Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 29 (Advanced Financial Models)

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

Take bank-account values $B_k=R^k$ with $R>0$, and let the stock multiply independently by $u$ or $d$ at each step, where $0<d<R<u$. Under the physical probability measure $P$, the up probability is $p\in (0,1)$; the filtration records the successive moves. These assumptions specify the standard binomial market. If $R$ lies outside the strict interval $(d,u)$, a long or short stock position financed through the bank account gives an arbitrage.

The risk-neutral probability in a binomial market is fixed by $qu+(1-q)d=R$, givingFor successor claim values $V_u,V_d$ at a node with current stock price $S$, solve the two replication equations. The current stock holding and current cash value areThe cost of this replicating portfolio in a binomial market isApply this backward option pricing recursion from any terminal contingent claim $H$, including claims depending on the whole sequence of moves. It constructs a self-financing portfolio and proves market completeness by induction on the remaining number of periods. Taking successive conditional expectations under the measure $Q$ with independent up probability $q$ gives the unique risk-neutral pricing valueAny other price would give an arbitrage by buying the cheaper of the claim and its replica and selling the dearer.

If $K$ is the number of up moves, a particular path has probabilities $p^K(1-p{)}^{n-K}$ under $P$ and $q^K(1-q{)}^{n-K}$ under $Q$. Therefore the binomial-market probability density, namely the Radon-Nikodym derivative, isThis last expression makes $Z$ a function of the terminal stock price. Summing over paths gives ${\mathbb{E}}_{P}Z=1$, and strict positivity proves equivalence of the two probability measures.

For unrestricted terminal portfolio wealth $H$, its initial replication cost is $R^{-n}{\mathbb{E}}_{Q}H$. Hence feasible wealth satisfies the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. For the stated exponential utility, $v^{\prime }(x)=e^{-ax}$ and $v^{\prime \prime }(x)=-a{e}^{-ax}<0$. The binomial exponential-utility terminal wealth iswhereIt satisfies the budget, and its marginal utility is $v^{\prime }(H^{\ast })=\lambda Z$ with $\lambda =\exp (-a{R}^n{w}_0-{\mathbb{E}}_Q\log Z)>0$. The concave supporting-tangent inequality givesTaking expectations under $P$ makes the last term zero by the budget. Strict concavity makes equality possible only when $H=H^{\ast }$ in every positive-probability state. The already constructed complete market supplies its replicating strategy, so this proves global optimality and uniqueness. The formula allows negative final wealth, as permitted by the unrestricted trading problem.

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.

Write $T=t_0$, $B_t=e^{-\rho (T-t)}$, and let $V_t=f(S_t,t)$ with $f(x,t)=xg(x,t)+B_{t}h(x,t)$. The portfolio holds $g(S_t,t)$ shares of stock and $h(S_t,t)$ units of the zero-coupon bond. It is self-financing when all changes of holdings are funded within it, which is the gain identityUnder the physical probability measure, write $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, where $\sigma >0$ and $d{B}_t=\rho B_{t}dt$. The Itô formula givesMatching the Brownian motion coefficients forces $f_x=g$. Since $f_x=g+x{g}_x+B_t{h}_x$, this is preciselyDifferentiating $f_x=g$ gives $f_{xx}=g_x$. Since $f_t=x{g}_t+\rho B_{t}h+B_t{h}_t$, matching the drift coefficients now givesThese are the self-financing conditions for smooth stock and bond holdings. Conversely, substituting both identities into the Itô formula gives the gain identity, so they are sufficient as well. Smoothness and the positive lognormal distribution density of $S_t$ extend the identities along the price process to every $x>0$, with boundary times obtained by continuity.

