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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 $c{S}_0{e}^{-m+v/2}$.

Let ${\varphi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$ be the centered normal probability density with variance $t$, and take $t>0$. For a zero-drift Brownian motion $B$, reflect the path after its first hit of $a>0$. The Brownian reflection principle identifies endpoints $y<a$ on paths that have hit $a$ with endpoints $2a-y>a$. Thus the endpoint density on paths staying below $a$ is ${\varphi }_t(y)-{\varphi }_t(2a-y)$ for $y<a$.

To introduce drift $\nu$, use the Girsanov theorem with likelihood $e^{\nu B_t-{\nu }^{2}t/2}$. The coordinate path then has the law of $W_s^{\nu }=B_s+\nu s$. Since the likelihood depends only on the endpoint, the joint event in question has probabilityCompleting the two squares givesConsequently the joint endpoint and maximum law for drifted Brownian motion isThe endpoint and the maximum have no relevant atom at the boundary, so the same integral includes $x=a$.

For the down-and-in claim, use non-dividend-paying Black-Scholes model dynamics under the risk-neutral measure. Put $T=t_0$ andLet ${\tau }_b$ be the first time the stock reaches $b$. If $Y_T\le \ell$, continuity ensures ${\tau }_b\le T$, so these terminal values contribute the payoff $f(S_T)1_{S_T\le b}$. For $y>\ell$, lower-barrier reflection followed by the same exponential tilt gives the endpoint density on paths that have hit the barrier:Indeed at zero drift the reflected density is ${\varphi }_T(y-2\ell )$, and multiplying by $e^{\nu y-{\nu }^{2}T/2}$ gives the displayed expression.

Thus the remaining, undiscounted payoff expectation isSet $z=y-2\ell$ andThen $e^{2\nu \ell }={\kappa }^{-\nu /\sigma }$, $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, and $z>-\ell$ is equivalent to $S_0{e}^{\sigma z}>\kappa b$. The integration variable $z$ has the ordinary terminal log-price density. Adding the first contribution therefore proves the static terminal-payoff representation of a down-and-in claim:This holds for nonnegative measurable payoffs, and for signed payoffs whenever the expectations are absolutely integrable. It is an equality of prices, not equality of pathwise terminal payments.

For the call, write $C(s,c,\tau )$ for the ordinary European call option value, with remaining lifetime $\tau >0$ and strike $c>b$. Before activation take current spot $s>b$. The first term in $g$ is zero because the call pays nothing below $b$. The second term reduces to ${\kappa }^{-\nu /\sigma -1}(x-\kappa c{)}^{+}$, since $\kappa c>\kappa b$. The homogeneity $C(s,\kappa c,\tau )=\kappa C(s/\kappa ,c,\tau )$ therefore gives the inactive down-and-in claim valueAt $s=b$ its value is $C(b,c,\tau )$, as required by activation. Its pre-activation option delta, however, has boundary limitwhereas the active contract's delta hedge holds $C_s(b,c,\tau )$ shares. The stock holding consequently jumps byThis jump is strictly positive, even for negative interest rates. To verify that there is no exceptional cancellation, put $z=\log (s/b)/\sigma$ and $y=\log (S_T/b)/\sigma$. The density for log-price paths killed at the lower barrier isThe down-and-out claim is the ordinary call minus $I$. Differentiate its value at $z=0+$: the bracket vanishes there and its derivative is $2y{\varphi }_{\tau }(y)/\tau$. Consequently its derivative with respect to the spot at $b$ isThis derivative is exactly the displayed delta jump. Thus on reaching the barrier before maturity the number of shares changes discontinuously. The portfolio value remains continuous; the bank position changes to finance the share adjustment. Hitting precisely at maturity has probability zero and does not alter this conclusion. This is the delta jump at activation of a down-and-in call.

