With constant coefficients, $\lambda=(\mu-r)/\sigma$ is constant, so the <stochastic exponential> $Z_t=e^{rt}Y_t$ is a true <martingale>. Under its <equivalent martingale measure> $\mathbb Q$, the <stock> follows the <Black-Scholes model> $dS_t=S_t(rdt+\sigma dW_t^{\mathbb Q})$. The Gaussian exponential <moment> gives the value of the <power option>
$$
V_t=e^{-r(T-t)}\mathbb E_{\mathbb Q}[\sqrt{S_T}\mid\mathcal F_t]=\sqrt{S_t}\exp\!\left(-\left(\frac r2+\frac{\sigma^2}8\right)(T-t)\right).
$$
The <option delta> is $\partial_sV=V/(2s)$, so the <replicating strategy> holds
$$
\boxed{\theta_t=\frac{V_t}{2S_t},\qquad\eta_t=\frac{V_t}{2B_t},\qquad V_0=\sqrt{S_0}\exp\!\left(-\left(\frac r2+\frac{\sigma^2}8\right)T\right).}
$$
Half the wealth value is in the <stock> and half in the <bank account>, with continuous rebalancing. To check the <self-financing portfolio> property under the original measure, the <Itô formula> gives $dV_t=\frac12(r+\mu)V_tdt+\frac12\sigma V_tdW_t=\theta_tdS_t+\eta_tdB_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the <Black-Scholes model> has the necessary finite Gaussian <moments>, and the explicit strategy attains $V_0=\mathbb E[Y_T\sqrt{S_T}]$. The <supermartingale> bound from part (b) therefore proves its minimality. The cost is independent of the physical <drift> $\mu$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 211 / 6 / d / Solution

{title2=$S_T^p$}

A power option is a <European contingent claim> whose payoff is a fixed power of the terminal underlying price. In the <Black-Scholes model>, its value is $S_t^p\exp((p-1)r(T-t)+\frac12p(p-1)\sigma^2(T-t))$ when the required <moments> are finite.

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