Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 39 / 6

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.

An explicit counterexample to the stronger reading uses a three-dimensional Bessel process $R$ with $R_0=1$ and $dR=dW+R^{-1}dt$, which is a positive strong solution in the Brownian filtration. Take $B=1$ and $S=R$. Then $r=0$, $\mu =R^{-2}$ and $\sigma =R^{-1}$ are continuous and $\sigma >0$. The unique local candidate is $Z=R^{-1}$, with $ZS=1$. The reciprocal three-dimensional Bessel strict local martingale is not a true martingale. To verify the loss of expectation, use the standard Bessel transition densityIntegrating $p_t(1,y)/y$ gives $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$, where $\Phi$ is the standard normal cumulative distribution function. A true pricing density for the constant bank account would have expectation one. Hence the printed hypotheses establish the local deflator statement, while a true-density reading needs an additional condition and is false as stated.

Let $h_t$ and $k_t$ be predictable holdings in the stock and account, with the stochastic integrability needed for a self-financing portfolio. Then $V=hS+kB$ and $dV=hdS+kdB$. The Itô product rule and self-financing identity giveSince $V=hS+kB$ and $\sigma \lambda =\mu -r$, the finite-variation term is zero. Thereforeand the deflated wealth is a local martingale.

For zero-capital nonnegative wealth under a local deflator, a nonnegative local martingale is a supermartingale, by localization and the Conditional Fatou lemma. Under the required nonnegative-wealth condition, $ZV\ge 0$ and starts at zero, so $0\le \mathbb{E}(Z_t{V}_t)\le 0$. Thus $Z_t{V}_t=0$ almost surely at every fixed $t$. Strict positivity of $Z$ gives $V_t=0$ almost surely. Taking a countable intersection over rational times and then using continuous wealth paths strengthens this toThis argument only needs the local deflator, so it remains valid without promoting $ZB$ to a true martingale.

Here the bank account is continuous and of finite variation, but for $t>0$ its rate is $r_t=t^{-1/2}$. This is integrable at zero, so the account itself is well defined. The stock satisfies $d{S}_t=S_{t}d{W}_t$. The previous continuity hypothesis on the rate no longer holds at the initial time.

Suppose a positive normalized state-price density existed; even a local martingale deflator would suffice for a contradiction. Then $ZB$ would be a continuous local martingale by the Brownian martingale representation theorem. Dividing by the positive continuous account gives continuous $Z$, with $Z_0=1$, andThe Itô product rule for $ZS$ forces its drift to vanish, so ${\eta }_t=Z_t{t}^{-1/2}$ for almost every positive time. Continuity and $Z_0=1$ imply that each path has a positive interval on which $Z_t\ge 1/2$. But thencontradicting the local square-integrability required for the stochastic integral. ThusThis is the singular initial market-price-of-risk obstruction: the necessary market price of risk is $-t^{-1/2}$, whose squared integral diverges at zero. On an interval starting at a strictly positive time this particular obstruction disappears; the initial-time normalization is essential.