In the Bachelier model, I have difficulties with a certain step. I want to figure out the distribution of $S_T$, which is the price process in the Bachelier model.

So far I could state that ($\mathbb{Q}$ is the EMM): \begin{eqnarray} dS_t = r S_t dt + \sigma W^\mathbb{Q}_t \label{SDE2} \end{eqnarray} and with that \begin{eqnarray} S_T = S_0 e^{rT} + \int\limits_{0}^{T}\sigma e^{r(T-s)} dW^\mathbb{Q}_s \end{eqnarray} Now I have found a book that states that $S_T$ has distribution: \begin{eqnarray} S_T \sim \mathscr N \left(S_0 e^{rT}, \sqrt{\frac{\sigma^2-\sigma^2e^{-2rT}}{2r}} \right) \end{eqnarray}

I do not understand why this should be, maybe my skills in stochastic integration are not sufficient.

Thank you for taking your time!

  • 4
    $\begingroup$ This is a direct application of Ito's isometry: en.m.wikipedia.org/wiki/It%C3%B4_isometry it gives you the mean (= 0) and variance ($= \int f^2(u)du$) of a Wiener integral $\int f(u) dW(u)$. $\endgroup$
    – byouness
    Jun 26, 2019 at 8:00

1 Answer 1


As explained by @byouness, using Itô's Isometry, we get: $$\begin{align} V(S_T)&=V^{\mathbb{Q}}\left(\int_0^T\sigma e^{r(T-s)} dW^\mathbb{Q}_s\right) \\[9pt] &=E^{\mathbb{Q}}\left(\left(\int_0^T\sigma e^{r(T-s)} dW^\mathbb{Q}_s\right)^2\right)-{\underbrace{E^{\mathbb{Q}}\left(\int_0^T\sigma e^{r(T-s)} dW^\mathbb{Q}_s\right)}_{=\int_0^T\sigma e^{r(T-s)} E^{\mathbb{Q}}(dW^\mathbb{Q}_s)=0}}^2 \\[-9pt] &=E^{\mathbb{Q}}\left(\int_0^T\sigma^2 e^{2r(T-s)} ds\right) \end{align}$$ The remaining integral is deterministic, thus: $$V(S_T)=\sigma^2\left[-\frac{e^{2r(T-s)}}{2r}\right]_{s=0}^{s=T}=\sigma^2\left(\frac{e^{2rT}-1}{2r}\right)$$ Note that your result is correct up to a minus sign. This is probably because the Bachelier dynamics for the stock price are also known as an Ornstein–Uhlenbeck process, which is normally defined with a minus sign in the drift, i.e.: $$dS_t = \color{red}{-}r S_t dt + \sigma W^\mathbb{Q}_t$$ in which case the volatility is given by the expression in your original post.


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