Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 41 (Advanced Financial Models)

We prove the positive state-price density alternative using finite-dimensional convex cones. Put $w=(1+\Vert A_1\Vert {)}^{-1}$ and defineAll displayed expectations are finite, since $w\Vert A_1\Vert \le 1$. The set $D$ is a convex cone, so $K$ is a closed convex cone. A vector $x$ is nonnegative on $K$ exactly when $x^{\top }{A}_1\ge 0$ almost surely. Indeed, one direction follows by taking every nonnegative bounded test weight $f$; for the other direction, if $x^{\top }{A}_1<0$ on a set of positive probability, take its indicator as $f$ to obtain a strictly negative expectation. This also identifies $K$ as the closed cone generated by the support of a probability distribution of $A_1$.

The three cases $A_0\notin K$, $A_0\in K\setminus \mathrm{ri}K$, and $A_0\in \mathrm{ri}K$ are treated below. They give respectively a negative-cost separating portfolio, a supporting zero-cost portfolio with a strictly positive payoff on an event of positive probability, or the required positive pricing weight. Thus the alternatives are exhaustive.

They are mutually exclusive. If $A_0=\mathbb{E}[\nu A_1]$ with $\nu >0$ and a portfolio has nonnegative payoff, then $x^{\top }{A}_0=\mathbb{E}[\nu x^{\top }{A}_1]\ge 0$. A negative initial cost is impossible. If the payoff is strictly positive with positive probability, this expectation is strictly positive, so nonpositive initial cost is impossible too. The given weighted integrability permits taking the expectation of this scalar product.

In a one-period market, $x^{\top }{A}_0$ is the portfolio's initial cost and $x^{\top }{A}_1$ its terminal value. The first two cases are investment-consumption arbitrage: nonnegative wealth can be taken out at both dates, with a strict benefit at one of them. If a strictly positive bank account is traded, any negative initial cost can be invested in it, converting the opportunity into a zero-cost arbitrage with nonnegative terminal payoff and positive payoff with positive probability.

In the third case, $\nu$ is a strictly positive state-price density. If a traded numéraire has price $B_0>0$ and terminal value $B_1>0$, its pricing identity is $\mathbb{E}[\nu B_1]=B_0$. ConsequentlyThe density has mean one and is strictly positive, so $Q$ is equivalent to the physical measure. Thus discounted asset prices are martingales in the one-period sense. This is the fundamental theorem of asset pricing formulation of absence of arbitrage; normalization to a probability measure uses the positive traded numéraire, not an unsupported assumption that $\mathbb{E}\nu =1$.

Suppose $A_0$ lies on the relative boundary of $K$ in its span $L$. There is a nonzero $x\in L$ with $x^{\top }y\ge 0$ for every $y\in K$ and $x^{\top }{A}_0=0$. Here is a direct construction of the supporting vector. Choose points $z_n\in L\setminus K$ converging to $A_0$, let $p_n$ be their nearest points in $K$, and set $x_n=(p_n-z_n)/\Vert p_n-z_n\Vert$. The minimization inequality for a nearest point gives $x_n^{\top }y\ge 0$ on $K$ and $x_n^{\top }{p}_n=0$. A subsequence of the unit vectors converges to a unit vector $x\in L$. Since $0\le x_n^{\top }{A}_0\le \Vert A_0-z_n\Vert$, it has the asserted support properties.

The test-weight characterization of $K$ now gives $x^{\top }{A}_1\ge 0$ almost surely. This random payoff cannot vanish almost surely: otherwise $x$ would annihilate every element of $D$, hence all of $L$, contradicting $x\in L$ and $\Vert x\Vert =1$. ThereforeThis supplies the positive-terminal-payoff alternative. If $K$ has zero-dimensional span, its relative boundary is empty, so that case requires no supporting vector.

