Square-root stock claim ($\sqrt{S_T}$)

The payout $\sqrt{S_T}$ is a concave stock payoff. Its conditional expected value has a discount-like dependence on future variance. The square-root stock supermartingale explains the negative variance drift.

For a square-root stock claim priced by $C(t,T)=\mathbb{E}[\sqrt{S_T}\mid {\mathcal{F}}_t]$, the constant-volatility geometric Brownian motion price is $\sqrt{S_t}\exp (-{\sigma }^2(T-t)/8)$. If $t<T$, $S_t>0$ and $0<C(t,T)\le \sqrt{S_t}$, its unique nonnegative inverse volatility is the displayed $\Sigma (t,T)$. For bounded predictable volatility in $dS=S\sigma dW$, bounds $0\le a\le \sigma \le b$ give $a\le \Sigma \le b$, without independence from the Brownian motion. At zero stock price the claim price is also zero and no volatility is identified by this equation; a separate convention is needed.

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.

If $\mathbb{E}[\sqrt{S_T}\mid {\mathcal{F}}_t]=\sqrt{S_t}\exp (-{\int }_t^T{f}_t(u)du)$ and $d{f}_t(T)=A_t(T)dt+B_t(T)d{W}_t$, the stochastic Fubini theorem and Itô formula give $f_t(t)={\sigma }_t^2/8$ and $A_t(T)=B_t(T)({\int }_t^T{B}_t(u)du-{\sigma }_t/2)$. The final term comes from the product cross-variation. Continuity extends the drift equality to the specified continuous versions.

In the joint filtration of Brownian history and an independent volatility history, conditioning on the entire volatility path and then using the tower property gives $\mathbb{E}[\sqrt{S_T}\mid {\mathcal{F}}_t]=\sqrt{S_t}\mathbb{E}[\exp (-\frac{1}{8}{\int }_t^T{\sigma }_u^{2}du)\mid {\mathcal{F}}_t]$. Removing the outer conditional expectation requires the integrated variance to be known at time $t$. Independence alone does not give this measurability; a volatility parameter disclosed later supplies a counterexample.