Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 39 / 2

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.