If $A_0\notin K$, let $p$ be a nearest point of the closed convex cone $K$ to $A_0$. Existence follows by minimizing the continuous squared distance on a sufficiently large compact ball. The first-order minimization inequality is $(A_0-p{)}^{\top }(y-p)\le 0$ for every $y\in K$. Set $x=p-A_0$. Taking $y=0$ and $y=2p$, both in the cone, gives $x^{\top }p=0$. ConsequentlyThe cone's test-weight characterization again yields $x^{\top }{A}_1\ge 0$ almost surely. This is separation from a closed convex cone with the sign needed forThe strict benefit is already available at the initial date, even if the terminal payoff happens to be zero.

Suppose $A_0\in \mathrm{ri}K$. We first justify the finite-dimensional fact $\mathrm{ri}K\subseteq D$, since $D$ itself need not be closed. Around any relative-interior point $z$, choose a small simplex in $L$ whose vertices lie in $K$ and which contains $z$ strictly inside. Approximate its vertices by points of $D$, possible because $\overline{D}=K$. For close enough approximations, the barycentric coordinates of $z$ remain strictly positive; hence $z$ is a convex combination of points of $D$ and belongs to $D$. If $L=\{0\}$, the assertion is immediate.

Let $m=\mathbb{E}[w{A}_1]\in L$. Relative openness lets us choose $\epsilon >0$ such that $A_0-\epsilon m\in \mathrm{ri}K\subseteq D$. There is therefore a bounded nonnegative measurable $f$ satisfying $A_0-\epsilon m=\mathbb{E}[wf{A}_1]$. SetThen $\mathbb{E}[\nu A_1]=A_0$ and $\mathbb{E}\Vert \nu A_1\Vert \le \Vert f{\Vert }_{\infty }+\epsilon <\infty$. This proves existence with strict positivity even when the unweighted asset vector is not integrable.

It also proves the strictly positive barycentre cone lemma. Every strictly positive weighted barycentre must lie in $\mathrm{ri}K$: otherwise the supporting vector constructed for the relative boundary would have zero value at that barycentre while its strictly positive weighted payoff had positive expectation. Thus the relative interior is exactly the set of such barycentres.

Consider a binomial market with dates $0,\dots ,n$, bank account $B_k=b^k$ for $b>0$, and stock price $S_{k+1}=S_{k}u$ or $S_{k}d$, where $0<d<u$. Both moves have positive physical probability, and the information at date $k$ is the history of the first $k$ moves. Absence of arbitrage requires and is implied by $d<b<u$. If $b\le d$, borrowing cash to buy stock has nonnegative payoff with a strict gain in one state; if $b\ge u$, the reverse trade does. In the interior range, the risk-neutral probability of an up move isIt makes ${\mathbb{E}}_Q[S_{k+1}\mid {\mathcal{F}}_k]=b{S}_k$, so $S_k/B_k$ is a martingale. This proves absence of arbitrage by pricing discounted gains under the equivalent measure.

At a node with stock price $s$, let the desired next-period portfolio values be $V_u,V_d$. Solve the two replication equations $\Delta su+\beta B_{k+1}=V_u$ and $\Delta sd+\beta B_{k+1}=V_d$. The replicating portfolio in a binomial market hasIts current value isRebalance at the next node using exactly the value supplied by the previous holdings, so the strategy is self-financing. Backward induction from any terminal contingent claim $H$ gives its hedge and the unique priceThus the finite binomial market is a complete market. The construction works for path-dependent claims on the full history tree, even when their values do not recombine merely as functions of the current stock price. For $H=f(S_n)$, independence under $Q$ gives the explicit time-zero priceThe stock hedge is a difference quotient of successor values, the discrete analogue of the delta hedge.

Suppose under the physical measure $P$ the up indicators are independent with probability $p\in (0,1)$. If $N_k$ is the number of up moves through date $k$, a particular history has probability $p^{N_k}(1-p{)}^{k-N_k}$ under $P$ and $q^{N_k}(1-q{)}^{k-N_k}$ under $Q$. Thus the binomial-market probability density, the Radon-Nikodym derivative on ${\mathcal{F}}_k$, isIt is strictly positive. Its next multiplier has conditional expectation $p(q/p)+(1-p)((1-q)/(1-p))=1$, so $L_k$ is a mean-one martingale and $L_k={\mathbb{E}}_P[L_n\mid {\mathcal{F}}_k]$. These facts verify normalization and equivalence of the two probability measures.

