# Foward-start option pricing

Consider a probability filtred space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfing the habitual conditions and is generated by $1 d$- Brownian Motion (with $\mathcal F_T = \mathcal F$).

Also, consider a financial market where the interest rate is nul, $r=0$, and the dynamics of the risky asset $S$ is given by $$S_t= S_0 + \int_0^t \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t \geq 0$$

where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.

Suppose that there is a measure $\mathbb Q \sim \mathbb P$ and a $\mathbb Q$-brownian motion $W^{\mathbb Q}$ such that $$S_t= S_0+\int_0^t \sigma_s ~dW_s^{\mathbb Q}\quad , t \leq T$$

We want to evaluate and hedge a forward-start option, whose payoff is $(S_T-\kappa S_1)^+$ where $\kappa >0$ (we suppose T>1). Let's admit that it can be perfectly hegded and let's note

$$p(t,x) := \mathbb E^{\mathbb Q} \left [ (S_T-\kappa S_1)^+ | S_t=x \right] \quad \text{for} \ (t,x) \in [0,1]\times(0,\infty)$$

Show that

$$p(t,x) = x F(1,\kappa, \int_0^T \sigma_s ^2 ds) \quad \text{if} \ t\in [0,1]$$

where, for $y, K, \gamma^2 >0$ $$F(y,K,\gamma^2) = \mathbb E \left [ (ye^Y -K)^+ \right] \quad \text{with} \ Y \sim \mathcal N(-\gamma^2/2, \gamma^2)$$

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Is this homework? –  Brian B Apr 9 '13 at 15:00
@BrianB : No, it was a problem wich was proposed in a old exam. –  Paul Apr 9 '13 at 15:12

for all $t<1$ and $x \in (0, +\infty)$, since $S$ is a $\mathbb Q$ -martingale. Now, if we include the aditional condition on $\sigma$ that $\sigma_t := \tilde\sigma(S_t) ~S_t$, we can conclude that \begin{align} p(t,x) &= x F(1,\kappa, \int_1^T \tilde\sigma(S_s)^2 ds), \end{align}
where $F$ is defined as in the question BUT under $\mathbb Q$ and not under $\mathbb P$ as it says.
There are two things that I don't understand in your arguments. - The conditional expectation is on $S_t=x$, how can you "split out" $S_1=x$ - Even assuming $S_t$ is a martingale, when it multiply with some other random variable, it's no longer to be sure that is a martingale, and I don't see how you can separate the martingale part from the whole, i.e separating $S_1$ from $S_1(\frac{S_T}{S_1} - K)^+$ –  ctNGUYEN Feb 18 at 12:40