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For a fixed twice-differentiable terminal payoff with sufficient growth control, differentiate its lognormal distribution pricing integral. Gaussian integration by parts gives the displayed identities. Convexity implies $p_{xx}\ge 0$, so increasing spot volatility cannot lower the price. The interest rate sensitivity depends on the sign of the cash position $p-x{p}_x$. Calendar-time sensitivity is $p_t=\rho (p-x{p}_x)-\frac{1}{2}{\sigma }^2{x}^2{p}_{xx}$. Thus a convex payoff with a negative cash position has a nonincreasing price at fixed spot when $\rho \ge 0$. Negative rates invalidate that conclusion: $f(x)=x-K$ gives $p=x-K{e}^{-\rho (T-t)}$ and $p_t=-\rho K{e}^{-\rho (T-t)}>0$ when $\rho <0$.

A European call option gives its holder the right, without an obligation, to buy one unit of an underlying financial asset at the strike price $K$ on the expiry date $T$. European describes exercise timing, not geography: exercise is allowed only at $T$. Here the underlying is one stock share, $S_T$ is its price at expiry, and the financial payoff is expressed as a cash amount. Physical exercise delivers a share against payment of $K$; immediate resale gives the same net amount under the assumed frictionless market.

The holder exercises only when buying at the strike price is advantageous. Writing $x^{+}=\max (x,0)$ for the positive part, the terminal financial payoff isThe European call option is a reusable building block for claim replication: its financial payoff is zero below the strike price and rises by one unit for each unit of stock price above it. The corner at $K$ lets combinations of European call options construct other payoff shapes, as in static replication on a finite terminal support.

For strike price $K=100$, the following are deterministic calculations conditional on each displayed terminal stock price. They assign no probabilities to the three scenarios. The European put option has the opposite exercise right and the financial payoff $(K-S_T{)}^{+}$.

A financial payoff excludes the initial contract price. If the illustrative European call option costs $C_0=10$, its financial payoff at $S_T=120$ is $20$, and subtracting its initial cost gives $10$. If that cost is instead borrowed for one year at the constant continuously compounded interest rate $r=0.05$, the terminal net amount is $20-10e^{0.05}\approx 9.48728904$. At $S_T=80$ or $100$, the corresponding net amount is $-10e^{0.05}\approx -10.51271096$. Keeping all cash amounts at the same date avoids confusing a financial payoff with a financed gain.

Robert Shiller's Options Markets lecture, Yale ECON 252 (2011) introduces these exercise rights and payoff shapes.

A European put option gives its holder the right, without an obligation, to sell one unit of an underlying financial asset at the strike price $K$ only on the expiry date $T$. For one stock share, its terminal financial payoff is the positive partFor strike price $K=100$, terminal stock prices $80$, $100$, and $120$ give financial payoffs $20$, $0$, and $0$, respectively. At $S_T=80$, selling for $100$ rather than the market price $80$ supplies the difference $20$; at $S_T=120$, the holder can let the right expire. Equality at $S_T=100$ gives zero net exercise value. The European call option example plots both financial payoffs on the same axes.

Combining a European put option with the underlying stock share produces the terminal amount $S_T+(K-S_T{)}^{+}=\max (S_T,K)$. Its floor at the strike price is why a European put option can describe protection against a fall in terminal stock value. The protection has an initial cost and lasts to the specified expiry; the terminal floor alone says nothing about the financial return. Put-call parity replicates this same amount with a European call option and a zero-coupon bond.

Using the put-call parity example's price $P_0=10+100e^{-0.05}-100\approx 5.12294245$, the cost financed for one year at interest rate $r=0.05$ grows to $P_0{e}^{0.05}=100-90e^{0.05}\approx 5.38560133$. The terminal net amount is therefore approximately $14.61439867$ when $S_T=80$, and $-5.38560133$ when $S_T=100$ or $120$. These computations use the unrounded price and distinguish the financial payoff from the financed gain. Unlike an American put option, a European put option supplies no right to exercise before $T$.

