interval, the shifted current curve determines the deterministic part and future Brownian increments are independent of past information. This proves that the curve-valued process is time-homogeneous Markov. General observation-time-dependent deterministic volatility instead gives a time-inhomogeneous Markov curve. A low-dimensional state requires more: the maturity functions carrying the randomness must close under the shift evolution. Finitely many Brownian drivers by itself does not ensure that the short rate is Markov.

For a precise obstruction, take two independent stationary Ornstein-Uhlenbeck processes $Z^1,Z^2$ with distinct positive decay parameters $a_1,a_2$, and let $r_t=m+Z_t^1+Z_t^2$. Its centered covariance is $C(h)=c_1{e}^{-a_1|h|}+c_2{e}^{-a_2|h|}$ with $c_1,c_2>0$. A stationary scalar Gaussian Markov process must satisfy the covariance criterion for a stationary Gaussian Markov process:Indeed Gaussian conditioning gives $\mathbb{E}[r_{s+t}-m\mid r_s]=C(t)(r_s-m)/C(0)$; if the present is a Markov state, conditioning the future also on the earlier $r_0$ cannot change that mean. Multiplication by $r_0-m$ and taking expectations yields the equation. For the two-factor covariance the difference of its two sides isSo the short rate alone is not Markov, although the two-dimensional factor process is. Distinct exponentially decaying HJM volatility components realize such a multi-factor Gaussian curve model.

Stationarity is an additional condition, not a consequence of Gaussianity or time-homogeneity. A stationary Gaussian forward curve for the time-homogeneous model can be constructed whenis finite and the deterministic mean satisfies $m^{\prime }(x)+A(x)=0$. Initialize $g_0$ with mean $m$ and covariance $K$, independently of future noise. The mild solution has the same mean, since $m(x+t)+{\int }_0^{t}A(x+s)ds=m(x)$. Its covariance isThus its one-time Gaussian curve distribution is invariant, and the time-homogeneous transition law makes all time-shifted finite-dimensional distributions invariant as well. The initial field can explicitly be built from an independent noise on negative times using the same square-integrable kernels.

For example, ${\sigma }_j(x)={\eta }_j{e}^{-a_{j}x}$ with $a_j>0$ giveswhere $d{Z}_t^j=-a_j{Z}_t^{j}dt+{\eta }_{j}d{W}_t^j$ and the factors start in their stationary normal laws. This is a finite-dimensional Markov realization when the sum is finite. In contrast, a nonzero constant maturity volatility makes $K(x,x)$ infinite, excluding a finite-variance stationary curve of this form. A deterministic initial curve is generally nonstationary even when a stationary distribution exists. These distinctions separate Gaussianity, Markov state sufficiency, and stationarity.