Bachelier model option pricing Consider a Brownian motion $W_t$ and Bachelier model $S_t = 1 + \mu t + \sigma W_t$ for the stock price $S_t$. Find the value of an option that pays $1(S_1 > 1)$.
Attempt: As I understand it, the answer is $P(S_1 > 1)$, which yields a big integral with parameter $\mu$. I suspect there should be a simple answer. Any help is greatly appreciated!
Update:
Kurt and Maximilian, thank you for your help. However, I don't believe the answers given are correct. As I understand it now, you are supposed to use risk neutral valuation argument by constructing delta-hedging portfolio, not evaluate the probability directly.
Interestingly, in Black-Scholes or Bachelier models, when evaluating options with risk-neutral valuation, the result usually doesn't depend on drift term $\mu$.
 A: The other two answers assume your model is already risk-neutral; however, your notation and your edit seem to indicate that in fact you have provided us with a model under the empirical measure. So to price this option we should do a change of measure to the risk-neutral measure. To do so, I assume a constant risk-free rate $r > 0$.
Let $X_t = D_t S_t$ be the discounted asset price, where $D_t$ is the discount factor ($dD_t = -rD_tdt$). Observe that the physical dynamics in the Bachelier model are given by $dS_t = \mu dt + dW_t$. We wish to argue that under the risk-neutral measure, $X_t$ is a martingale. We compute the dynamics of $X_t$ using the Itô product rule: $$dX_t = D_t dS_t + S_t dD_t = D_t(\mu dt+ dW_t) + -rD_tS_tdt = D_t\left[ (\mu - rS_t) dt + dW_t \right] = D_t d\widetilde{W}_t$$
Where we have performed the change of measure $d\widetilde{W}_t = (\mu - rS_t) dt + dW_t$. $\widetilde{W}_t$ is a Brownian motion under the risk-neutral measure. You may check this is a valid change of measure using the Novikov criterion. Replacing this in the dynamics for $S_t$, we get: $$dS_t = \mu dt + dW_t = \mu dt + d\widetilde{W}_t - (\mu - rS_t) dt  = rS_t dt + d\widetilde{W}_t$$
By using a stochastic integration factor (e.g. see here), you may check that this SDE is satisfied by the following process: $$S_T = S_te^{r(T-t)} + e^{rT}\int_t^T e^{-rs}d\widetilde{W}_s$$
By identifying this as a Gaussian process and using the Itô isometry, this gives us the following conditional distribution under the risk-neutral measure:
$$S_T\, | \, S_t \sim N \left( S_t e^{r(T-t)}, \frac{1}{2r}\left(e^{2r(T-t) - 1} \right)\right)$$
Using the above formula, the fact that the price of the option is the discounted risk-neutral expectation, and the fact that $\mathbb{Q}(S_1 > 1, S_2<1) = \mathbb{Q}(S_2<1 | S_1 > 1) \mathbb{Q}(S_1>1)$, can you compute the required option price?
A: Your suspicion is correct: $W_1$ and $W_2/\sqrt{2}$ are two standard normals with correlation $1/\sqrt{2}\,.$ The integral you got can therefore be evaluated using the CDFs of the univariate and the bivariate normal distribution:
\begin{eqnarray*}
P(S_1>1,S_2<1)&=&P(S_2<1)-P(S_1<1,S_2<1)\\
&=&\Phi\big(-\mu\,\sqrt{2}\big)-\Phi\Big(-\mu,-\mu\,\sqrt{2},1/\sqrt{2}\Big)\,.
\end{eqnarray*}
A: I am not able to find the simple answer (if there is one). Therefore, let's get to the integrals (which also do not lead to a satisfying answer, by the way):

More generally, if $$S_t=1-f(t)+W_t$$ for any function $f:[0,2]\to\mathbb R$, then $$\mathsf P(S_1>1\text{ and }S_2<1) =\mathsf P(W_1>f(1)\text{ and }W_2<f(2)).$$
Now, let $X=W_1$ and $Y=W_2-W_1$. By Definition of a Brownian motion, $X,Y$ are independent and $\mathcal N(0,1)$-distributed. (I.e. standard normal.)
We are looking for $$\mathsf P(X>f(1)\text{ and } X+Y<f(2)).$$
Now, by the Transformationssatz for measures, using Iverson brackets in the notation, $(\Omega,\mathcal A,\mathsf P)$ as the probability space, $ \cdot_\#$ for the pushforward,
\begin{equation*}
\begin{split}
\mathsf P(X>f(1)\text{ and }X+Y<f(2))&=\int_\Omega [X>f(1)][X+Y<f(2)]\,\mathrm dP
\\
&=\int_{\mathbb R^2} [x>f(1)][x+y<f(2)]\,\mathrm d(X,Y)_\#\mathsf P(x,y).
\end{split}
\end{equation*}
By independence of $(X,Y)$, we have ($\otimes$ is the product measure) $(X,Y)_\#\mathsf P=(X_\# \mathsf P)\otimes(Y_\#\mathsf P)$, so that the above equals
\begin{equation*}
\begin{split}
\int_{f(1)}^\infty\,\mathrm dX_\#\mathsf P(x)\left(\int_{-\infty}^{f(2)-x}\,\mathrm dY_\#\mathsf P(y)\right)
\end{split}
\end{equation*}
Defining the error function through
\begin{equation*}
\operatorname{erf}(z)=\frac{2}{\sqrt\pi}\int_0^z\exp(-t^2)\,\mathrm dt
\end{equation*}
for $z\in\mathbb R$ gives, since $X,Y\sim\mathcal N(0,1)$,
\begin{equation*}\begin{split}
\int_{f(1)}^\infty\,\mathrm dX_\#\mathsf P(x)\left(\int_{-\infty}^{f(2)-x}\,\mathrm dY_\#\mathsf P(y)\right)
&=\frac1{2\sqrt{2\pi}}\int_{f(1)}^\infty\left(\operatorname{erf}\left(\frac{f(2)-x}{\sqrt 2}\right)+1\right) \exp\left(-\frac{x^2}2\right)\,\mathrm dx.
\end{split}\end{equation*}
I am not sure how to simplify this further. Note that in your case, $f(2)=2 f(1)=-2\mu$ for a fixed $\mu\in\mathbb R$.