For a stock with observed dynamics $d{S}_t/S_t=\sigma d{\hat{W}}_t+\sigma m_{t}dt$ and constant riskless interest rate $r$, the finite-horizon pricing density for Gaussian drift learning with $k=r/\sigma$ gives ${\zeta }_t=e^{-rt}{D}_t$. The terminal logarithmic utility condition $1/w_T^{\ast }=\lambda {\zeta }_T$ and the state-price budget constraint give $\lambda =1/w_0$. Replication then gives $w_t^{\ast }=w_0/{\zeta }_t$ at every time. The Itô formula for $1/\zeta$ makes the optimal dollar stock fraction $(m_t-r/\sigma )/\sigma$. Its finite expected log return follows from finite posterior second moments on finite horizons.

The derivative of a claim price with respect to the interest rate, holding the payoff and other model parameters fixed. For a dividend-free Black-Scholes model with remaining time $\tau$, it is $\tau (x{p}_x-p)$, so its sign is opposite the bank-account value in the replicating strategy.

The derivative of an option's value with respect to calendar time at fixed current underlying price and other parameters. In the dividend-free Black-Scholes model, the Black-Scholes equation givesThe first term is the interest earned on the bond value of the replicating portfolio, while the second uses option gamma. A convex payoff with a short bond position has $\Theta \le 0$ when $\rho \ge 0$; a concave payoff with a long bond position has $\Theta \ge 0$. Negative interest rates need not obey these signs: the affine payoff $x-K$ has $\Theta =-\rho K{e}^{-\rho (T-t)}$. For a fixed payoff function, remaining-maturity sensitivity is $-\Theta$.

Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.

Let $T=t_0$, $\tau =T-t$, and let $N$ have the standard normal distribution. For a twice differentiable payoff with enough growth control for the following expectations and derivatives, risk-neutral pricing in the dividend-free Black-Scholes model givesThe conditional discounted payoff is a martingale under the risk-neutral measure. Applying the Itô formula to $e^{-\rho s}V(S_s,s)$ and setting its drift to zero therefore gives the Black-Scholes equationThe delta hedge holds $\Delta =V_x$ shares; the remaining bond value is $\beta =V-x\Delta$, or $h=\beta /e^{-\rho (T-t)}$ units of the zero-coupon bond. Substitution in the Itô formula and the displayed equation gives $dV=\Delta dS+hdB$, proving replication and self-financing.

Differentiating the pricing expectation in the current stock price gives its option delta and option gamma:Thus a nondecreasing payoff has a nondecreasing price; a nonincreasing payoff has a nonincreasing price. A convex payoff has $\Gamma \ge 0$, and a concave payoff has $\Gamma \le 0$. This is convexity preservation in Black-Scholes pricing.

The Black-Scholes parameter sensitivities follow directly from the same expectation. Differentiation in the interest rate gives the option rhoFor spot volatility, differentiation first givesThe Gaussian integration by parts identity $\mathbb{E}[N\psi (N)]=\mathbb{E}[{\psi }^{\prime }(N)]$, applied to $\psi (N)=f^{\prime }(xM)xM$, cancels the term containing $f^{\prime }$ and gives the option vegaConsequently increasing spot volatility increases a convex payoff's value and decreases a concave payoff's value, with weak inequalities in affine or degenerate cases. A short bond position gives positive option rho, while a long bond position gives negative option rho. The physical stock drift $\mu$ does not enter these prices: the risk-neutral measure replaces it by $\rho$.

The option theta is the calendar-time derivative at fixed current stock price. The Black-Scholes equation givesFor $\rho \ge 0$, a convex payoff with $\beta <0$ therefore has $V_t\le 0$, while a concave payoff with $\beta >0$ has $V_t\ge 0$. The signs are strict when $\rho >0$. At $\rho =0$, an affine payoff can give equality. Remaining-maturity sensitivity is $V_T=-V_t$ when the payoff is independent of $T$. All these time comparisons hold at fixed current stock price.