A one-factor short-rate model uses a single scalar state, normally the short rate itself. Under the physical probability measure writeThe second process is the continuous-time bank account. Specifying the physical diffusion alone does not determine bond prices: it must be accompanied by the market price of risk $\lambda (t,r)$, or equivalently by pricing-measure dynamics. If the Girsanov theorem applies to the likelihood $\mathcal{E}(-\int \lambda dW)$, then under the risk-neutral measureChanging measure alters drift, not quadratic variation. The coefficients need conditions for a nonexplosive solution and for the relevant discounted prices and measure-change density to be true martingales.

A unit zero-coupon bond maturing at $T$ has priceThe Markov property makes current time and rate sufficient state variables. Apply Itô formula to $P(t,r_t;T)/B_t$. Its drift must vanish, giving the short-rate bond pricing equationConversely, under the appropriate boundary and integrability conditions the Feynman-Kac formula identifies its solution with the displayed discounted expectation. For a terminal rate-dependent payoff $H(r_T)$, the same equation has terminal condition $H$ instead of one. A coupon bond is a sum of such unit-bond prices weighted by the coupon payments.

The common market price of risk can also be derived from trading, instead of being introduced arbitrarily for each maturity. Under the physical measure a bond's diffusion exposure is $\eta P_r$ and its excess drift above bank financing is $P_t+\mu P_r+{\eta }^2{P}_{rr}/2-rP$. For two nondegenerate traded bonds, combine their holdings to cancel their Brownian exposure. Absence of arbitrage requires that combination to earn the bank rate; hence their excess-drift-to-exposure ratios must coincide. Calling that common ratio $\lambda$ yields the pricing equation with ${\mu }_Q=\mu -\eta \lambda$ for every maturity. A derivative $V(t,r)$ is then hedged locally by holding $V_r/P_r$ units of a nondegenerate bond and financing the remaining value through the bank account. Their Brownian exposures agree, and substituting both pricing equations proves the drift gain identity. This is short-rate diffusion hedging; it requires $\eta P_r\ne 0$, rather than a bare assertion that one factor automatically gives completeness everywhere.

The entire yield curve is generated consistently from these bond prices. The continuously compounded yield and the instantaneous forward rate areThese encode different summaries: the yield averages the forward rate over the remaining lifetime. Bond prices must satisfy $P(t,t)=1$ and be positive; they need not be below one in models permitting negative rates.

The Vasicek model has constant pricing dynamicsSolving the linear equation gives $r_{t+u}=b+(r_t-b)e^{-au}+\eta {\int }_t^{t+u}{e}^{-a(t+u-v)}d{W}_v^Q$. Consequently, with $\tau =T-t$ and $B(\tau )=(1-e^{-a\tau })/a$,This conditional normal random variable has variance ${\eta }^2{\int }_0^{\tau }B(u{)}^{2}du$. Its normal exponential moment yieldsEvaluating the elementary integral givesThis explicit form makes estimation and valuation convenient. The rate is Gaussian, its conditional mean reverts to $b$, and its long-run variance is ${\eta }^2/(2a)$. Its disadvantage, when nonnegative rates are desired, is a positive probability of negative rates.

The CIR model replaces constant volatility by square-root volatility:The standard solution stays nonnegative. The Feller positivity condition for the CIR model is $2ab\ge {\eta }^2$ in this parameterization; with a positive initial rate it makes zero inaccessible. Below this threshold zero can be reached, although the process does not cross into negative rates. For an affine diffusion bond pricing ansatz $P=A(\tau )e^{-B(\tau )r}$, substitution into the bond equation and matching its constant and rate coefficients givesWriting $d=\sqrt{a^2+2{\eta }^2}$, factor the quadratic in the first equation and integrate from $B=0$; integrating the second equation then gives the CIR bond pricing formulasThe square-root form produces an affine term structure while respecting nonnegative rates. These parameters are pricing-measure parameters; historical mean-reversion estimates alone do not determine them without a risk-premium specification.

