0
$\begingroup$

I have been on this for hours and it's not getting me anywhere. Any help is so highly and deeply appreciated.

A call option with strike $K$ and expiration $T$ pays $C_{T}=\left(S_{T}-K\right)^{+}$ at time $T$.

$C_{t}=e^{-r(T-t)} E_{Q}\left(C_{T} \mid S_{t}\right)$

I need to find the option pricing formula for $S_{t}=S_{0}+\mu t+\sigma B_{t}$ where r = 0.

This is my attempt

By Girsanov's theorem $\exists$ EMM Q such that $S_{t}=S_{0}+\sigma \hat{B}_{t} .$

$\begin{aligned} C_{t} &=E_{Q}\left(C_{T} \mid S_{t}\right) \\ & \left.=E_{Q}\left(S_{T}-k\right) S_{t}=x\right) \end{aligned}$

$=E_{Q}\left(S_{t}+\sigma \hat{B}_{t}-k \mid S_{t}=x\right)$ $=E Q\left(x+\sigma \hat{B}_{t}-k\right)$

$g(x)=\left\{\begin{array}{cc}x-k & x>k \\ 0 & x \leq k\end{array}\right.$ $g^{\prime}(x)=\left\{\begin{array}{cc}1 & x>k \\ 0 & x \leq k\end{array}\right.$

$\begin{aligned} \frac{\partial C}{\partial x}(x, t) &=P(x+z>k) \\ &=P(z>k-x)=\\ & P\left(N(0,1)>\frac{k-x}{\sigma \sqrt{T-t}}\right) \\ &=\Phi\left(\frac{x-k}{\sigma \sqrt{T-T}}\right) \end{aligned}$

$a_{t}=\frac{\partial c}{\partial x}\left(S_{t}, t\right)=\Phi\left(\frac{S_{t}-K}{\sigma \sqrt{T-t}}\right)$

I am so lost after this. I am not sure if what I am doing is right or wrong either.

$\endgroup$
1
$\begingroup$

Dropping the "hat-notation" on the Brownian motion:

$$S_t=S_0+\sigma B_t$$

Therefore:

$$C_0=\mathbb{E}\left[\frac{\left(S_t-K\right)^{+}}{e^{rt}}\right]=e^{-rt}\mathbb{E}\left[\left(S_t-K\right)I_{S_t>K}\right]=e^{-rt}\mathbb{E}\left[S_t I_{S_t>K}\right]-e^{-rt}\mathbb{E}\left[K I_{S_t>K}\right]$$

Now:

$$\mathbb{E}\left[K I_{S_t>K}\right]=K\mathbb{P}\left(S_t>K\right)=K\mathbb{P}\left(S_0+\sigma B_t>K\right)=K\mathbb{P}\left(B_t>\frac{K-S_0}{\sigma}\right)=K\mathbb{P}\left(Z>\frac{K-S_0}{\sigma \sqrt{t}}\right)=KN\left(\frac{S_0-K}{\sigma \sqrt{t}}\right)$$

Above, $N(.)$ is the normal CDF.

Now:

$$\mathbb{E}\left[S_t I_{S_t>K}\right]=\int_{h=K}^{\infty}hf_{S_t}(h)dh=\frac{1}{\sigma\sqrt{2\pi}}\int_{h=K}^{\infty}he^{\frac{-(h-S_0)^2}{2\sigma^2}}dh$$

So the option price is:

$$C_t=\frac{e^{-rt}}{\sigma\sqrt{2\pi}}\int_{h=K}^{\infty}he^{\frac{-(h-S_0)^2}{2\sigma^2}}dh-e^{-rt}KN\left(\frac{S_0-K}{\sigma \sqrt{t}}\right)$$

The integral above can be simplified, will try to amend later.

$\endgroup$
1
  • $\begingroup$ Is the first term of $C_t$ the pdf of a normal? Also the $e^{-r t}$ disappears right because r= 0? So when I tried this the answer I got was: $\phi(K) - K \Phi \frac{S_0 - K}{\sigma \sqrt{t}}$ where $\phi$ is the pdf of normal and $\Phi$ is the cdf of normal. $\endgroup$ May 15 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.