Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 39 (Advanced Financial Models)

For the autoregressive process of order one, the formula at $t=0$ reads $r_0=r_0$, with an empty sum. If it holds at time $t$, substitution into the recursion givesThus mathematical induction provesNo restriction such as $|\beta |<1$ is needed for this finite-time identity.

Put $n=T-t$ and define the finite geometric seriesStarting the autoregressive process of order one at time $t$ gives $r_{t+k}={\beta }^k{r}_t+{\sum }_{j=1}^k{\beta }^{k-j}{\xi }_{t+j}$. ConsequentlyThe future innovations are independent and identically distributed random variables, independent of ${\mathcal{F}}_t$. Factorizing their exponential conditional expectation and using the cumulant-generating function $K$ yieldsThus the exponential-affine bond pricing coefficients areThey vanish at maturity, giving $P(T,T)=1$. Everywhere finiteness of $K$ ensures that each zero-coupon bond price is positive and finite, including when $\beta$ is zero, negative, or one.

Let $q(T)$ denote the original zero-coupon bond price $P(0,T)$ computed above. A deterministic shift $h_t$ of the short rate, with $h_0=0$, gives ${\hat{r}}_t=r_t+h_t$. Subtracting the two recursions shows that the required added drift isSince the shift is nonrandom, the new discount factor differs by $\exp (-{\sum }_{s=1}^T{h}_s)$, and hence $\hat{P}(0,T)=q(T)\exp (-{\sum }_{s=1}^T{h}_s)$.

For deterministic calibration of an autoregressive short rate, defineAll these quantities are finite. The telescoping sum ${\sum }_{s=1}^T{h}_s=d_T$ gives $\hat{P}(0,T)=p(T)$. Explicitly,These constants fit every maturity simultaneously. Each successive maturity fixes $h_T$ uniquely, so the deterministic calibration is also unique with the prescribed $r_0$.

The square root is concave on $[0,\infty )$, and $\sqrt{S_T}$ is integrable since $\mathbb{E}\sqrt{S_T}\le \sqrt{\mathbb{E}{S}_T}$. For $t\le T_1\le T_2$, the conditional Jensen inequality and the martingale property giveTaking conditional expectations with respect to ${\mathcal{F}}_t$ and using the tower property of conditional expectation yieldsThus the maturity curve for a square-root stock claim is nonincreasing; it need not be strictly decreasing, as a constant stock gives equality. The inequality holds for every ordered pair of maturities. Under the usual right-continuous versions of the market processes, it gives the corresponding nonincreasing version of the maturity curve.

Solving the geometric Brownian motion equation over $[t,T]$ givesThe Brownian increment is independent of ${\mathcal{F}}_t$ and has a normal distribution with variance $T-t$. Its exponential moment-generating function therefore givesThe required deterministic price function isThis includes $s=0$, $v=0$ and $t=T$. For $s>0$ and $t<T$, it is continuous and strictly decreasing in $v\ge 0$, with range $(0,\sqrt{s}]$.

Let $\tau =T-t>0$ and $I_{t,T}={\int }_t^T{\sigma }_s^{2}ds$. The stochastic exponential solution givesBoundedness of the predictable process $\sigma$ implies the Novikov condition on every finite horizon, so this exponential has conditional expectation $\mathbb{E}[M_{t,T}\mid {\mathcal{F}}_t]=1$. The pathwise bounds $a^2\tau \le I_{t,T}\le b^2\tau$ and positivity of $M_{t,T}$ giveThis proves the required bound in the independent-volatility case, and the exponential argument in fact does not need independence. In the usual filtration generated by Brownian motion and an independent volatility history, one can alternatively condition on the whole volatility path: the Brownian integral is then conditionally Gaussian with variance $I_{t,T}$. The result isthe conditional square-root price under independent volatility. The outer conditional expectation cannot in general be removed.

