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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.

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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 $e^{-r(T-t)} E^\mathbb{Q}[g(F_{T,T})| \mathcal{F}_t]$.

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

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