The state-price density for terminal cash is ${\zeta }_n=L_n/B_n=b^{-n}{L}_n$. In particular $V_0={\mathbb{E}}_P[{\zeta }_{n}H]$. At an intermediate date the conditional change-of-measure identity givesIf physical up probabilities depend on the past, replace the constant-probability expression by the product over periods of $q/p_k$ on an up move and $(1-q)/(1-p_k)$ on a down move. Each multiplier still has conditional mean one, provided $0<p_k<1$, and the pricing measure retains conditional up probability $q$.

In a complete market, terminal-wealth optimization can first be solved over terminal payoffs and then implemented by replication. Let initial wealth be $x_0>0$, and take an increasing differentiable strictly concave utility function $U$ on $(0,\infty )$ satisfying the Inada conditions. The state-price budget constraint for a terminal wealth $X$ is ${\mathbb{E}}_P[{\zeta }_{n}X]\le x_0$. Monotonicity spends the whole budget at the optimum.

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. Choose $y>0$ to satisfy ${\mathbb{E}}_P[{\zeta }_{n}I(y{\zeta }_n)]=x_0$. In a finite tree with positive state probabilities, this left side is continuous and strictly decreasing from infinity to zero, so $y$ exists uniquely. The complete-market terminal utility optimizer isTo prove optimality, concavity gives $U(X)\le U(X^{\ast })+U^{\prime }(X^{\ast })(X-X^{\ast })=U(X^{\ast })+y{\zeta }_n(X-X^{\ast })$. Taking expectations and using the budget proves $\mathbb{E}U(X)\le \mathbb{E}U(X^{\ast })$. Strict concavity gives uniqueness of the optimal terminal wealth. The finite-market setting avoids additional infinite-state existence and integrability issues.

Replicate $X^{\ast }$ using the backward recursion of part (i). Its wealth process is $V_k=B_k{\mathbb{E}}_Q[X^{\ast }/B_n\mid {\mathcal{F}}_k]$, and the two-successor difference quotient gives its stock holdings. This self-financing strategy starts with $x_0$ by the budget equality and remains positive at every node, since it is the discounted conditional expectation of a strictly positive terminal wealth.

For logarithmic utility, $U(x)=\log x$, $I(z)=1/z$, the multiplier is $y=1/x_0$ and $X^{\ast }=x_0/{\zeta }_n=x_0{B}_n/L_n$. For CRRA utility with risk aversion $\gamma >0$, $\gamma \ne 1$, the normalized payoff isConstraints such as prohibiting short sales change the attainable terminal-payoff set and require a constrained optimization; the unconstrained complete-market formula uses the stated assumptions.

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 $\rho$.

The binomial risk-neutral pricing formula converges to $e^{-\rho T}{\mathbb{E}}_Q[f(S_T)]$ for bounded continuous terminal payoffs by convergence in distribution. For ordinary call and put payoffs, uniform stock second-moment bounds give the required uniform integrability and hence convergence despite unboundedness. Bounded payoffs with discontinuities only at events of zero limiting probability are handled by the same weak-convergence criterion. Corresponding continuous path functionals can be treated at the process level. Arbitrary discontinuous path claims do not follow from weak convergence without further conditions. In the smooth limiting model, the discrete hedging difference quotient becomes the delta hedge $V_x$, and its value satisfies the Black-Scholes equation.

