3
$\begingroup$

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!

$\endgroup$
  • 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 at 8:00
4
$\begingroup$

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.

$\endgroup$

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.