Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 34 (Advanced Financial Models)

Use the riskless financial asset as numéraire, normalized to have initial value one and terminal value $R>0$. Write $S_0\in {\mathbb{R}}^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price random vector. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An arbitrage is such a portfolio with $h\cdot X\ge 0$ almost surely and $\mathbb{P}(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless financial asset to obtain a zero-cost arbitrage.

An equivalent martingale measure is a probability measure $Q$ with exactly the same null events as the physical probability measure $P$, under which discounted asset prices are integrable and satisfy ${\mathbb{E}}_Q[S_1/R]=S_0$. Equivalently, ${\mathbb{E}}_{Q}X=0$. Such a measure rules out arbitrage: every zero-cost portfolio has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.

For the converse, the probability space need not be finite, and the payoffs need not be integrable under $P$. Here is a direct Gaussian-damped martingale density construction. SetThe set $N$ is a linear subspace. Choosing a finite basis of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus \{0\}$, then $h\cdot X$ is not almost surely zero. Absence of arbitrage implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.

All the analytical steps in minimizing $F$ can be checked directly. On $|h|\le H$ the integrand is bounded by $e^{H^2/4}$, since $-|X{|}^2+H|X|\le H^2/4$. Its first two directional derivatives are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j{e}^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the Taylor theorem remainder bound, integrated over $P$, provesThus differentiation under the integral does not require an unproved physical moment assumption.

To prove coercivity, suppose $|h_k|\to \infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional compactness follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\le -\epsilon ,|X|\le K\}$ has positive probability: the negative-gain event is the countable union of such events with $\epsilon =1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\le -\epsilon /2$ on $A$, and henceIf coercivity failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_{\ast }$ a minimizer. If $L=\{0\}$, simply take $h_{\ast }=0$. In every case the directional derivatives vanish at $h_{\ast }$. The vector ${\mathbb{E}}_P[X{e}^{-|X{|}^2-h_{\ast }\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Thereforeis a positive mean-one density defining an equivalent martingale measure. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes absence of arbitrage if and only if an equivalent martingale measure exists, without quoting a financial duality theorem or assuming a finite state space.

Now suppose the probability space is finite. Discard null states and enumerate its positive-probability atoms by $1,\dots ,m$. Every measurable terminal payoff is a vector in ${\mathbb{R}}^m$. Let $A$ be the $m\times (s+1)$ matrix whose columns are the discounted terminal payoffs of the traded financial assets, with first column $1$, and let $p=(1,S_0)$ be their initial price vector. A portfolio with holding vector $z$ replicates the discounted payoff $Az$, and an equivalent martingale measure is exactly a vector $q$ satisfyingThe first pricing equation includes ${\sum }_j{q}_j=1$. A complete market means $\mathrm{im}A={\mathbb{R}}^m$. If $q,q^{\prime }$ both price the financial assets, then $(q-q^{\prime })\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q^{\prime }$.

Conversely, if the financial market is incomplete, there is a nonzero vector $v$ with $A^{T}v=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\mathrm{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives ${\sum }_j{v}_j=0$. Existence of an equivalent martingale measure was already proved, so choose one strictly positive vector $q$. Forboth $q+\epsilon v$ and $q-\epsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm \epsilon v)=p$. Thus uniqueness fails. On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.

Let the bank account be $B_r=R^r$, where $R>0$ is the one-period riskless gross return. In the binomial market, $S_{r+1}$ is either $u{S}_r$ or $d{S}_r$, with $0<d<1<u$ and both successors having positive physical probability. The information at time $r$ consists of the first $r$ moves. A trading strategy chooses its holdings using only this information. A self-financing portfolio funds rebalancing from its current value; there is no external contribution.

Absence of arbitrage requires $d<R<u$. If $R\le d$, borrowing to buy the stock yields a nonnegative gain, strictly positive on the up branch; if $R\ge u$, shorting the stock and investing its price yields the reverse arbitrage. When $d<R<u$, define the risk-neutral probability in a binomial marketGiving each successive up move conditional probability $q$ defines an equivalent martingale measure $Q$, since every finite path has positive probability and ${\mathbb{E}}_Q[S_{r+1}\mid {\mathcal{F}}_r]=R{S}_r$.

For a contingent claim with successor values $V_u,V_d$ at a node of current stock value $s$, solve the two replication equations. The number of shares and current cash value areThen $\Delta us+RC=V_u$ and $\Delta ds+RC=V_d$, so the replicating strategy costsHold $C/B_r$ bank-account units over that period. Starting with the terminal payoff and applying these equations backwards supplies an adapted self-financing strategy replicating every path-dependent contingent claim. This proves market completeness. A different claim price would permit an arbitrage by buying the cheaper claim or replica and selling the dearer one. Iterating the backward pricing relation gives the unique risk-neutral pricing formulaThe physical up probability affects real-world frequencies but does not enter this replication price.