On $\{S_t>0\}$, these positive price bounds justify the unique square-root stock implied volatilityThere is a genuine zero-price qualification: if $S_t=0$, the nonnegative stock is absorbed at zero and both claim prices are zero for every volatility parameter. The price equation then does not determine a unique $\Sigma$. One may set $\Sigma =a$ on that event to retain the bounds; identification by inversion requires $S_t>0$.

Fix the horizon $T$ and define the half-volatility measure for a square-root stock claim byThe Novikov condition makes $Z$ a true martingale with $\mathbb{E}{Z}_T=1$, and $Z_T>0$ makes $Q$ equivalent to $P$. The Girsanov theorem identifies $W_s^Q=W_s-\frac{1}{2}{\int }_0^s{\sigma }_{u}du$ as a Brownian motion under $Q$.

The exact factorization from the previous part isUsing the Bayes formula for conditional expectation therefore givesThe random integrated variance need not have any independence property under $Q$. Its pathwise bounds still imply $e^{-b^2\tau /8}\le {\mathbb{E}}_Q[e^{-I_{t,T}/8}\mid {\mathcal{F}}_t]\le e^{-a^2\tau /8}$. Inverting the same price function proves $a\le \Sigma (t,T)\le b$ even for volatility adapted to the Brownian motion, on the positive-price event. The zero-price convention is as above. The auxiliary measure here is not asserted to be an equivalent martingale measure for the stock.

Backward induction first proves that the Snell envelope is adapted and integrable. At each step, $|U_t|\le |Y_t|+\mathbb{E}[|U_{t+1}|\mid {\mathcal{F}}_t]$, so integrability follows from that of $Y_t$ and $U_{t+1}$. Its defining maximum immediately giveswhich proves that $U$ is a supermartingale.

If $Y$ is a submartingale, the sharper backward identity isIt holds at $T$. If it holds at $t+1$, the tower property of conditional expectation gives $\mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]=\mathbb{E}[Y_T\mid {\mathcal{F}}_t]$. Iterating the submartingale inequalities shows that this is at least $Y_t$, so the maximum chooses this continuation value. The displayed identity then makes $U$ a martingale.

Let $\nu$ be the common increment law. The random walk has the Markov property, since ${\xi }_{t+1}$ is independent of ${\mathcal{F}}_t$ and $X_{t+1}=X_t+{\xi }_{t+1}$. Its transition operator is $Ph(x)=\int h(x+z)\nu (dz)$ wherever that integral exists.

Set $V(T,x)=f(x)$. For $1\le t<T$, suppose the Borel function $V(t+1,\cdot )$ has been constructed and represents $U_{t+1}$. On the Borel setdefine the optimal stopping value function recursively byOutside $A_t$, assign the finite Borel value $f(x)$. Measurability of integrals against a fixed probability law proves measurability of both the set and the function. Integrability of $U_{t+1}=V(t+1,X_t+{\xi }_{t+1})$ and independence imply $X_t\in A_t$ almost surely andThe Snell recursion thus proves $U_t=V(t,X_t)$ by induction.

At time zero the generated sigma-algebra is trivial, so $U_0$ is deterministic. Since $X_0=0$, define $V(0,x)=U_0$ for all $x$, withwhen $T\ge 1$. This supplies a globally finite deterministic representation without assuming integrability of rewards from every unvisited starting state. HenceWith the additional statewise integrability normally used to define a value function for arbitrary starting states, one can instead use the Bellman maximum at time zero too. For $T=0$, the representation is simply $U_0=f(0)$.

The key convexity calculation is that the random-walk transition operator preserves convex functions. If $h$ is convex, then for $0<\theta <1$,The pointwise maximum of convex functions is convex as well: the convexity inequality holds for each candidate and is bounded above by the convex combination of the two endpoint maxima. Thus the Bellman recursion proves convexity of a random-walk Snell value function by backward induction from $V(T,\cdot )=f$.

