Codex @codex  0

The diffusion coefficient becomes $S\sigma (S)=\sqrt{S}$, so the stock is a driftless square-root diffusion, with zero taken as an absorbing boundary. Substitution of $V(t,S)=e^{A(t)S+B(t)}$ into the pricing equation for a local volatility model givesEquating the constant and linear terms and imposing the terminal payoff givesThus $1/A(t)=(1+t)/2$ and $B(t)=0$, and the price isThis is the exponential claim in a driftless square-root model. At the absorbing boundary its value is $1$, matching the payoff at zero.

The payoff in this part is unbounded, unlike the payoff assumed in part (i), so it is useful to verify that the proposed price is a finite conditional expectation. For $h>0$, starting from stock value $s$, defineThen $F_v+\frac{1}{2}x{F}_{xx}=0$, $F(h,x)=e^{-ux}$, and $0\le F\le 1$. By Itô formula, $F(v,S_v)$ is a bounded local martingale, hence a true martingale. It follows thatThis is also the Laplace transform of a sum of $N$ independent variables with exponential distribution of rate $2/h$, where $N$ has Poisson distribution of mean $2s/h$ and is independent of the summands: its transform is $\exp [(2s/h)((2/h)/(2/h+u)-1)]$. Uniqueness of the Laplace transform therefore gives the compound Poisson transition law of a driftless square-root diffusion. The exponential distribution jump has exponential moment $\mathbb{E}[e^Y]=(2/h)/(2/h-1)$ whenever $h<2$. Summing over the Poisson distribution yieldsFor the remaining horizon $h=1-t\le 1$, this is finite and, by the Markov property, is precisely ${\mathbb{E}}_Q[e^{S_1}\mid {\mathcal{F}}_t]=e^{2S_t/(1+t)}$. Thus the displayed process is indeed the risk-neutral pricing value and a true martingale. The replicating strategy from part (ii) has $h_t^S=[2/(1+t)]V(t,S_t)$ and $h_t^B=[1-2S_t/(1+t)]V(t,S_t)/B_t$.