# Bachelier model call option pricing formula

Does anybody have the Bachelier model call option pricing formula for $r > 0$?

All the references I've read assume $r = 0$. I don't speak French, so I can't read Bachelier's original paper.

• The dissertation has been translated into English. – Dave Harris Mar 7 '17 at 4:41
• (1) IIRC Bachelier did not include non-zero interest rates in his model. (2) Bachelier's paper was translated into English starting on Page 17 of the book 'The Random Character of Stock Market Prices' (3) It should not be difficult to discount the payoff at a chosen interest rate – noob2 Mar 7 '17 at 4:53

We assume that, under the risk-neutral measure, the stock process $$\{S_t, t \ge 0\}$$ satisfies an SDE of the form \begin{align*} dS_t = r S_t dt + \sigma dW_t, \end{align*} where $$r$$ is the constant interest rate, $$\sigma$$ is the constant volatility, and $$\{W_t, t \ge 0\}$$ is standard Brownian motion. For $$0 \le t \le T$$, \begin{align*} S_T = S_t e^{r(T-t)} + \sigma\int_t^T e^{r(T-s)}dW_s. \end{align*} That is, \begin{align*} S_T \mid S_t &\sim N\left(S_t e^{r(T-t)},\, \frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right) \right)\\ &\sim S_t e^{r(T-t)} + \sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}\,\xi, \end{align*} where $$\xi$$ is standard normal random variable. Then \begin{align*} C_t &= e^{-r(T-t)}E\left(\left(S_T-K\right)^+ \mid \mathcal{F}_t \right)\\ &=e^{-r(T-t)}E\left(\left(S_t e^{r(T-t)} + \sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}\,\xi-K\right)^+ \mid \mathcal{F}_t \right)\\ &=e^{-r(T-t)}\sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}E\left(\left(\xi -\frac{K-S_t e^{r(T-t)}}{\sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}}\right)^+ \mid \mathcal{F}_t \right)\\ &=e^{-r(T-t)}\left(S_t e^{r(T-t)}-K\right)\Phi\left(\frac{S_t e^{r(T-t)}-K}{\sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}}\right) \\ &\qquad + e^{-r(T-t)}\sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}\,\phi\left(\frac{S_t e^{r(T-t)}-K}{\sqrt{\frac{\sigma^2}{2r}\left(e^{2r(T-t)}-1 \right)}}\right), \end{align*} where $$\Phi$$ is the cumulative distribution function of a standard normal random variable, and $$\phi$$ is the corresponding density function.

Let $$K^*=e^{-r(T-t)}K,$$ and $$v^2(t, T) = \frac{\sigma^2}{2r}\left(1-e^{-2r(T-t)}\right).$$ Then, we can re-express the price as \begin{align*} C_t &= \left(S_t-K^*\right)\Phi\left(\frac{S_t-K^*}{v(t, T)}\right) +v(t, T)\,\phi\left(\frac{S_t-K^*}{v(t, T)}\right). \end{align*} See also Section 3.3 of the book Martingale Methods in Financial Modeling; however, note that there are a few typos in this book.

One other possibility is to assume that \begin{align*} S_t = e^{rt}(S_0 + \sigma W_t). \end{align*} Then the corresponding option price can be similarly obtained. See also the book mentioned above.

• Can U explain the last equality before the comment. I can't see why u can go from $E(....)^+$ to write it in terms of the $\Phi$ and $\phi$ – Sanjay Mar 20 '18 at 21:42
• @Sanjay: Can you compute the expectation $E\big((\xi-\alpha)^+\big)$, where $\alpha>0$ and $\xi$ is standard normal? – Gordon Mar 21 '18 at 13:51
• That is exactly the computation i do wrong (I think?) $E[ (\xi-a)1_{Z>a}] = E[\xi1_{Z>a}]-E[a1_{Z>a}] =\int_a^ \infty x\phi(x)dx-a\Phi(-a)$. I can't rewrite this term into the desired result. If I can show that: $\int_a^ \infty x\phi(x)dx=\phi(-a)$, then I think I will be able to understand your proof. – Sanjay Mar 21 '18 at 14:39
• Note that $$\int_a^{\infty} x \phi(x) dx = \frac{1}{\sqrt{2\pi}}\int_a^{\infty} x e^{-x^2/2} dx = \frac{1}{\sqrt{2\pi}} (- e^{-x^2/2})|_a^{\infty} = \phi(a).$$ – Gordon Mar 21 '18 at 15:40

It's pretty simple to derive with basic knowledge of stochastic calculus. But since you are looking for the easy answer here it is:

$$C_t=e^{-r(T-t)}\sigma\sqrt{T-t} (D \Phi(D)+\phi(D))$$ where $D=\frac{F_{t,T}-K}{\sigma \sqrt{T-t}}$ and $\Phi(\cdot)$ and $\phi(\cdot)$ are respectively the normal cdf and pdf. $F_{t,T}=S_te^{r(T-t)}$ is the forward price.

• Shouldn't the interest rate appear in your expression for $D$? I.e. I would have expected $D = \left( S_0 e^{r (T - t)} - K \right) / \sigma \sqrt{T - t}$. – LocalVolatility Mar 7 '17 at 9:31
• Indeed. Corrected – NSZ Mar 7 '17 at 9:42
• If $S=r dt + \sigma dW$ under risk neutral, then I would expect the forward to satisfy $dF = \sigma dW$, so $F_{t,T} = S_{t} + r(T-t)$, as opposed to $F_{t,T}=e^{r(T-t)}S_t$. Am I missing something? – Olaf Mar 7 '17 at 10:10
• No because the drift would be $rS_tdt$ so that the discounted process is a martingale under the risk neutral measure. – NSZ Mar 7 '17 at 10:12
• What is the dynamics of your stock process? I have difficulty to re-produce your results with both stock process assumptions in my answer above. – Gordon Mar 8 '17 at 21:44

You might want to differentiate between the growth rate $\mu$ and the discount rate $r$.

Gordon's solution is the most logical thing to do.

NSZ's solution amounts to assuming a process $$dF = \sigma dW$$ with a discount rate $r$ and $F(t,T) = S(t) e^{r(T-t)}$.

We apply Ito's Lemma to $f(t,F) = F e^{r(t-T)}$ to obtain in terms of $S$: $$dS = r S dt + \sigma e^{r(t-T)} dW\,.$$