The stated time-monotonicity conclusion needs the nonnegative-rate condition. For example, the convex affine payoff $f(x)=x-K$ hasThus with negative interest rates it provides an explicit counterexample even though the replica is short in bonds.

Use the usual frictionless option model with nonnegative deterministic interest rates and no stock dividends. Let $B(t,T)\le 1$ be the price of a unit zero-coupon bond, $C_E$ the European call price, and $C_A$ the American call price. A call payoff dominates $S_T-X$, so no-arbitrage pricing gives $C_E\ge S_t-XB(t,T)$, and $C_A\ge C_E$. If $S_t>X$, thenThe last term is the immediate exercise payoff. Waiting retains downside optionality and defers payment of the strike. For positive interest and $X>0$ the inequality is strict while exercise has positive payoff. If $S_t\le X$, exercising for a zero payoff provides no advantage over keeping the right. Thus early exercise of a non-dividend-paying call is never strictly beneficial under these assumptions, and $C_A=C_E$. For completeness, discounted European call value is a true martingale in the standard pricing model. For any exercise stopping time $\tau \le T$, its pointwise dominance of immediate payoff gives ${\mathbb{E}}_Q[B(t,\tau )(S_{\tau }-X{)}^{+}\mid {\mathcal{F}}_t]\le {\mathbb{E}}_Q[B(t,\tau )C_E(\tau )\mid {\mathcal{F}}_t]=C_E(t)$. Taking the supremum over exercise rules proves $C_A\le C_E$, and waiting to maturity gives the reverse inequality. This proves no early exercise of a call without dividends.

The contrast for a put is the interest earned on a strike received sooner. The European put has payoff at most $X$, so $P_E\le XB(t,T)$. When interest is positive and $S_t<X(1-B(t,T))$, immediate exercise gives $X-S_t>XB(t,T)\ge P_E$. Hence committing never to exercise before maturity is worse than an available early-exercise strategy in such a deep-in-the-money state. This does not say that every put should be exercised early, only that a general prohibition is not rational.

The interest-rate hypothesis is essential to the printed claim. If $r<0$ and the stock is deterministic, $S_u=S_t{e}^{r(u-t)}$ with no dividends, choose spot sufficiently large that it stays in the money through $T$. The European call price is $S_t-X{e}^{-r(T-t)}<S_t-X$, so immediate exercise is better. At zero interest, early exercise can be indifferent in degenerate cases; it is the usual positive-interest setting that makes the non-exercise comparison strict.

Assume the stated cash dividend is the only dividend before maturity and interest rates are nonnegative. Before its ex-dividend date, compare exercise now with retaining the option until any later deterministic date $u<t_1$. There is no dividend between these dates, and retaining the option has value at least that of a European call expiring at $u$, namely at least $S_t-XB(t,u)$. This is at least the immediate in-the-money payoff, and is strictly larger with positive interest. Thus exercise earlier in the open pre-dividend interval is not optimal; the only possible dividend-induced early-exercise opportunity is the last cum-dividend instant, conventionally denoted $t_1-$.

At that instant write $S_{-}=S_{t_1-}$ and $S_{+}=S_{-}-D$, the ex-dividend price. Immediate in-the-money exercise pays $S_{-}-X$. Continuing across the dividend and retaining the call to maturity gives at least $S_{+}-XB(t_1,T)$, since no further dividend remains. Therefore continuation minus immediate exercise is bounded below byConsequently the dividend-date call-exercise criterion isIf the pre-dividend call is out of the money, exercising gives no benefit in the first place. After the dividend, part (a) excludes later early exercise. The comparison is a sufficient condition: failing it does not prove exercise is optimal, because continuation also includes put/time value not present in this lower bound.

The date convention matters. A dividend can make exercise just before the ex-dividend date optimal. Thus a literal assertion excluding $t_1-$ under all dividends would be false; the pre-dividend non-exercise argument excludes earlier dates, leaving this last cum-dividend opportunity to be tested. Further dividends would require separate tests at their own dates.

