Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 29J (Stochastic Financial Models)

A standard Brownian motion starts at zero, has continuous sample paths and independent increments, with $W_t-W_s\sim N(0,t-s)$ for $0\le s<t$.

Assume $S_0,K,T>0$ and $\sigma >0$; changing the sign of $\sigma$ just changes the sign of Brownian motion. Itô formula gives $d{S}_t=S_t[(\mu +{\sigma }^2/2)dt+\sigma d{W}_t]$. In the complete Black-Scholes model, the risk-neutral pricing theorem prices an integrable payoff $Y$ by $e^{-rT}{\mathbb{E}}_{Q}Y$, where the discounted asset is a martingale. Girsanov theorem therefore replaces the logarithmic drift by $r-{\sigma }^2/2$:Let $d_2=[\log (S_0/K)+(r-{\sigma }^2/2)T]/(\sigma \sqrt{T})$ and $d_1=d_2+\sigma \sqrt{T}$. The exercise event is $Z>-d_2$. Completing the square in the Gaussian integral yieldsConsequently ${\mathbb{P}}_Q(S_T>K)=\Phi (d_2)$ and ${\mathbb{E}}_Q[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Thus the European call option price isThe physical drift $\mu$ disappears because the price is obtained under the martingale measure. If $\sigma =0$ in an arbitrage-free deterministic model, $S_t=S_0{e}^{rt}$ and the formula is interpreted by its zero-volatility limit.

Let $G_T=\exp (T^{-1}{\int }_0^T\log S_{u}du)$, the continuous Geometric Asian option average. Under the same risk-neutral measure,The Brownian integral is Gaussian with mean zero and varianceThus $\log G_T\sim N(m,v)$, where $m=\log S_0+(r-{\sigma }^2/2)T/2$ and $v={\sigma }^{2}T/3$. DefineThe truncated lognormal distribution calculation from (i) givesFor comparison, a lognormal call written in terms of its mean $F$ and log-variance $v$ is $c(F,v)=F\Phi (d_1)-K\Phi (d_2)$. Differentiation gives ${\partial }_{F}c=\Phi (d_1)>0$ and ${\partial }_{v}c=F\phi (d_1)/(2\sqrt{v})>0$, where $\phi$ is the standard normal density. The terminal stock has $F_S=S_0{e}^{rT}$ and $v_S={\sigma }^{2}T$, while $v_G=v_S/3$. Thereforesince this condition gives $F_G\le F_S$ and the log-variance is strictly smaller. In particular the requested inequality holds for the usual nonnegative interest rate.

The PDF gives no sign restriction on $r$, so its unconditional comparison needs qualification. If $r<-{\sigma }^2/6$, then $F_G>F_S$, and as $K\downarrow 0$ the two prices tend to $e^{-rT}{F}_G$ and $e^{-rT}{F}_S$. By continuity, sufficiently small positive strikes reverse the claimed inequality. Also strictness needs nonzero volatility or another source of a strict mean inequality. These qualifications do not affect the pricing formula.