Let $M_t={\sup }_{s\le t}{W}_s$ and write ${\varphi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$ for the Brownian terminal density. By the Brownian reflection principle, for $y<a$ the reflected paths that hit $a$ and end at $y$ have terminal density ${\varphi }_t(2a-y)$. HenceThe Girsanov theorem changes zero drift into drift $\nu$ using the finite-horizon density $e^{\nu W_t-{\nu }^{2}t/2}$. The finite-horizon maximum of Brownian motion with drift therefore has distribution functionIn the first integral term, complete the square to get a normal density with mean $\nu t$. In the second, substitute $z=2a-y$ and complete the square with mean $-\nu t$; the prefactor is $e^{2a\nu }$. Here $\Phi$ is the standard normal distribution function, $a>0$ and $t>0$.

Let $T_{a,b}=\inf \{s\ge 0:W_s=a+bs\}$, with $b$ real. Survival means that $W_s-bs$ stays below $a$. Tilt by $e^{\theta W_t-{\theta }^{2}t/2}$; under the tilted probability, $W_s$ has drift $\theta$. Applying the maximum formula with drift $\theta -b$ gives the exponential terminal moment below a linear Brownian boundaryThis holds for every real $\theta$. The maximum distribution has no atom at the positive barrier, so strict or weak survival inequalities give the same probability here.

For the hitting-time moment, first use drift $-b$ in the maximum formula and take the complementary probability:If $z_1=(-a-bt)/\sqrt{t}$ and $z_2=(-a+bt)/\sqrt{t}$, their standard normal densities satisfy $\varphi (z_1)=e^{-2ab}\varphi (z_2)$. Differentiating the two terms, with $z_1^{\prime }=(a-bt)/(2t^{3/2})$ and $z_2^{\prime }=(a+bt)/(2t^{3/2})$, yields the drifted Brownian first-passage densityIts total mass need not be one, because the upward-sloping boundary may never be reached.

For $2\theta \le b^2$, put $\kappa =\sqrt{b^2-2\theta }\ge 0$. Multiplication by $e^{\theta s}$ gives $e^{\theta s}{f}_b(s)=e^{a(\kappa -b)}{f}_{\kappa }(s)$. Integrating to $t$ produces truncated discounted Brownian first passage, with negative discount rate allowed when the square root remains real:This includes the equality case $\kappa =0$, where it becomes $2e^{-ab}\Phi (-a/\sqrt{t})$. The expression integrates only finite hitting times up to $t$; no value is assigned to an exponential at an infinite hitting time.

For the cash-at-hit digital call, use the risk-neutral measure in the Black-Scholes model and denote expiry by $T=t_0$. ThenHitting $c>S_0$ corresponds to $W_s^Q=a+bs$, where $a=\log (c/S_0)/\sigma >0$ and $b=({\sigma }^2/2-\rho )/\sigma$. The unit-cash payment is made at the hitting time, so risk-neutral valuation discounts at that time, not at expiry. Set $\theta =-\rho$ andThe identity on the right verifies the square-root condition even for negative interest rates. The time-zero price isThis is in pounds for the unit-pound contract. It differs from a maturity-paid digital call option, whose discounting and exercise event are different.

Write $T=t_0$ and $B_t=e^{-\rho (T-t)}$ for the risk-free asset paying one at $T$. Under the physical measure, $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, with $\sigma >0$, while $d{B}_t=\rho B_{t}dt$. A self-financing portfolio with predictable stock and bond holdings $g_t,h_t$ has value $V_t=g_t{S}_t+h_t{B}_t$ and gainsRebalancing therefore adds no outside money and withdraws none. This gains equation, rather than merely the value decomposition, defines self-financing.

Let $p(x,t)=xg(x,t)+B_{t}h(x,t)$. The Itô formula givesThe prescribed portfolio gains instead have diffusion coefficient $\sigma S_{t}g$ and drift $\mu S_{t}g+\rho B_{t}h$. Equality of diffusion coefficients requires $p_x=g$, equivalentlyUnder this condition, $p_{xx}=g_x$. Equality of drifts is then $p_t+\frac{1}{2}{\sigma }^2{x}^2{g}_x=\rho B_{t}h$. Since $p_t=x{g}_t+B_t{h}_t+\rho B_{t}h$, this becomesThese conditions are also sufficient by the same Itô formula. Necessity initially holds along the stock process. Its positive transition density on $(0,\infty )$ for every positive time, together with continuity of the coefficient functions, extends the identities to all $x>0$ and then to the time endpoints by continuity.