Both preceding constant-parameter models constrain the family of initial curves. The Hull-White model allows a deterministic time-dependent drift:For a smooth desired initial forward curve $f_0(t)$, its exact forward-curve fit in the Hull-White model is obtained by takingTo derive this rather than merely assert calibration, write $r_t=x_t+\varphi (t)$, where $d{x}_t=-a{x}_{t}dt+\eta d{W}_t^Q$, $x_0=0$. The conditional Gaussian-integral calculation at time zero givesso its forward curve is $\varphi (T)-{\eta }^{2}B(T{)}^2/2$. Choose $\varphi (T)=f_0(T)+{\eta }^{2}B(T{)}^2/2$. Differentiating this shift and adding $a\varphi (t)$ gives exactly the displayed $\beta (t)$. Thus $P(0,T)=\exp (-{\int }_0^T{f}_0(u)du)$ for every maturity, while $a$ and $\eta$ govern the stochastic fluctuations. This Gaussian model still allows negative rates.

Finally, one-factor models are economical and often give explicit bond prices, but their single Brownian driver is restrictive. Bond diffusion exposures $v_i=\eta P_{i,r}$ give the instantaneous covariance matrix $v{v}^T$, of rank at most one. Nondegenerate maturity changes are instantaneously perfectly correlated, with correlation $+1$ or $-1$, though their finite-horizon correlations need not equal one. A single factor therefore cannot independently describe level, slope and curvature shocks to the whole yield curve. Multiple factors or a full forward-rate model offer greater flexibility. A valid one-factor model must specify the pricing drift, fit the initial curve to the required accuracy, and handle boundary and martingale conditions; naming a rate diffusion alone is not a complete pricing model.

Use $\tilde{\mathbb{P}}$ for the reference measure under which $X$ is a Brownian motion. Pathwise, $B_t=X_t-{\int }_0^t{X}_{s}ds$. Put $h_s=X_s{1}_{\{s\le T\}}$ and defineThe stopped integrand has absolute value at most one. The Novikov condition holds on every finite horizon, so $Z$ is a density martingale. The Itô formula also gives the more useful identityThus $Z$ is uniformly integrable over the entire time axis, not just a true martingale on each finite horizon. Under the reference measure, $\tilde{\mathbb{E}}(t\wedge T)=\tilde{\mathbb{E}}{X}_{t\wedge T}^2\le 1$, so $T<\infty$ almost surely. Hence $Z_t\to Z_T>0$ and $\tilde{\mathbb{E}}{Z}_T=1$.

Define the probability measure on the original sigma-algebra byIts restriction to ${\mathcal{F}}_t$ has density $Z_t$. The Girsanov theorem states that subtracting the integrated density integrand from the reference Brownian motion gives a Brownian motion under the new measure. Thereforeis a Brownian motion under $\mathbb{P}$, and $B_{t\wedge T}=W_{t\wedge T}^{\mathbb{P}}$. This is exactly the requested Brownian property until $T$. The bounded Girsanov density for exit of an unstable linear diffusion also proves global absolute continuity, so a mere collection of unspecified finite-horizon measures is unnecessary.

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.

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.

Use ${\theta }_t$ for the dollar amount invested in the stock; the number of shares is ${\theta }_t/S_t$. The bank account contains $w_t-{\theta }_t$ dollars. With no intermediate consumption, the self-financing portfolio equation isIf instead $\theta$ denotes share holdings, replace each dollar ${\theta }_t$ here by $S_t{\theta }_t$. The equation says that trading only reallocates existing portfolio wealth; gains come from cash interest and stock returns.

For the standard unconstrained complete-diffusion formulation, assume the completed natural Brownian filtration, initial wealth $w_0>0$, and nonnegative admissible wealth. Define the market price of risk ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. The boundedness assumptions make $\kappa$ bounded. The state-price density normalized at zero isIts money-market-adjusted process $Z_t=e^{{\int }_0^t{r}_{s}ds}{\zeta }_t$ is a true martingale by the Novikov condition. Under $d\mathbb{Q}=Z_{T}d\mathbb{P}$, the Girsanov theorem gives $W_t^{\mathbb{Q}}=W_t+{\int }_0^t{\kappa }_{s}ds$ as a Brownian motion, and the discounted stock has zero drift.