For a terminal payoff $f(S_n)$, write $k=n-r$. Its current price is the restriction to the available nodes of the functionIf $f$ is convex, each summand is convex in $s$ and each weight is positive; summing their defining convexity inequalities proves that $g_r$ is convex. This provides an explicit convex extension between the actual stock nodes, so “convex on the nodes” is unambiguous. If $f$ is strictly convex, each positive scaling preserves strict convexity, so $g_r$ is strictly convex as well.

For the replicating portfolio in a binomial market, the stock holding isFor $r+1<n$, let $F=g_{r+2}$. The next down and up holdings are the secant slopes of $F$ on the adjacent intervals with endpoints $d^{2}s,uds,u^{2}s$:For any convex function and $a<b<c$, its defining inequality at $b=(1-t)a+tc$ rearranges to $(F(b)-F(a))/(b-a)\le (F(c)-F(b))/(c-b)$. The inequality is strict for a strictly convex function. Applying this to the three stock values orders the two next holdings. Meanwhile the backward pricing relation gives, by subtracting its values at $us$ and $ds$,The coefficients are strictly positive and sum to one. Therefore the current share holding lies between the next two holdings:Both inequalities are strict for a strictly convex payoff. For an affine payoff the share holding is constant, explaining the required weak interpretation. These are numbers of shares, not the fraction of portfolio wealth invested in the stock.

A useful finite-horizon form of the Girsanov theorem is as follows. Let $W$ be a Brownian motion and $\eta$ an adapted, progressively measurable process with ${\int }_0^T{\eta }_s^{2}ds<\infty$ almost surely. Suppose its stochastic exponentialis a true martingale with mean one. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\eta }_s^{2}ds)<\infty$ is a sufficient condition. Under the equivalent probability measure $dQ=Z_{T}dP$, the process $W_t^Q=W_t+{\int }_0^t{\eta }_{s}ds$ is a Brownian motion.

For the proof sketch, the Itô formula gives $dZ=-Z\eta dW$. Integration by parts cancels the drift in $d(Z{W}^Q)$, since $Z\eta dt$ from $d{W}^Q$ cancels $d[Z,W^Q]=-Z\eta dt$. After localization, the Bayes formula for conditional expectation makes $W^Q$ a continuous local martingale under $Q$. Its quadratic variation is still $t$, because adding a finite-variation process does not change quadratic variation. One can see the Brownian conclusion directly: for real $z$, the Itô formula makes $\exp (iz{W}_t^Q+z^{2}t/2)$ a local martingale. Its modulus is bounded on the fixed horizon, so it is a true martingale. ThusThis is the conditional characteristic function of a centered normal distribution with variance $t-s$, independent of the past. Continuity and these independent normal increments give the defining properties of Brownian motion.

Now put ${\varphi }_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$ and $\Phi (z)={\int }_{-\infty }^z{\varphi }_1(x)dx$, the standard normal distribution function. At zero drift the endpoint density of paths staying below $a$ isIndeed, reflect the portion of a path after its first hitting time of $a$. By continuity and the independent symmetric Brownian motion increments after this stopping time, reflection preserves its probability law and maps a crossing path ending at $x<a$ to a path ending at $2a-x>a$. This subtracts exactly ${\varphi }_t(2a-x)$ from the unrestricted endpoint density; it is the Brownian reflection principle.

Taking the constant drift tilt $dQ/dP=e^{\mu W_t-{\mu }^{2}t/2}$ in the Girsanov theorem makes $W$ under $Q$ have the same law as $W+\mu s$ under $P$. Therefore the desired finite-horizon maximum of Brownian motion with drift has probabilityCompleting squares gives $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(x)={\varphi }_t(x-\mu t)$ and $e^{\mu x-{\mu }^{2}t/2}{\varphi }_t(2a-x)=e^{2\mu a}{\varphi }_t(x-2a-\mu t)$. HenceThe maximum has no atom at a positive level, so strict and weak barrier inequalities agree. This also follows from continuity of the displayed distribution function. At $\mu =0$ the formula reduces to $2\Phi (a/\sqrt{t})-1$.