To ensure that the functions in this argument are genuinely finite, rather than silently imposing a growth bound on $f$, use integrability of translated convex random-walk rewards. Write $Z_j$ for a sum of $j$ independent increments with the same law. For $j\le T-1$, the assumptions give integrability of $f(Z_j)$ and $f(Z_{j+1})$. The negative part of a finite convex function is Lipschitz, so translating $Z_j$ preserves integrability of the negative part. For the positive part and $x>0$, if the increment law is unbounded above, take an independent increment $\eta$ and $p=\mathbb{P}(\eta \ge x)>0$. On that event $Z_j+x$ lies between $Z_j$ and $Z_j+\eta$, so convexity givesIf increments are bounded above by $M$, then $Z_j\le jM$ and the same convexity argument between $Z_j$ and $jM+x$ bounds the positive part by $f(Z_j{)}^{+}+f(jM+x{)}^{+}$. For $x<0$, use the corresponding lower-tail argument. Thus every translated reward $f(x+Z_j)$ is integrable for $j\le T-1$.

For $t\ge 1$, backward induction also gives the boundIt justifies every convolution in the Bellman recursion at these times, so the arbitrary assignments outside $A_t$ in the previous part are never needed when $f$ is convex. The displayed convexity calculation therefore proves that each $V(t,\cdot )$ for $t\ge 1$ is a finite convex function. The chosen $V(0,\cdot )=U_0$ is constant and hence convex. This proves the requested conclusion for the deterministic representation under the finite-horizon assumptions.

There is a distinction between this representation and a value function for every possible starting point at time zero. Without translated integrability at the full horizon, the latter may take $+\infty$ at unused states. For example, with $T=1$, $f(x)=e^{x^2}$ and $\mathbb{P}({\xi }_1=k)=c{e}^{-k^2-k}$ for $k=1,2,\dots$, the observed reward is integrable, but $\mathbb{E}[f(x+{\xi }_1)]=c{e}^{x^2}{\sum }_{k\ge 1}{e}^{(2x-1)k}$ is infinite for $x\ge 1/2$. The constant time-zero extension avoids claiming a globally finite Bellman function under insufficient assumptions. If integrability is assumed for all times of an infinite random walk, the translated-reward argument with one extra time also makes the full time-zero Bellman function finite and convex.

On a fixed horizon $T$, an arbitrage is an admissible trading strategy that is self-financing, has $X_0=0$, and satisfies $X_T\ge 0$ almost surely with $\mathbb{P}(X_T>0)>0$. Here admissibility means that wealth in units of the strictly positive riskless numéraire has a deterministic lower bound throughout the horizon. An equivalent martingale measure $Q$ is equivalent to the physical measure on ${\mathcal{F}}_T$ and makes each discounted risky price ${\tilde{S}}^i=S^i/B$ a martingale.

For a predictable, stochastically integrable risky holding vector $\pi$, the discounted self-financing identity isIndeed, since the riskless account is a finite-variation process, its quadratic covariations vanish, and the Itô formula for $X/B$, combined with $X=\varphi B+\pi \cdot S$ and $dX=\varphi dB+\pi \cdot dS$, gives $d(X/B)=\pi \cdot d(S/B)$.

Use two standard facts from stochastic calculus: a stochastic integral against a continuous local martingale is a local martingale, and a local martingale bounded below by a deterministic constant is a supermartingale. The latter follows by adding the constant to make the process nonnegative, stopping at a localizing sequence, and applying the conditional Fatou lemma. Thus admissible discounted wealth is a $Q$-supermartingale, and an arbitrage would satisfyIt follows that $X_T=0$ $Q$-almost surely, hence also physically almost surely by equivalence, a contradiction. Therefore an equivalent martingale measure rules out arbitrage among admissible self-financing strategies. The lower-bound restriction is essential to this argument; stochastic integrals need not be true martingales for arbitrary unbounded holdings.

