 Solution

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 $S_t>0$.