# Pricing forward contract on a stock

Please tell me where I've gone wrong (if I did in fact make a mistake). I'm pricing a long forward on a stock. The usual setup applies:

• This has payoff $S(T) - K$ at time $T$.
• We are at $t$ now.
• $S(T) = S(t)e^{(r-\frac12 \sigma^2)(T-t)+\sigma(W(T)-W(t))}$.
• $W(t)$ is a Wiener process.
• $K \in \mathbb{R}_+$.
• $Q$ is the risk-neutral measure.
• $\beta(t) = e^{rt}$ is the domestic savings account, a tradable asset. $r$ is the constant riskless rate.

My Attempt:

$f(t,S) = E^Q[\frac{\beta(t)}{\beta(T)}(S(T)-K)|\mathscr{F}_t]$

$= E^Q [\frac{\beta(t)}{\beta(T)}S(T)|\mathscr{F}_t] - E^Q [\frac{\beta(t)}{\beta(T)}K|\mathscr{F}_t]$

$= E^{P_S}[\frac{\beta(t)}{\beta(T)}S(T) \frac{\beta(T)S(t)}{\beta(t)S(T)}|\mathscr{F}_t] - \frac{\beta(t)}{\beta(T)}K$

$= S(t) - K\frac{\beta(t)}{\beta(T)}$

$= S(t) - Ke^{-r(T-t)}$

I am increasing in confidence that this is correct because I get the same answer when I work with measure $P^*$ associated with taking the growth optimal portfolio as the numeraire. –  Jase Nov 23 '12 at 14:45
You should probably make clear what $r$ (a constant, I guess) and $r(t)$ is. The easiest would be constant interest- ie. $r$ everywhere, right? –  Richard Nov 23 '12 at 20:34
In my mind you are simply right: you arrive at $$f(t,S) = S(t) - K e^{-r(T-t)}.$$ Assume that $t=0$, so we are at the inception of the contract, then $$f(0,S) = S(0) - Ke^{-r T}.$$ If you choose $K = S(0) e^{r T}$ then the contract value at inception is zero. This simply means that the fair price for the forward is given by $K= S(0) e^{r T}$ which is the formula that you find in text books. Does this answer your question?