On a finite horizon $[0,T]$, let $W$ be a standard $d$-dimensional Brownian motion and let $\theta$ be a predictable process with ${\int }_0^T|{\theta }_s{|}^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T|{\theta }_s{|}^{2}ds)<\infty$ is a sufficient condition. The Girsanov theorem says that under $dQ=Z_{T}dP$,Because $Z_T>0$, the two measures are equivalent on the terminal sigma-algebra.

For the proof sketch, Itô formula gives $d{Z}_t=-Z_t{\theta }_t\cdot d{W}_t$. Apply the product rule to $Z_t{W}_t^{Q,i}$. Its finite-variation term $Z_t{\theta }_t^{i}dt$ is cancelled by $d[Z,W^i{]}_t=-Z_t{\theta }_t^{i}dt$. After localization the product is a martingale under $P$, and the conditional change-of-measure formula makes $W^{Q,i}$ a local martingale under $Q$. Adding finite variation leaves its covariations unchanged: $[W^{Q,i},W^{Q,j}{]}_t={\delta }_{ij}t$. The Lévy characterization of multidimensional Brownian motion completes the proof. Localization is essential if the integrands are not bounded; the true-martingale hypothesis on $Z$ permits the final measure change.

For the deal, write $T=t_0$ and use constant interest rate $\rho$, volatility $\sigma >0$ and no stock dividends. Under the risk-neutral measure, the Black-Scholes model isFor example this follows from the preceding theorem with $\theta =(\mu -\rho )/\sigma$ when the physical stock drift is $\mu$. The delivered shares have terminal cash value $S_T\max (r,c/A)$, so its initial value isThe factor $S_T$ must be included: the deal is a stock delivery, not a cash payment of the random number of shares.

Use the stock-numeraire measure in the Black-Scholes model, whose density isThe Girsanov theorem gives $W_t^S=W_t^Q-\sigma t$ as Brownian motion under $Q^S$. ThereforeThe discrete geometric average under the stock-numeraire measure has normally distributed logarithm withThe last identity uses the printed descending order of sampling times. Each $t_i$ is the minimum in exactly $2i-1$ ordered pairs. In particular $v>0$.

Assume the ordinary positive deal parameters $r,c>0$ and put $k=\log (c/r)$, $d=(m-k)/\sqrt{v}$. The change of numeraire gives $V_0=S_0{\mathbb{E}}_{Q^S}\max (r,c{e}^{-Y})$, where $Y=\log A$. Split the integral at $k$. Completing the square in $e^{-y}$ times the normal probability density givesThus the stock-delivery option with a geometric average has valueHere $\Phi$ is the standard normal distribution function. Delivery time $T$ does not enter separately once the sampling dates are fixed: the stock numeraire cancels the final stock factor and discounting. As a check, for one sample at $T$, the payment is $\max (r{S}_T,c)$ and the formula becomes the value of $r$ shares plus a put on their terminal value. If $c=0$ the value is $r{S}_0$; if $r=0$ the value is $c{S}_0{e}^{-m+v/2}$.

If the value is independent of spot, both spatial derivatives vanish. The Black-Scholes equation reduces to $V_t-rV=0$, so $\boxed{V(S,t)=A{e}^{r(t-T)}}$, where $A$ is an arbitrary constant. A constant terminal payoff is therefore discounted at the interest rate $r$; its separation constant is $\lambda =0$.

