Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 35 (Advanced Financial Models)

Normalize by a riskless bank account with $B_0=1$, $B_1=R>0$. For $d$ risky assets let $Y=S_1/R-S_0$ be the vector of discounted gains, and let $H$ denote a discounted square-integrable contingent claim. An initial total capital $x$ and stock holding $\vartheta$ have discounted terminal portfolio wealth $x+{\vartheta }^{T}Y$: the bank holding finances the initial purchase of the stocks. Thus the space of attainable claims isIt is finite-dimensional and hence closed. The market is complete market when every square-integrable claim is in this span; with finitely many assets this is also equivalent to replication of every bounded claim.

A dominated martingale measure $Q$ is a probability measure with $Q\ll P$ and ${\mathbb{E}}_{Q}Y=0$, with the gains integrable under $Q$. Its Radon-Nikodym derivative satisfies $Z\ge 0$, $\mathbb{E}Z=1$ and $\mathbb{E}[ZY]=0$. An equivalent martingale measure has $Z>0$ almost surely. A signed martingale measure obeys the same normalization and pricing equations without requiring nonnegativity. The distinction matters for quadratic hedging. For the completeness assertion use the usual no-arbitrage hypothesis, expressed here by existence of an equivalent martingale measure $Q_0$.

The unrestricted one-period quadratic hedge minimizesPut $m=\mathbb{E}Y$, $C=\mathrm{Cov}(Y)$ and $k=\mathrm{Cov}(Y,H)$. Remove redundant gains so that $C$ is invertible. Differentiating in $x$ gives $x=\mathbb{E}H-{\vartheta }^{T}m$. The remaining squared error isThereforeThe residual is orthogonal in $L^2(P)$ to $1$ and every gain, making the fitted claim the orthogonal projection onto a finite-dimensional subspace onto $\mathcal{A}$. Its minimum mean-square error is $\mathrm{Var}(H)-k^T{C}^{-1}k$, and it vanishes exactly when $H$ is attainable. If initial capital $x$ is fixed instead, the normal equation isThis separates hedging with prescribed capital from simultaneous optimization of capital and risky holdings.

The minimal martingale measure in a one-period market is characterized by leaving unchanged the expectation of every centered square-integrable $L$ orthogonal to the martingale part $Y-m$. For square-integrable pricing densities this forces $Z-1$ into the span of $Y-m$: the orthogonal complement consists exactly of those $L$. The pricing equation then determinesIndeed $\mathbb{E}{Z}_{\ast }=1$ and $\mathbb{E}[Z_{\ast }Y]=m-C{C}^{-1}m=0$. For every such orthogonal $L$, $\mathbb{E}[Z_{\ast }L]=0$. Conversely a normalized density preserving all these $L$ must have the displayed form, since its coefficients must satisfy $C\beta =-m$. The least-squares initial capital is consequently $x_{\ast }=\mathbb{E}[Z_{\ast }H]$.

For any other square-integrable signed pricing density $Z$, put $L=Z-Z_{\ast }$. Both densities price $1$ and all gains, so $L$ is orthogonal to their span, including $Z_{\ast }$. HenceEquality holds only when $Z=Z_{\ast }$. Densities with infinite second moment cannot improve this finite minimum. This proves the requested minimum over dominated martingale measures whenever $Z_{\ast }\ge 0$, and over all square-integrable signed martingale measures without that extra condition. If $Z_{\ast }>0$, it is also an equivalent martingale measure.

Absence of arbitrage alone does not guarantee that the minimal density is a probability density. For a concrete example, let a single discounted gain take values $-1,1,10$ with probabilities $1/10,4/5,1/10$. Then $m=17/10$, $C=801/100$ and $Z_{\ast }(10)=-610/801$. Nevertheless pricing probabilities $(109/200,89/200,1/100)$ are strictly positive and give mean gain zero, so an equivalent martingale measure exists. This is negative minimal density in an arbitrage-free one-period market. If the term martingale measure is reserved for probabilities, the printed minimum assertion needs the extra nonnegativity hypothesis; otherwise it is the signed minimum-norm assertion just proved. Minimizing over the nonnegative pricing densities instead is a different constrained projection problem.

Finally, prove completeness and uniqueness of dominated martingale measures. If the market is complete market, every event indicator can be replicated. Under any dominated martingale measure, its expectation equals its initial discounted cost. Consequently any two such measures agree on every event and are equal; existence is supplied by $Q_0$.

Conversely suppose $\mathcal{A}$ has dimension $k$ and is a proper subspace of the whole payoff space. Choose $k+1$ disjoint events $A_i$ of positive probability. Such events exist whenever $L^2(P)$ has dimension greater than $k$: otherwise the space has at most $k$ atoms and dimension at most $k$. Let $f_1,\dots ,f_k$ be a basis of $\mathcal{A}$. The $k+1$ vectorsare linearly dependent. Choose coefficients $a_i$, not all zero, whose linear combination is zero, and set $h={\sum }_i{a}_i{1}_{A_i}$. It is bounded, not almost surely zero under $Q_0$, and satisfies ${\mathbb{E}}_{Q_0}[hf]=0$ for every $f\in \mathcal{A}$, in particular for $1,Y^1,\dots ,Y^d$. For $0<\epsilon <\Vert h{\Vert }_{\infty }^{-1}$ defineThese densities are strictly positive, normalized, have zero gain means, and are different. Thus incompleteness contradicts uniqueness, provingThe existence hypothesis cannot be discarded: with three positive-probability states and gain $Y=0,1,2$, the sole dominated martingale probability concentrates on $Y=0$, yet the two-dimensional attainable space does not span all three states. That market permits arbitrage.

