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For bounded predictable $\sigma$ on $[0,T]$, the stochastic exponential $Z_s=\exp (\frac{1}{2}{\int }_0^s{\sigma }_{u}d{W}_u-\frac{1}{8}{\int }_0^s{\sigma }_u^{2}du)$ is a positive martingale by the Novikov condition. Under $dQ/dP=Z_T$, the Bayes formula for conditional expectation givesIndeed, the square-root stock ratio is $(Z_T/Z_t)\exp (-\frac{1}{8}{\int }_t^T{\sigma }_u^{2}du)$. This exact identity gives the bounds on square-root stock implied volatility by bounding integrated variance pathwise. The Girsanov theorem gives $W^Q=W-\frac{1}{2}\int \sigma du$. The measure is an auxiliary pricing identity for this payoff, not a claim that it is an equivalent martingale measure for the original stock.

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 $S_t>0$.