Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 202 (Stochastic Calculus with Applications to Finance)

Apply Itô formula to $Z_t^{(A)}=\exp (A{X}_t-A^{2}t/2)$. Its semimartingale decomposition isThe second term is a continuous finite-variation process. Since $Z^{(A)}$ is assumed to be a local martingale, uniqueness of the semimartingale decomposition makes this term identically zero. Both $A$ and $Z^{(A)}$ are nonzero, so the quadratic variation of $X$ is $[X{]}_t=t$.

The Lévy characterization of Brownian motion now says that $X_t-X_0$ is a Brownian motion. Consequentlyis a constant multiple of an exponential Brownian martingale. It is therefore a true martingale for every $a\in \mathbb{R}$.

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Set $K_t=\sqrt{H_t}$ and define the stochastic integralStrict positivity and predictability of $H$ make the integrand locally admissible. The process $W$ is a continuous local martingale starting from zero, and the quadratic variation of a stochastic integral givesBy the Lévy characterization of Brownian motion, $W$ is a Brownian motion. The associativity of stochastic integration then yieldswhich is the required representation.

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Use first the test function $f(x)=x$. The assumed martingale problem says thatis a continuous local martingale. Next use $f(x)=x^2$ to see thatis a local martingale. On the other hand, Itô formula applied to $X=X_0+\int b(X_s)ds+Y$ shows thatis a local martingale. Their difference is both a continuous local martingale and a finite-variation process, so

Part (b), with $H_s=\sigma (X_s{)}^2>0$, supplies a Brownian motion $W$ such thatThereforeso $X$ is a weak solution of a stochastic differential equation to the stated equation.

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Let $(X_n)$ be Cauchy in the normChoose a subsequence $(X_{n_k})$ for whichTonelli theorem impliesalmost surely. The subsequence therefore converges uniformly on $[0,\infty )$, outside one null event, to a continuous process $X$. For each $t$, $X_t$ is the almost-sure limit of ${\mathcal{F}}_t$-measurable variables; completeness of the filtration makes the chosen version adapted. The same summable bound and monotone convergence theorem show that $\mathbb{E}{\sup }_t|X_t|<\infty$ and that $X_{n_k}\to X$ in norm.

Since the original sequence is Cauchy, the usual triangle argument upgrades convergence of the subsequence to $X_n\to X$ in norm. Thus the space of indistinguishability classes is a Banach space.

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Fix a deterministic horizon $T$ and $\epsilon >0$. For every localization index $N$,by Markov inequality. Because $T_N\uparrow \infty$ almost surely, the first term tends to zero as $N\to \infty$. For fixed $N$, the second tends to zero as $n\to \infty$. Taking first the limit superior in $n$ and then $N\to \infty$ provesin probability. This is precisely uniform convergence on compacts in probability.

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Use the continuous semimartingale decomposition $X=X_0+M+A$, where $M$ is a continuous local martingale and $A$ is a continuous adapted finite-variation process. Pointwise limits preserve predictability, so $H$ is predictable; it is bounded by the common bound for the $H_n$.

Localize so that $[M{]}_{\infty }$ and the total variation $V(A{)}_{\infty }$ are bounded. The Doob L2 maximal inequality and the Itô isometry giveby the dominated convergence theorem. For the finite-variation part,almost surely, again by dominated convergence, now for each sample path. Hence the two integrals converge uniformly in probability after every localization. Part (b) removes the localization and provesu.c.p.

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PutThen $f_n\to |\cdot |$ uniformly, $f_n^{\prime }(x)=\tanh (nx)$, and $f_n^{\prime \prime }(x)=n{\mathrm{sech}}^2(nx)$. Itô formula givesThe bounded predictable integrands $\tanh (nX)$ converge pointwise to $\mathrm{sgn}(X)$, with $\mathrm{sgn}(0)=0$. Part (c) therefore makes the stochastic integrals converge u.c.p. The left-hand side converges u.c.p. to $|X|$, so the increasing continuous processesalso converge u.c.p. Their limit $A$ has a continuous increasing version: extract almost-sure locally uniform convergence from each compact interval and use a diagonal argument. We obtainthe Tanaka formula with $A=L_t^0(X)$. It expresses $|X|$ as a continuous local martingale plus a continuous finite-variation process, so $|X|$ is a continuous semimartingale.

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For the level-$N$ dyadic partition, writeThe dyadic partitions are nested, so the triangle inequality makes $V_N(t)$ nondecreasing in $N$, and $\Vert f_t\Vert ={\lim }_N{V}_N(t)$.

Fix $t_1<t_2$. As the mesh tends to zero, the last dyadic point before $t_1$ approaches $t_1$. Refining from there to $t_2$, the triangle inequality says that the added variation is at least $|f(t_2)-f(t_1)|$ minus the two endpoint errors, which tend to zero by continuity. ThereforeSince $\Vert f\Vert <\infty$, both limits are finite and may be subtracted, giving

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Every dyadic partition is among the finite partitions on the right-hand side, so the displayed supremum is at least $\Vert f\Vert$. For the converse, the claim is immediate if $\Vert f\Vert =\infty$. If it is finite, apply part (a) to every interval of an arbitrary partition $0\le t_0<\cdots <t_n=1$:Summing telescopes and givesTaking the supremum proves that the dyadic definition equals the usual total variation of a function.

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If $f$ is continuously differentiable, then for every finite partition the fundamental theorem of calculus and the triangle inequality givePart (b) therefore implies $\Vert f\Vert <\infty$.

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For $k\ge 1$, setThen $t_k\downarrow 0$ and $f(t_k)=(-1{)}^k{t}_k$. Consecutive values have opposite signs, soFinite partitions containing $t_N,t_{N-1},\dots ,t_1$ therefore have variation at leastwhich diverges with $N$ by comparison with the harmonic series. Part (b) now gives $\Vert f\Vert =\infty$.

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Factor the square root of the stochastic exponential asThe stochastic exponential in this expression is a positive local martingale and hence a supermartingale, so its expectation is at most one. If $[M{]}_T\ge a$ almost surely, then

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If $[M{]}_T\le b$, the Novikov condition holds for $(M-M_0)/2$ becauseIts stochastic exponential is therefore a true martingale with expectation one. Using the factorization from part (i),

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With zero interest rate, the bank account is constant. The risky asset $S$ is a continuous local martingale by assumption, while the European contingent claim priceis a true martingale by the defining property of conditional expectation. Thus the original probability measure is an equivalent local martingale measure for all traded discounted prices. The fundamental theorem of asset pricing then excludes arbitrage for admissible self-financing strategies.

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In the Black-Scholes model,Conditioning on ${\mathcal{F}}_t$ and using the moment-generating function of the independent Gaussian increment $W_T-W_t$ givesThe function $c$ satisfies the zero-rate Black-Scholes equation, so Itô formula leaves only its stochastic term:Consequently the required delta hedge is

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For this square-root payoff, the time-zero Black-Scholes model price at volatility $\hat{\sigma }$ isParts (a)(i) and (a)(ii), together with $a\le [M{]}_T\le b$, giveThe exponential is strictly decreasing, so comparison with the defining Black-Scholes price givesThus the Black-Scholes implied volatility lies between the lower and upper realized-variance bounds.

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