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Deriving the Bachelier Price Distribution with Interest Rates

Article Quant Q&A · Author: cruiser0223

Summary

The document derives the terminal price distribution for a Bachelier-style process with proportional drift and constant volatility. Solving the stated stochastic differential equation gives a deterministic mean of the initial price grown at the interest rate, plus a stochastic integral with an exponential weighting function.

It applies Itô’s isometry to find the variance of that integral, since its integrand is deterministic and the Brownian integral has zero mean. For positive drift, the variance is proportional to the difference between the exponential growth factor and one, divided by twice the rate. The expression in the question instead corresponds to a negative drift, as in a mean-reverting Ornstein–Uhlenbeck process. The discussion focuses on checking the sign and variance; it does not assess whether either model is suitable for a particular asset or pricing task.

Key ideas

  • The stochastic integral in the terminal price has zero mean.
  • Itô’s isometry reduces its variance to the integral of the squared deterministic integrand.
  • With positive proportional drift, the variance grows with the exponential factor minus one.
  • A negative drift produces the alternate variance expression associated with mean reversion.

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# Problem at deriving Bachelier formula with interest rates


# Problem at deriving Bachelier formula with interest rates












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!

## Answer by Daneel Olivaw (score 4, accepted)

https://quant.stackexchange.com/a/46303

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.