Subtract the boundary slope from the Brownian motion: $X_t=B_t-bt$. Then $T_{a,b}$ is the first-passage time of $X$ to the fixed level $a$. The zero-drift Brownian reflection principle givesHere $\Phi$ is the standard normal distribution function. To change the drift to $-b$, use the Girsanov theorem on each finite horizon, with density $L_t=\exp (-b{B}_t-b^{2}t/2)$. On $\{T_a\le t\}$, conditional expectation at the stopped time gives $\mathbb{E}[L_t\mid {\mathcal{F}}_{T_a}]=L_{T_a}$, because the density is a true martingale. Since $B_{T_a}=a$, the drifted Brownian first-passage density is consequentlyThere may be additional mass at infinite hitting time; the displayed density only describes finite passage. With $e^{-\theta \infty }=0$, apply the given zero-drift Brownian first-passage Laplace transform:The zero-discount limit is $1$ for $b\le 0$ and $e^{-2ab}$ for $b>0$, so the hitting probability is less than one precisely when the line moves away with positive slope.

For a direct verification of the linear-boundary Brownian first-passage distribution, defineBoth terms tend to zero as $t\downarrow 0$. Let $\varphi$ be the standard normal density. The identityfollows by expanding the two squares. Differentiating the two normal probabilities then givesSince this is the previously derived hitting density and both distribution functions start at zero,This proof applies to every real $b$, including the defective distribution when $b>0$.

For the barrier digital put, use the dividend-free Black-Scholes model with constant interest rate $\rho$ and spot volatility $\sigma >0$. Under the risk-neutral measure,Set $T=t_0$ andReaching the barrier $c$ is equivalent to $W_t^Q$ reaching $a+bt$. The contingent claim pays one only on the survival event $\{T_{a,b}>T\}$. Risk-neutral pricing therefore givesThis is a path-survival price, not the price of a terminal digital put option. Continuity of the stock paths and the absence of an atom in the finite-horizon maximum imply that alternative touching conventions change no price. As $c\to \infty$, the value tends to $e^{-\rho T}$; as $c\downarrow S_0$, it tends to zero. Both checks agree with the survival interpretation.

Take a non-dividend-paying stock with physical dynamics $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$ and continuous-time bank account $B_t=e^{\rho t}$. Write $T=t_0$ and assume the candidate claim price $p(x,t)$ is $C^{2,1}$ before maturity, with enough growth control for the hedge to be admissible. The terminal condition is $p(x,T)=f(x)$.

The Itô formula gives the claim's price changeA self-financing portfolio worth $p$ with $\Delta$ shares has bank-account value $p-\Delta S_t$. Its gain isTo replicate the claim, the Brownian coefficients must agree, forcing the delta hedge $\Delta =p_x$. Equating the remaining drift coefficients cancels the physical drift $\mu$ and yieldsThe cancellation explains why the Black-Scholes equation contains the interest rate, not the physical stock drift. Conversely, if an admissible solution of this equation is given, holdings $p_x$ in the stock and $(p-x{p}_x)/B_t$ in the bank account have value $p$ and gains exactly $dp$. They are therefore a self-financing portfolio replicating the terminal payoff. Absence of arbitrage forces the claim to have this price. Merely decomposing the current value into arbitrary holdings would not justify self-financing; the gain identity and delta hedge are essential.

Now suppose the claim holder receives the cash rate $k(S_t,t)$ before $T$. The price $p$ is the ex-dividend value, so its total gain is $dp+kdt$. The replicating portfolio pays the same amounts out of its wealth, with no external injections. Thus its gain identity isAgain the Brownian coefficients force $\Delta =p_x$. Comparing drifts now gives the Black-Scholes equation with claim dividends:In particular the cash-flow source has a positive sign in the left-hand side. It is not an underlying-stock dividend yield and does not replace the drift $\rho x$ by a dividend-adjusted drift.

For another justification, under the risk-neutral measure $Q$, discount the claim's cum-dividend gain. The Itô formula shows thatexactly when the displayed equation holds. Subject to the usual integrability making this a true martingale, conditional expectation at maturity givesPositive interim payments therefore add positive value, which also checks the source sign. For instance, a constant payment rate $k_0$ and zero terminal payoff have price $k_0(1-e^{-\rho (T-t)})/\rho$, interpreted as $k_0(T-t)$ at $\rho =0$; substitution satisfies $p_t-\rho p+k_0=0$.

