European Call Pricing under Arithmetic Brownian Motion
Summary
The discussion seeks a European call value when the underlying follows arithmetic Brownian motion, with zero interest rates. The questioner changes measure to remove the drift and tries to use the probability of finishing above the strike to derive the option’s sensitivity to the underlying. The answer instead decomposes the payoff into the expected underlying value above the strike minus the discounted strike conditional on exercise, then expresses these terms using a normal distribution and a tail integral.
This outlines a route to pricing from the terminal distribution, but the posted derivation is incomplete and contains inconsistent time and discounting notation. The answer leaves its integral unsimplified and does not present a final closed-form price. The model is also an arithmetic process, unlike the geometric Brownian motion commonly used for stock prices, so its suitability depends on the underlying and assumptions. Readers should verify the measure and pricing setup before relying on the formula.
Key ideas
- The call payoff can be split into an underlying-value term and a strike term over the exercise event.
- Under arithmetic Brownian motion, the terminal underlying value has a normal distribution.
- The exercise probability can be expressed with the standard normal cumulative distribution function.
- The response leaves the pricing integral unsimplified and includes notation inconsistencies.
- The arithmetic price model has different implications from a geometric stock-price model.
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Full text
# option pricing formula for $S_{t}=S_{0}+\mu t+\sigma B_{t}$ where r = 0
# option pricing formula for $S_{t}=S_{0}+\mu t+\sigma B_{t}$ where r = 0
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.
## Answer by Jan Stuller (score 1)
https://quant.stackexchange.com/a/63986
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.