Take a riskless gross return $R>0$ per period and a stock whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in (0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,A self-financing portfolio chooses each stock holding from information available before the next move, and invests the remainder in the bank account. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the binomial market complete.

The unique risk-neutral probability of an up move is determined by the discounted-stock martingale equation:The risk-neutral measure assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p{)}^{n-K}$ and pricing probability $q^K(1-q{)}^{n-K}$. Its Radon-Nikodym derivative is thereforeSince $S_n=S_0{d}^n(u/d{)}^K$, eliminate $K$ to obtain the requested terminal-price expression:The pricing density depends only on the terminal stock price even though different paths to that price have different intermediate histories.

For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfiesThe utility function has derivative $v^{\prime }(x)=x^{-(\gamma -1)/\gamma }$. Put $\eta =\gamma /(\gamma -1)>1$. Its first-order condition makes $v^{\prime }(X)$ proportional to $Z$, so $X=C{Z}^{-\eta }$. Choosing $C$ to satisfy the pricing budget constraint gives the binomial power-utility terminal wealthAll states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence givesCombining these two formulas with the terminal-price expression for $Z$ gives $X_{\ast }$ entirely as a function of $S_n$.

To prove global optimality, strict concavity gives $v(X)-v(X_{\ast })\le v^{\prime }(X_{\ast })(X-X_{\ast })$. Because $v^{\prime }(X_{\ast })=C^{-1/\eta }Z$, the expectation of the right side is zero for any other payoff satisfying the same pricing budget constraint. Equality requires $X=X_{\ast }$ almost surely. Backward replication supplies its self-financing strategy, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^n{w}_0$.

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

A self-financing portfolio changes its holdings using only its existing wealth. If it holds $g_t$ shares and $h_t$ units of a traded bond or bank account $B_t$, its value $V_t=g_t{S}_t+h_t{B}_t$ obeys the gain identityThe changes in holdings do not add or withdraw capital. Set $T=t_0$, $B_t=e^{-\rho (T-t)}$, so $d{B}_t=\rho B_{t}dt$, and let $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, with $\sigma >0$. Write $V_t=p(S_t,t)$ and suppress the arguments of the smooth holdings $g,h$.

Differentiate $p=xg+B_{t}h$:The Itô formula minus the proposed portfolio gains givesUniqueness of the local-martingale and finite-variation decomposition shows that self-financing requires $p_x=g$. Since $S_t$ has full support on $(0,\infty )$ for each $t>0$, smoothness gives the corresponding functional identity for every positive $x$, and continuity extends it to the interval endpoints. ThusDifferentiating this identity once more in $x$ gives $g_x+x{g}_{xx}+B_t{h}_{xx}=0$, so $p_{xx}=g_x$. Substitute both identities into the drift condition to obtainConversely these equations give $p_x=g$, $p_{xx}=g_x$ and zero drift and diffusion in the displayed gain difference. They therefore prove the claimed self-financing conditions for smooth stock and bond holdings in both directions.

Using $B_{t}h=p-xg$ in the gain identity and setting $g=p_x$, self-financing impliesThis is the Black-Scholes equation, independent of the physical drift $\mu$. Conversely, if a smooth $p$ satisfies it, define its delta hedge byThe preceding Itô formula verifies that these holdings are self-financing.

For prescribed arbitrary holdings, the value equation alone is insufficient. The corrected equivalence is the Black-Scholes value equation needs delta-compatible holdings: a specified portfolio is self-financing if and only if its value solves the equation and $g=p_x$. Indeed $p=0$, $g=1$, $h=-x/B_t$ satisfies the value equation but its proposed gains are $d{S}_t-\rho S_{t}dt$, which have nonzero Brownian part while $dp=0$. Thus the printed final equivalence is valid as an existence assertion for the delta hedge, not for every decomposition of the same value. No change to the valid two holdings equations is needed.

With continuous underlying stock dividends, the stock's gain is $d{S}_t+\theta S_{t}dt$, so self-financing becomes $d{V}_t=g_t(d{S}_t+\theta S_{t}dt)+h_{t}d{B}_t$. Under the risk-neutral measure the ex-dividend stock drift is $(\rho -\theta )S_t$. Matching its diffusion still gives $g=p_x$, and matching its gain drift gives the Black-Scholes equation with continuous stock dividendsOnly the coefficient of $x{p}_x$ changes. The $-\rho p$ discount term remains, because the claim is financed at the bank rate; there is no separate source term for a dividend paid by the claim itself. The dividends here are paid by the underlying stock.

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.