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.


1 Answer 1


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

$$S_t=S_0+\sigma B_t$$


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


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

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


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

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.

  • $\begingroup$ 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, 2021 at 2:48

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