The Itô product rule applied to the wealth equation givesFor nonnegative admissible wealth this is a nonnegative local martingale, hence a supermartingale. Therefore every affordable terminal payoff $X=w_T$ satisfies the state-price budget constraint $\mathbb{E}[{\zeta }_{T}X]\le w_0$. Nonzero volatility and the Brownian martingale representation theorem allow replication of nonnegative claims with finite state-price cost.

For a differentiable strictly concave utility function with the usual conditions ensuring an interior integrable optimizer, put $I=(U^{\prime }{)}^{-1}$, the inverse marginal utility. Pointwise maximization of $U(x)-y{\zeta }_{T}x$ givesThe second equation determines the positive constant $y$. To prove optimality, use the concave supporting-tangent inequality:Taking expectations and using the budget bound proves that the candidate dominates every admissible affordable payoff, whenever the utility expectations are well defined. This is the complete-market terminal utility optimizer.

The actual wealth and dollar portfolio are as explicit as general adapted coefficients permit. LetThe Brownian martingale representation theorem, with localization if only $L^1$ integrability is initially known, gives $h$. The dollar portfolio from a deflated wealth martingale isThe bank holds $w_t^{\ast }-{\theta }_t^{\ast }$ dollars. Equivalently, with $B_t=e^{{\int }_0^t{r}_{s}ds}$, $w_t^{\ast }=B_t{\mathbb{E}}^{\mathbb{Q}}[w_T^{\ast }/B_T\mid {\mathcal{F}}_t]$. These formulas need no unjustified differentiability of a value function with random coefficients.

The printed assumption that $U$ is only increasing and concave does not guarantee that the interior formula exists, or even that the optimal value is finite. In general replace the inverse formula by an affordable selectionwhere $\mathcal{D}$ is the wealth domain; boundary choices or supporting slopes handle nondifferentiability. Existence, finite cost and replication still have to hold. A concrete obstruction is the unbounded linear terminal-wealth utility: take $U(x)=x$, $r=0$, $\mu =\sigma =1$, and the Brownian filtration. Then ${\zeta }_T=e^{-W_T-T/2}$. For $A_n=\{{\zeta }_T<1/n\}$, the bounded nonnegative claimis replicable, costs $w_0$, and has $\mathbb{E}{X}_n>n{w}_0$. Thus there is no finite optimal expected utility under the literal utility hypotheses alone. Also, bounded coefficients do not ensure market completeness in an arbitrarily enlarged filtration containing untraded randomness; the full replication formula uses the Brownian market information assumption stated above.

For the Cox-Ross-Rubinstein diffusion approximation, fix calendar maturity $T$, take $h=T/n$, and choose $u=e^{\sigma \sqrt{h}}$, $d=e^{-\sigma \sqrt{h}}$, and the one-step bank factor $b=e^{\rho h}$. For sufficiently fine mesh, $d<b<u$. The risk-neutral probability expands asEach stock log-increment is $\pm \sigma \sqrt{h}$. Its mean under $Q$ is $(\rho -{\sigma }^2/2)h+O(h^2)$ and its variance is ${\sigma }^{2}h+O(h^2)$. Independent increments, their vanishing maximum size, and the Donsker invariance principle for this triangular array give the path limitExponentiation gives the geometric Brownian motion of the Black-Scholes model, and the bank account tends to $e^{\rho t}$.

The physical model is matched by taking $p_h=(e^{\mu h}-d)/(u-d)$. Its corresponding limit has log drift $\mu -{\sigma }^2/2$, so $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t^P$. The likelihood ratios from part (ii) have the matching continuous-time limit. With $\lambda =(\mu -\rho )/\sigma$, a Taylor expansion of their two possible log-multipliers giveswhere ${\epsilon }_k$ is $+1$ or $-1$. Thus the limiting Radon-Nikodym derivative is the Girsanov theorem density $\exp (-\lambda W_T^P-{\lambda }^{2}T/2)$, turning the physical stock drift into $\rho$.