For the up-and-out power claim, write $T=t_0>0$ and use the risk-neutral measure for the Black-Scholes model, under whichwith $\sigma >0$. The barrier survives exactly when ${\sup }_{t\le T}(W_t+m_{0}t)<A$. Its risk-neutral pricing value isThe factor inside the expectation is a mean-one Exponential martingale for Brownian motion. Under the probability tilted by it, $W_t-2\sigma t$ is a Brownian motion, so the barrier process has drift $m=m_0+2\sigma =(\rho +3{\sigma }^2/2)/\sigma$. Applying the preceding survival formula yieldsAs $c\to \infty$, the bracket tends to one and the price becomes the discounted second moment $S_0^2{e}^{(\rho +{\sigma }^2)T}$. As $c\downarrow S_0$, it tends to zero. If $\sigma =0$, use the deterministic path instead: the price is $S_0^2{e}^{\rho T}$ when $S_0\max (1,e^{\rho T})<c$, and zero otherwise. At $T=0$, with $c>S_0$, the price is simply $S_0^2$.

Write $T=t_0$ and $b(t)=e^{-\rho (T-t)}$ for the price of the unit zero-coupon bond maturing at $T$, so $db=\rho bdt$. Let the stock satisfy $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ with $\sigma >0$. Assume the value function $f(x,t)=xg(x,t)+b(t)h(x,t)$ is $C^{2,1}$ and the holdings have enough smoothness for the calculations below. The self-financing portfolio condition is the gain identityThe Itô formula expresses the left side asMatching the diffusion terms gives $f_x=g$, or $x{g}_x+b{h}_x=0$. Differentiating $f_x=g$ gives $f_{xx}=g_x$. Also $f_t=x{g}_t+b{h}_t+\rho bh$. Matching the remaining drift terms consequently gives the two necessary conditionsThey are sufficient because substituting them into the Itô formula restores exactly the gain identity. Necessity first holds along the stock process; its positive lognormal density for every $t>0$, together with continuity of the coefficient functions, gives the displayed identities for all $x>0$ and $0<t<T$, and continuity extends them to the endpoints.

Equivalently, the smooth value of any self-financing portfolio satisfies the Black-Scholes equationIndeed, the gain identity after imposing $g=f_x$ has drift $\mu x{f}_x+\rho (f-x{f}_x)$, which gives this equation. Conversely, if $f$ satisfies this equation, takeThe Itô formula then verifies that these holdings form a self-financing portfolio of value $f$. This converse asserts existence of the correct holdings. It does not certify an arbitrary decomposition of $f$ into stock and bond holdings: for example, $f=0$, $g=1$, $h=-x/b$ solves the value equation but fails $g=f_x$. The Black-Scholes value equation needs delta-compatible holdings precisely for this reason.

For a constant-proportion portfolio, the stock value must satisfy $xg=\gamma f$. Since $g=f_x$, integration of $x{f}_x=\gamma f$ gives $f(x,t)=C(t)x^{\gamma }$. Substituting into the Black-Scholes equation givesThus, for initial portfolio wealth $V_0$ and stock price $S_0$, the complete explicit solution isHere $g_t$ counts shares and $h_t$ counts maturity-$T$ bonds; the initial cost is $V_0$. Directly, the self-financing gain equation is $d{V}_t/V_t=[\rho +\gamma (\mu -\rho )]dt+\gamma \sigma d{W}_t$, which also verifies the displayed value. Constant proportions concern the values of the holdings, not constant numbers of shares or bonds. The nondegenerate volatility assumption is necessary for the preceding diffusion matching; with $\sigma =0$, one only obtains a gain condition along the deterministic stock path.

Write $T=t_0$, and specify the physical Black-Scholes model as $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with constant $\sigma >0$ and bank rate $\rho$. The physical drift $\mu$ is needed for an expected utility calculation; a pricing model specified only under a risk-neutral measure does not determine the investor's preferences over risky positions. PutHere $\vartheta$ is the market price of risk, $Z_T$ is the density of the risk-neutral measure $Q$, and $\xi$ is the state-price density. The Girsanov theorem makes $W_t^Q=W_t+\vartheta t$ a Brownian motion under $Q$.

For a dollar holding ${\pi }_t$ in the stock and consumption rate $c_t$, the portfolio wealth equation isSince $d{\xi }_t=-\rho {\xi }_{t}dt-\vartheta {\xi }_{t}d{W}_t$, integration by parts gives the useful deflated wealth equation with consumptionFor strategies with a true deflated gain martingale, terminal wealth $X=V_T$ and consumption obey the state-price budget constraintFor nonnegative wealth and consumption, the admissibility condition can instead give a budget inequality; increasing utilities use the full budget at an optimum. Conversely, suitable integrable payoff and consumption plans with this budget are financeable in the Brownian complete market. To construct the strategy, setBy the Brownian martingale representation theorem, write $d{M}_t=H_{t}d{W}_t$. The required dollar stock holding is ${\pi }_t=(H_t/{\xi }_t+\vartheta V_t)/\sigma$; substitution in the deflated gain equation verifies financing. Remaining wealth goes into the bank account. For square-integrable data the representation holds directly, and localization handles the customary integrable admissible data.