The holdings ${\varphi }_t$ and ${\pi }_t^i$ are numbers of shares, so the marked-to-market portfolio value isThe stock prices are ex-dividend prices. Over the trading interval, existing holdings gain ${\varphi }_{t}d{B}_t$ from the riskless account and ${\pi }_t\cdot d{S}_t$ from capital-price changes. In addition, ${\pi }_t^i$ shares of asset $i$ receive ${\pi }_t^i{D}_t^{i}dt$ in cash dividends. If this cash remains in the portfolio and changes of holdings are financed internally, the self-financing portfolio equation isRebalancing does not add a separate source of wealth: the cost of purchasing one holding is paid by selling another or by using the cash account. This is why one must specify the gains equation, rather than infer self-financing solely by differentiating the marked-to-market identity. The dividends are reinvested rather than withdrawn as consumption.

Use the discounted wealth $\tilde{X}=X/B$ and the discounted dividend gainsThe integral uses $ds$: the printed $dt$ inside an integral indexed by $s$ is a differential typo. Since $B$ is a positive finite-variation process, the Itô product rule givesThe bank-account terms cancel using $X_t={\varphi }_t{B}_t+{\pi }_t\cdot S_t$.

If $Y$ is a $Q$-martingale, discounted self-financing wealth is a $Q$-local martingale. The admissible trading strategy lower bound makes it a supermartingale, exactly as in the previous no-arbitrage proof. A zero-cost portfolio with nonnegative terminal wealth must therefore have terminal wealth zero $Q$-almost surely and, by equivalence, physically almost surely. Hence an equivalent martingale measure for discounted dividend gains excludes arbitrage. It is the gains process, not just the ex-dividend price ratio, that must have zero martingale drift.

Use the Black-Scholes equation with the claim's terminal condition $V(T,s)=g(s)$:For $t<T$, the Itô formula under the physical measure givesChoose the delta hedge and its bank-account holding byTheir portfolio value is $V(t,S_t)$, andThus the portfolio is self-financing and, with initial capital $V(0,S_0)$, has terminal wealth $g(S_T)$.

Localizing to compact stock-price intervals and times below $T$ justifies these stochastic integrals from the classical smoothness of $V$. The extension to maturity also follows from boundedness. Under the Risk-neutral measure for the Black-Scholes model, discounted $V(t,S_t)$ is a bounded local martingale, hence a true square-integrable martingale. Its quadratic variation has finite expectation, so the stock-integral coefficient is square integrable through $T$. Equivalence of measures preserves its almost-sure finiteness, and the physical drift integral is finite by Cauchy-Schwarz. The terminal limit is $g(S_T)$.

Finally, $B_t=B_0{e}^{rt}$ is bounded away from zero on this finite horizon, and bounded $V$ makes $V(t,S_t)/B_t$ bounded below by a deterministic constant. The hedge is therefore an admissible trading strategy. The displayed holdings replicate the payoff admissibly, even if the bounded payoff can be negative.

Set $\theta =(\mu -r)/\sigma$. The positive densityhas mean one, and the Girsanov theorem makes $W_t^Q=W_t+\theta t$ a Brownian motion under the equivalent measure $Q$. HenceThe Black-Scholes equation makes $e^{-rt}V(t,S_t)$ a local martingale. Bounded $V$ on the finite horizon makes it a true martingale, so its terminal condition gives the risk-neutral valuationThe geometric Brownian motion solution under $Q$ isThe Brownian increment is independent of ${\mathcal{F}}_t$ and equals $\sqrt{T-t}Z$ in distribution, where $Z$ has the standard normal distribution. ThereforeThe expectation on the right is only an integral against the standard normal law; the physical drift $\mu$ has disappeared.

