Forward-Start Call Valuation Under Proportional Volatility

# Foward-start option pricing


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

I would appreciate any advice. Thanks in advance.

## Answer by Paul (score 1, accepted)

https://quant.stackexchange.com/a/7769

Note that \begin{align} p(t,x) &:= \mathbb E^{\mathbb Q} \left [ S_1 (S_T/ S_1-\kappa)^+ | S_t=x \right] \\&= x \mathbb E^{\mathbb Q} \left [ (S_T/x-\kappa )^+ | X_t=x \right] \end{align}

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.

## Answer by clarkmaio (score 1)

https://quant.stackexchange.com/a/37116

Ciao! Let me say that I think you gave the wrong dynamic for $S_t$ (for example because it is a financial asset with Normal dynamic...so it could be negative!!(?) ).

I'll give a solution for both case (normal and log-normal) but only the log-normal one leads to the right solution.

Log-normal case

In this case $S_T$ has the following dynaic in $\mathbb{Q}$: $$ dS_t = S_t \sigma_t dW_t. $$ Integrating $S_t$ starting from $1$ you get:

$$ S_T = S_1\exp\left[-\frac{1}{2}\int_1^T\sigma_s^2 ds + \int_1^T\sigma_s dWs\right] $$

so that $S_T /S_1$ is independent from $S_t \ \forall t$ (this will be important in a minute). Morover notice that: $$ S_T/S_1 \sim e^Y $$ where $Y \sim \mathcal N \left( -\frac{1}{2}\int_1^T \sigma_s^2 ds, \int_1^T \sigma_s^2 ds \right) = N \left( -\frac{1}{2}\gamma^2 , \gamma^2 \right) $

Now we can do the explicit computation. Tee main idea is to use the tower property and divide the integral in two part. After that we will use the previous observation and the fuct that the first integral doesn't depend on $S_1$: $$ \begin{align} \mathbb{E} \left[ (S_T - kS_1)^+ | S_t = x \right] & = \mathbb{E} \left[S_1\left. \left( \frac{S_T}{S_1} - k\right)^+ \right| S_t = x \right] \\ & =\mathbb{E} \left. \left[ \mathbb{E} \left. \left[S_1\left( \frac{S_T}{S_1} - k\right)^+ \right| S_1\right] \right| S_t = x\right]\\ & =\mathbb{E} \left. \left[ S_1 \mathbb{E} \left[\left.\left( \frac{S_T}{S_1} - k\right)^+ \right| S_1\right] \right| S_t = x\right] \\ & =\mathbb{E} \left. \left[ S_1 C(k, 1, \Sigma, 0) \right| S_t = x\right] \end{align} $$

$C(k, 1, \Sigma, 0)$ is the price of a call option with strike $k$, starting point $1$, rate $0$ and volatility the volatility of $S_T/S_1$.

The Call option has starting value $1$ and it doesn't depend on $S_1$ so that we can put it outside the expected value. Now we use the fact that $S_1$ is a martingale in $\mathbb{Q}$ and we obtain:

$$ \mathbb{E} \left[ (S_T - kS_1)^+ | S_t = x \right] = C(k, 1, \Sigma, 0) S_t $$

which is the final result.

Normal case

Here the problem is clear: $S_T/S_1$ depends on $S_1$ so that we cannot use the same technique of the previous case. I have no smart idea so that I will follow the "samurai way" of the brutal computation.

We have: $$ S_T = S_1 + \int_1^T \sigma_s dW_s \sim \mathcal{N}(S_1, \int_1^T \sigma^2_s ds) $$ so that:

$$ \begin{align} \mathbb{E} \left[\left. (S_T - kS_1)^+ \right|S_t\right] &= \mathbb{E} \left[ \left. \mathbb{E} \left[\left. (S_T - kS_1)^+ \right| S_1\right] \right|S_t \right] \\ & = \mathbb{E} \left[\left. \frac{1}{\sqrt{2\pi}\Sigma} \int_0^{+\infty} x\exp\left(\frac{-(x-\mu)^2}{2\Sigma^2} \right) dx \right| S_t\right] \end{align} $$

where $\mu = (1-k)S_1$ and $\Sigma = \int_1^T \sigma_s^2 ds$.

Let me focus on the integral:

$$ \begin{align} \frac{1}{\sqrt{2\pi}\Sigma} \int_0^{+\infty} x\exp\left(\frac{-(x-\mu)^2}{2\Sigma^2} \right) dx & = \frac{1}{\sqrt{2\pi}} \int_{\frac{\mu}{\Sigma}}^{+\infty} (\Sigma \xi + \mu)\exp\left(\frac{-\xi^2}{2} \right) d\xi \\ & = \frac{\Sigma}{\sqrt{2\pi}} \exp \left(- \frac{\mu^2}{2\Sigma^2} \right) + \frac{\mu}{\sqrt{2\pi}}\Phi\left( \frac{\mu}{\Sigma} \right) \end{align} $$

where $\Phi$ is the cumulative function of the standard gaussian. At his point, using the fact that: $$ S_1 = S_t + \int_t^1 \sigma_s dW_s \sim \mathcal{N} \left( S_t, \int_t^1 \sigma_s^2 ds \right) $$ we have to solve the following integral:

$$ \begin{align} \mathbb{E} \left[ \frac{\Sigma}{\sqrt{2\pi}} \exp \left(- \frac{\mu^2}{2\Sigma^2} \right) + \frac{\mu}{\sqrt{2\pi}}\Phi\left( \frac{\mu}{\Sigma} \right) \right] & = \frac{\Sigma}{\sqrt{2\pi}}\int_\mathbb{R} \exp \left( -\frac{(1-k)^2x^2}{2 \Sigma^2} \right) \exp\left( -\frac{(x-S_t^2)}{2\Gamma^2} \right) dx \\ & \quad + \int_\mathbb{R} (1-k)x \Phi\left( \frac{(1-k)x}{\Sigma} \right)\exp\left(- \frac{(x-S_t)^2}{2\Gamma^2} \right) dx \end{align} $$ where $\Gamma = \int_1^t \sigma_s^2 ds $.

The first integral should be easy to compute just by completing the square in the exponent (the result will be very bad looking but easy to do).

About the second one, it's seems more tricky...it could be usefull the following result:

Result $$ \int_\mathbb{R} x\Phi(x)e^{\frac{-x^2}{2}} = \int_\mathbb{R} e^{-x^2} = \sqrt{\pi} $$ (I'm still working on it)

Ciao!