The portfolio with constant stock value in bond units must have $xg(x,t)=\theta B_t$, so $g(x,t)=\theta B_t/x$. The first condition then gives $h_x=\theta /x$, henceThe second condition reduces to $c^{\prime }(t)=\theta ({\sigma }^2/2-\rho )$. Initial portfolio wealth requires $h(S_0,0)=w_0/B_0-\theta =w_0{e}^{\rho T}-\theta$. Therefore the holdings areThe corresponding portfolio wealth isUnder the risk-neutral measure, $\log (S_t/S_0)=(\rho -{\sigma }^2/2)t+\sigma W_t^Q$, so its value in zero-coupon bond units is $V_t/B_t=w_0{e}^{\rho T}+\theta \sigma W_t^Q$. This also verifies the self-financing gain equation after using the bond as numéraire.

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 $T$ for the fixed maturity and $s\le T$ for the observation time. The original PDF's discounting uses the instantaneous forward rate over $[s,T]$ and the short rate over $[0,s]$. Let $X(s,u)=F_{s,u}-{\mu }_{s,u}$ be the centered Gaussian forward-rate field, and defineAssume the covariance and mean have the regularity needed for these mean-square integrals and the maturity differentiation below. Splitting $Y$ at $u=s$ shows that the discounted zero-coupon bond price isThis integrated Gaussian forward-rate process has deterministic varianceIts independent increments hold relative to the entire specified filtration. Indeed, if $r\le s$ and $z\le r$, the covariance between $Y(s,T)-Y(r,T)$ and any earlier $X(z,w)$ is the integral ofSince uncorrelated jointly Gaussian variables are independent, the increment is independent of every finite family of earlier observations and hence of their generated sigma-algebra. The increment has normal distribution with mean zero and variance $v(s,T)-v(r,T)$. Also $Y(0,T)=0$ because $c(0,u,w)=0$.

The Gaussian moment-generating function now givesTherefore the Gaussian forward-rate covariance drift restriction is exactly the deterministic conditionIt turns $Z_{s,T}/Z_{0,T}$ into a Gaussian exponential martingale with deterministic variance. The subparts establish the requested equivalent descriptions from this common calculation.

If the discounted zero-coupon bond price is a martingale, its expectation is $Z_{0,T}=e^{-A(0,T)}$. The Gaussian moment-generating function applied to the representation above gives $\mathbb{E}{Z}_{s,T}=e^{-A(s,T)+v(s,T)/2}$, soDifferentiate in maturity $T\ge s$. The two mean derivatives are ${\mu }_{s,T}$ and ${\mu }_{0,T}$. Symmetry of the covariance in its last two arguments makes the two boundary terms in the derivative of the double integral equal. HenceWe obtain the Gaussian forward-rate covariance drift restrictionwhich proves that the martingale condition implies the mean restriction. Only maturity differentiation is used; a time derivative of the covariance is unnecessary.

Assume the displayed Gaussian forward-rate covariance drift restriction. Its diagonal instance givesUsing it on the two parts of the deterministic discounted exponent yieldsThe last equality uses covariance symmetry to reflect the triangular integration region across its diagonal. The independent increments of the integrated Gaussian forward-rate process and its conditional Gaussian moment-generating function then give, for $r\le s\le T$,All these prices are integrable because their exponents are Gaussian random variables with finite variance. Thus the mean restriction implies the discounted bond martingale condition, completing the equivalence of the first two descriptions.

Let $D_s=\exp (-{\int }_0^s{R}_{u}du)$, where $R_u=F_{u,u}$ is the short rate. At maturity the zero-coupon bond pays one, so $Z_{T,T}=D_T$. If the discounted bond price is a martingale, its terminal conditional expectation isThe positive factor $D_s$ is measurable with respect to the current filtration. Dividing it out givesConversely, this identity says $Z_{s,T}=\mathbb{E}[D_T\mid {\mathcal{F}}_s]$. The tower property gives $\mathbb{E}[Z_{s,T}\mid {\mathcal{F}}_r]=Z_{r,T}$ for every $r\le s$, establishing the martingale condition. The integrated short rate is a Gaussian random variable with finite variance, so the Gaussian moment-generating function proves that $D_T$ is integrable. Together with the preceding subparts, this proves all three descriptions equivalent.