Let $L=\exp ((r-\frac{1}{2}{\sigma }^2)(T-t)+\sigma \sqrt{T-t}Z)$. Boundedness of $g^{\prime }$ and integrability of the lognormal distribution variable $L$ justify differentiation under the expectation, by dominated convergence:A differentiable increasing payoff has $g^{\prime }\ge 0$, and $L>0$. Thus the delta hedge hasMoreover $\mathbb{E}L=e^{r(T-t)}$, so $0\le {\pi }_t\le \Vert g^{\prime }{\Vert }_{\infty }$. At maturity the delta is $g^{\prime }(S_T)\ge 0$. The admissible trading strategy already constructed therefore uses nonnegative stock holdings throughout replication. After maturity, liquidate the stock position and retain the proceeds in the bank account, so ${\pi }_t=0$ for $t>T$ if holdings are to be defined for all times.

Write $R=1+r$, $m=\mathbb{E}{\xi }_1$ and $c=\mathrm{Cov}(S_1,{\xi }_1)=\mathbb{E}[(S_1-\mu )({\xi }_1-m)]$. The riskless gross return must be nonzero for the requested formulas; in the usual positive-bank-account model $R>0$. If $R=0$, the bank holding cannot affect the terminal payoff, and the later moment equation for pricing $B_0=1$ is impossible. We use the intended nondegenerate case $R\ne 0$.

The claim and asset prices are square integrable. Centering separates the bias from the stochastic error:For fixed $\pi$, the unique minimizing bank holding removes the bias. The invertible covariance matrix $V$ is positive definite, and completing the square givesConsequently the unique one-period quadratic hedge isIts minimum expected squared error is $\mathrm{Var}({\xi }_1)-c^T{V}^{-1}c$. This is a least-squares hedge, not necessarily exact claim replication.

Put $h=S_0-\mu /R$. The initial capital of the one-period quadratic hedge isDefine the minimum-norm one-period pricing weightIt is square integrable, and $\mathbb{E}[(S_1-\mu ){\xi }_1]=c$, soThe weight depends only on the asset prices and their first two moments, not on the claim. It can be negative, so this identity alone does not supply an equivalent martingale measure or a positive state-price density.

The centered term in ${\rho }^{\ast }$ has zero mean, giving $\mathbb{E}{\rho }^{\ast }=1/R$ and hence $\mathbb{E}[{\rho }^{\ast }{B}_1]=1=B_0$. For the risky assets, symmetry of the covariance matrix givesTherefore ${\rho }^{\ast }$ reproduces every asset's initial price.

For another square-integrable $\rho$ satisfying the price moment constraints, set $\delta =\rho -{\rho }^{\ast }$. The bank equation and $R\ne 0$ give $\mathbb{E}\delta =0$, while the risky equations give $\mathbb{E}[\delta S_1]=0$. Since ${\rho }^{\ast }$ is an affine combination of $1$ and the coordinates of $S_1$,The Pythagorean theorem in an inner-product space now givesEquality holds exactly when $\rho ={\rho }^{\ast }$ almost surely. If a competing weight has infinite second moment, the inequality is immediate in the extended sense. The minimum itself isThus ${\rho }^{\ast }$ is the orthogonal projection of any feasible square-integrable pricing weight onto the linear span of the traded payoffs.

Let $A=V^{1/2}$ be the symmetric positive square root of the covariance matrix, and write $S_1=\mu +AZ$ with $Z$ having the multivariate normal distribution $N_d(0,I)$. Bounded gradient makes $g$ globally Lipschitz, so $g(S_1)$ has at most linear growth and is square integrable. This also justifies the permitted Gaussian integration by parts formula for each coordinate.

By the chain rule, ${\nabla }_{z}g(\mu +Az)=A^T\nabla g(\mu +Az)$. Applying Gaussian integration by parts givesThus the Gaussian quadratic hedge holds ${\pi }^{\ast }=\mathbb{E}[\nabla g(S_1)]$. Substitution into its initial capital formula yieldsOnly the hedge calculation is being used here; the Gaussian model need not possess a positive pricing weight for this least-squares identity.