= Solution
Let $\tau=T-t>0$ and $I_{t,T}=\int_t^T\sigma_s^2ds$. The <stochastic exponential> solution gives
$$
\sqrt{S_T}=\sqrt{S_t}\,M_{t,T}e^{-I_{t,T}/8},\qquad
M_{t,T}=\exp\left(\frac12\int_t^T\sigma_s\,dW_s-\frac18 I_{t,T}\right).
$$
Boundedness 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 $a^2\tau\leq I_{t,T}\leq b^2\tau$ and positivity of $M_{t,T}$ give
$$
\sqrt{S_t}\,e^{-b^2\tau/8}
\leq C(t,T)\leq\sqrt{S_t}\,e^{-a^2\tau/8}.
$$
This 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 is
$$
C(t,T)=\sqrt{S_t}\,\mathbb E[e^{-I_{t,T}/8}\mid\mathcal F_t],
$$
the <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 volatility>
$$
\Sigma(t,T)^2=-\frac8\tau\log\left(\frac{C(t,T)}{\sqrt{S_t}}\right),
\qquad\boxed{a\leq\Sigma(t,T)\leq b.}
$$
There 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 $S_t>0$.
Back to article page
