Let an asset follow a Brownian motion $$dS = \mu dt + \sigma dW$$ with $\mu$ and $\sigma$ constant. The constant interest rate is $r$. What process does $S$ follow in the risk-neutral measure? Develop a formula for the price of a call option and for the price of a digital call option.

In chapter 6, Mark Joshi states that $\mu = r$ if and only if the stock grows at a risk-neutral rate. Then in the solution by Mark Joshi, he states that since the $S_t$ grows at the same rate as a riskless bond so its drift must be $rS_t$.

I do not see how the drift must be $rS_t$.

Then the solution goes on with $F_t = e^{r(T-t)}S_t$ then

$$dF_t = e^{r(T-t)}\sigma dW_t$$

and then states that

$$F_T\sim F_0 + \overline{\sigma}\sqrt{T}N(0,1)$$

I do not understand where this comes from. I am having a hard time following his solution. Any suggestions are greatly appreciated. I can provide the full solution if needed.


1 Answer 1


Under the risk-neutral measure the discounted (under some numéraire) price process is a martingale. If we have a bank account with dynamics $dB_t = r B_t dt$ then the discounted asset $X_t = \frac{S_t}{B_t}$ will have the dynamics

\begin{equation} dX_t = \frac{dS_t}{B_t}- \frac{S_t dB_t}{B_t^2} = (\mu - r S_t) \frac{1}{B_t} dt + \frac{\sigma}{B_t} dW_t \end{equation}

Now we make an ansatz for the risk-neutral measure $\mathbb{Q}$ defined by $dW_t = dW^\mathbb{Q}_t + \frac{r S_t - \mu}{\sigma}dt$ and see that this indeed transform the discounted price into a martingale

\begin{equation} dX_t = \frac{\sigma}{B_t} dW^\mathbb{Q} _t \end{equation}

The asset price will now have the dynamics

\begin{equation} dS_t = rS_t dt + \sigma dW^\mathbb{Q} _t \end{equation}

This equation makes it somewhat inconvenient to compute derivative prices so we instead use a forward contract on the asset with the same maturity as the derivatives we wish to price. The price of a forward with maturity date $T$ is $F_{t,T} = e^{r(T-t)} S_t$ hence

\begin{equation} d F_{t,T} = e^{r(T-t)}\sigma dW^\mathbb{Q} _t \end{equation}

Integrating from $t=0$ to $T$ gives \begin{equation} F_{T,T} = F_{0,T} + \sigma \int_0^T e^{r(T-t)}dW^\mathbb{Q} _t \end{equation}

hence $F_{T,T} \sim N( F_{0,T} , \sigma^2 \int_0^T e^{2r(T-t)}dt ) $.

Since $F_{T,T}=S_T$ we can compute the price of any derivative with payoff $g(S_T)$ using $E^\mathbb{Q}[g(F_{T,T})| \mathcal{F}_t]$. Since forward contracts are paid at maturity we must discount this back to todays value and we get the price at time $t$ with $e^{-r(T-t)} E^\mathbb{Q}[g(F_{T,T})| \mathcal{F}_t]$.

  • $\begingroup$ The part I do not understand is when you say clearly then write $F_{T,T}$ $\endgroup$
    – Wolfy
    Commented Jan 19, 2018 at 23:46
  • $\begingroup$ Anyway you can simplify this answer, I honestly cannot follow it at all. $\endgroup$
    – Wolfy
    Commented Jan 20, 2018 at 0:00
  • $\begingroup$ Clarified it a bit, you simply integrate both sides. $\endgroup$
    – Freelunch
    Commented Jan 20, 2018 at 18:27
  • $\begingroup$ I got it now thanks it was the notation that threw me off $\endgroup$
    – Wolfy
    Commented Jan 20, 2018 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.