First take no consumption. For an increasing strictly concave differentiable utility function whose marginal derivative can be inverted, let $I_u=(u^{\prime }{)}^{-1}$. Choose $\lambda >0$ so thatThis candidate is not merely a first-order guess. The concave supporting-tangent inequality gives, pointwise,Taking expectations and using the budget proves $\mathbb{E}u(X)\le \mathbb{E}u(X^{\ast })$ for every feasible competitor. Finance the optimizer by its risk-neutral pricing value $V_t=e^{-\rho (T-t)}{\mathbb{E}}_Q[X^{\ast }\mid {\mathcal{F}}_t]$, using the preceding representation to find the holdings.

For a merely concave differentiable utility function, the general formulation is to select $X^{\ast }(\omega )$ from the maximizers of $u(x)-\lambda {\xi }_T(\omega )x$ over its wealth domain, and choose the multiplier to satisfy the budget. If increasing utility is not assumed, the equality-budget multiplier may have either sign; the positive-multiplier inverse formula above is for increasing utility. Flat portions allow multiple maximizers; boundaries require one-sided inequalities rather than equality of marginal derivatives. Existence of a multiplier, integrability and an admissible maximizer must be checked. Concavity alone does not guarantee a well-posed problem: with unrestricted dollar positions, linear utility $u(x)=x$ and $\mu >\rho$ permits unbounded expected terminal wealth. The explicit utilities below satisfy the needed conditions in their stated domains.

For exponential utility $u(x)=(1-e^{-ax})/a$ on $\mathbb{R}$, $u^{\prime }(x)=e^{-ax}$ and $I_u(y)=-\log y/a$. ThereforeThe budget is equivalently ${\mathbb{E}}_Q{X}^{\ast }=w_0{e}^{\rho T}$, since ${\xi }_T=e^{-\rho T}dQ/dP$. Under $Q$, ${\mathbb{E}}_Q{W}_T=-\vartheta T$, giving $\log \lambda =\rho T-{\vartheta }^{2}T/2-a{w}_0{e}^{\rho T}$. HenceMatching the Brownian coefficient of $dV$ with $\sigma {\pi }_t$ gives the finite-horizon exponential-utility portfolioHold ${\pi }_t^{\ast }/S_t$ shares and invest $V_t-{\pi }_t^{\ast }$ in the bank account. This is a dollar amount, not a fixed wealth fraction. The Gaussian exponential formula also gives the optimal expected utilityThis calculation uses unrestricted wealth and an admissibility class permitting the displayed Gaussian terminal wealth, with integrable deflated gains and the relevant exponential moment. A nonnegative-wealth constraint would change the optimizer; no such constraint is imposed on this exponential-utility example.

For both terminal and running utility function, choose $X$ and a nonnegative adapted consumption rate $c$ to maximize ${\mathbb{E}}_P[u(X)+{\int }_0^{T}v(c_t)dt]$ subject to the combined budget. Pointwise maximization of $u(x)-\lambda {\xi }_{T}x$ and $v(c)-\lambda {\xi }_{t}c$ gives, for interior strictly concave cases,Apply the concave supporting-tangent inequality to $u$ and to $v$ at every time, then integrate and take expectations. The difference in objectives is at most $\lambda$ times the difference in budget expenditures, which is nonpositive. This proves optimality, not just the marginal conditions. The financing value isThe question's consumption rate $R_t$ is denoted $c_t$ here to keep it distinct from interest-rate notation.

For logarithmic utility with $u(x)=a\log x$, $v(x)=b\log x$, assume $a,b,w_0>0$ and positive wealth and consumption. The marginal conditions and budget giveEach state-price-weighted payment is deterministic, so conditional valuation immediately gives $V_t=[a+b(T-t)]/(\lambda {\xi }_t)$. Put $A_t=a+b(T-t)$. Since the Itô formula gives $d(1/{\xi }_t)=(1/{\xi }_t)[(\rho +{\vartheta }^2)dt+\vartheta d{W}_t]$, this value satisfiesComparison with the wealth equation verifies the complete finite-horizon logarithmic investment and consumption strategy:Thus the optimal share holding is ${\pi }_t^{\ast }/S_t$, and the remaining value $V_t-{\pi }_t^{\ast }$ is in the bank account; borrowing or shorting is allowed if that value or the stock holding is negative. Terminal wealth is positive and consumption is positive throughout. Finally $\mathbb{E}\log {\xi }_t=-(\rho +{\vartheta }^2/2)t$, so the optimal objective, if desired explicitly, isFinite-horizon lognormal moments ensure the budget and expected utilities are finite, completing the feasibility and optimality checks.

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.