Let $T=t_0$, $\tau =T-t$, $x=S_t>0$, and $\sigma >0$. Use a dividend-free Black-Scholes model with constant interest rate $\rho$. The terminal payoff is assumed independent of the parameters being varied. In addition to twice differentiability, impose enough integrability to differentiate the pricing formula; for example, $f,f^{\prime },f^{\prime \prime }$ of polynomial growth suffice. Twice differentiability alone is not enough: the smooth payoff $f(x)=e^{x^2}$ has infinite expectation under every nondegenerate lognormal distribution.

Under the risk-neutral measure $Q$, future stock values conditional on $S_t=x$ have the formClaim replication, or equivalently the martingale property of discounted value in the complete market, givesThe replicating strategy holds the option delta $p_x$ shares and bank-account value $\beta =p-x{p}_x$. No physical drift parameter appears: the Girsanov theorem replaces the physical drift by $\rho$ under the pricing measure, and the replication argument cancels it directly. The risk-neutral pricing operator is linear in the payoff and preserves pointwise payoff inequalities.

Here is a verification of the Black-Scholes equation using the integral itself. Differentiation in spot givesDifferentiating with respect to $\tau$ givesFor $G(Z)=S_T{f}^{\prime }(S_T)$, the Gaussian integration by parts identity $\mathbb{E}[ZG]=\mathbb{E}[G^{\prime }]$ yieldsSubstitution cancels the $-{\sigma }^2/2$ term and provesSince $p_t=-p_{\tau }$, this is precisely the Black-Scholes equation. As $\tau \downarrow 0$, $M\to 1$ and growth control permits passage to the limit, giving $p(x,T)=f(x)$. The delta hedge then supplies the admissible replicating gains under the same regularity assumptions.

The Black-Scholes parameter sensitivities describe the dependence on the model parameters without assuming a particular payoff. If $f$ is increasing, then $p_x\ge 0$. If $f$ is convex, then $p_{xx}\ge 0$, so its price is convex in spot. The option gamma $p_{xx}$ measures the response of the option delta to spot changes.

For the interest rate, the distribution of the future stock also changes with $\rho$, so looking only at discounting gives the wrong general answer. Differentiation gives the option rhoThus a short bank-account position gives positive rate sensitivity, while a positive bank-account position gives negative rate sensitivity. For a cash payoff the result is negative; for the payoff $f(x)=x$ it is zero.

For spot volatility, differentiate the lognormal multiplier and then use Gaussian integration by parts:Consequently a convex payoff has nonnegative option vega. This monotonicity can also be understood by a mean-preserving increase of dispersion in the discounted stock. Strict positivity requires nonzero curvature seen with positive probability; an affine payoff has zero option vega.

Time enters through the remaining maturity $\tau$. Holding spot fixed, the Black-Scholes equation givesUnder the conventional assumption $\rho \ge 0$, a convex payoff and a short bond position $\beta <0$ imply $p_t\le 0$. The inequality is strict if $\rho >0$, or if $p_{xx}>0$ with positive volatility. Thus the price at fixed stock value is nonincreasing in calendar time, and nondecreasing in maturity, under these hypotheses. It is not a claim that the realized stochastic price path decreases: that path still has Brownian exposure $\sigma S_t{p}_{x}d{W}_t^Q$.

The printed conclusion needs the nonnegative-rate qualification; it is false for arbitrary negative rates. For $K>0$, the twice-differentiable convex payoff $f(x)=x-K$ hasThis is an exact counterexample, not a failure of the pricing formula. Also, with $\rho =0$ and an affine payoff, “decreasing” can only mean nonincreasing. Without payoff monotonicity or the specified bond-position sign, neither the spot nor rate nor time sensitivity has a universal sign. Any additional payoff parameter, such as a strike, is differentiated through $f$; pointwise payoff ordering, rather than a generic parameter rule, determines its price effect.

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.