For a self-financing portfolio, substitute $g=p_x$ and $B_{t}h=p-x{p}_x$ into the drift identity to obtain the Black-Scholes equationConversely, a smooth solution of this equation gives a self-financing portfolio by choosing the delta hedgeIts value is $p$, and both coefficient identities follow directly.

The printed deduction needs this holdings qualification: the Black-Scholes value equation needs delta-compatible holdings. A value function alone cannot make an arbitrary prescribed decomposition self-financing. As a counterexample, take $p=0$, $g=1$, $h=-x/B_t$. The value function solves the Black-Scholes equation, but this zero-value portfolio has gains $d{S}_t-\rho S_{t}dt$, which contain the nonzero term $\sigma S_{t}d{W}_t$. Thus for the originally specified $g,h$, the exact equivalence is the pricing equation together with $g=p_x$, not the pricing equation alone.

For a claim paying the holder at rate $k(S_t)$, let $p$ denote its remaining, ex-dividend value. A replicating portfolio pays out $k(S_t)dt$, so its net wealth equation is $dp=gdS+hdB-k(S_t)dt$. The delta hedge is still $g=p_x$, but the Black-Scholes equation becomesIf there is a terminal payoff $f$, set $p(x,T)=f(x)$; with no terminal payment set it to zero. The corresponding risk-neutral valuation isEach intervening payment is discounted from its own payment time.

Let $T=t_0$, $\Delta =T-t$ and assume the payoff has sufficient integrability for risk-neutral valuation. Under the risk-neutral measure $Q$, the Black-Scholes model has $d{S}_u=\rho S_{u}du+\sigma S_{u}d{W}_u^Q$, so conditionally on $S_t=x$,Absence of arbitrage gives the time-$t$ priceFor a regular payoff this is the solution of the Black-Scholes equation with terminal value $f$, and the delta hedge $g=V_x$ replicates it. Suitable limits or conditional-expectation representations extend the price formula beyond smooth terminal payoffs.

For the logarithmic stock payoff, put $Y=(\rho -{\sigma }^2/2)\Delta +\sigma \sqrt{\Delta }Z$. Its Gaussian moment-generating function is $\mathbb{E}{e}^Y=e^{\rho \Delta }$, while differentiating that moment gives $\mathbb{E}[Y{e}^Y]=(\rho +{\sigma }^2/2)\Delta e^{\rho \Delta }$. ThereforeThe option delta is $\log x+1+(\rho +{\sigma }^2/2)\Delta$, and the second stock derivative is $1/x$. The Black-Scholes equation can also be checked directly from $V_t=-x(\rho +{\sigma }^2/2)$.

For general convex $f$, let $Y_{+}=\exp ((\rho -{\sigma }^2/2)\Delta +\sigma \sqrt{\Delta }Z)>0$. For $0\le \lambda \le 1$, convexity givesTaking the positive discounted expectation proves that $V(\cdot ,t)$ is convex. Its derivative, where defined, is therefore nondecreasing, proving that the stock holding in the delta hedge rises weakly with the stock price. For twice differentiable $f$, differentiation gives $V_{xx}=e^{-\rho \Delta }\mathbb{E}[Y_{+}^2{f}^{\prime \prime }(x{Y}_{+})]\ge 0$. For a merely convex payoff, Gaussian smoothing at $t<T$ and the usual growth assumptions give the same conclusion; one-sided derivatives are monotone even without smoothing. Strict increase is not implied by convexity alone: an affine payoff has constant option delta.

For the random constant Gaussian interest-rate mixture, represent the rate as $\rho ={\rho }_0+\tau Y$, with $Y,Z$ independent standard normal variables. This independence simply represents averaging the constant-rate price function over its Gaussian parameter. Completing the square in $Y$ yields the identityApply it inside the time-zero price formula:The combined normal term has variance ${\tau }^2{T}^2+{\sigma }^{2}T$. Define ${\rho }_{\ast }={\rho }_0-{\tau }^{2}T/2$ and ${\sigma }_{\ast }^2={\sigma }^2+{\tau }^{2}T$. Then the discount factor is $e^{-{\rho }_{\ast }T}$, and the log-price mean is $({\rho }_{\ast }-{\sigma }_{\ast }^2/2)T$. Thus the expectation is exactly the constant-parameter Black-Scholes model price with those parameters:The argument applies to nonnegative payoffs by Tonelli theorem, and to signed payoffs whenever absolute integrability justifies the expectations. It averages constant-rate prices; it does not assume a stochastic short-rate process.

Use a risk-neutral measure $Q$, and let $P_{s,t}$ be the price at time $s$ of a zero-coupon bond paying one at its maturity $t$. The information available at $s$ is the filtration ${\mathcal{F}}_s$. For positive differentiable bond prices, the instantaneous forward rate iswith $P_{t,t}=1$. The short rate is $R_s=f(s,s)$. The continuous-time bank account $B_s=\exp ({\int }_0^s{R}_{u}du)$ is the discounting numéraire.

If $P_{s,t}/B_s$ is a true martingale for each fixed maturity, then its terminal value is $B_t^{-1}$ andMultiplication by the ${\mathcal{F}}_s$-measurable $B_s$ givesConversely this conditional-expectation formula makes the discounted bond a martingale by the tower property of conditional expectation. The equivalence uses integrability and true martingales, rather than merely local martingales.

Forward-rate modelling specifies the stochastic evolution of $f(s,t)$ as $s$ varies at fixed maturity. For example, with one Brownian driver, write $df(s,t)=\alpha (s,t)ds+\beta (s,t)d{W}_s^Q$. Differentiation of the integral representation and Itô formula give the bond's relative volatility $-{\int }_s^t\beta (s,u)du$ and drift $R_s-{\int }_s^t\alpha (s,u)du+\frac{1}{2}({\int }_s^t\beta (s,u)du{)}^2$. The discounted-bond condition therefore imposesor, after differentiation in maturity, $\alpha (s,t)=\beta (s,t){\int }_s^t\beta (s,u)du$. This illustrates how the no-arbitrage condition constrains forward-rate drift.

For the Gaussian short-rate model with constant coefficients, let $d{R}_s=\theta ds+\sigma d{W}_s^Q$ and fix $\Delta =t-s$. Future Brownian increments are independent of ${\mathcal{F}}_s$, givingThe last stochastic integral is Gaussian with mean zero and variance ${\Delta }^3/3$, by the Itô isometry. Its Gaussian moment-generating function therefore givesWhen $\sigma \ne 0$ and the initial rate is deterministic, the Brownian and short-rate filtrations coincide because $W_s^Q=(R_s-R_0-\theta s)/\sigma$. If $\sigma =0$, the same bond formula holds deterministically, while the asserted filtration equivalence is unnecessary.

For a deterministic drift function ${\theta }_v$, the same calculation givesLet the observed initial curve be $P_{0,t}^{\mathrm{obs}}>0$, normalized by $P_{0,0}^{\mathrm{obs}}=1$, and define $f_0(t)=-\frac{d}{dt}\log P_{0,t}^{\mathrm{obs}}$. For a differentiable initial forward curve chooseIndeed the model's initial forward rate is $R_0+{\int }_0^t{\theta }_{v}dv-{\sigma }^2{t}^2/2=f_0(t)$, so integrating it gives exactly the observed bond curve. This is forward-curve calibration of a shifted Brownian short rate, with $R_t=f_0(t)+{\sigma }^2{t}^2/2+\sigma W_t^Q$. Every finite collection of positive bond quotes can first be fitted by a smooth positive curve; an arbitrary entire curve requires the stated smoothness or an appropriate weaker integral interpretation. The freedom is in the deterministic drift, and the initial short rate must match the initial forward rate.