# 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? – Ric 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?

Richard nails it.

One needs to distinguish the forward price (or just "forward"), which is a number that denotes at which strike you can now enter a forward without upfront payment, and the value of a forward contract, which is typically zero at inception (if the strike chosen is indeed the forward price), but then varies over time, and ends up as $S(T) - K$ at T, with whatever strike K was chosen.

So, if there are no dividends and other cost of carry besides rates r, the forward price at 0 for expiry T is indeed $K = S(0) e^{rT}$, and thus the value at time $t$ of a forward contract expiring at time $T$ that was entered at time 0 is

$S(t) - S(0)e^{rt}$

which, incidentally, shows nicely that a forward has a delta of 1, at least in the absence of dividends and other distractions (which is why, incidentally, I think delta-one desks should be renamed to gamma-zero... :-)

• What if the stock paid a continuous dividend, $\delta$, during the time up until maturity? Would that give $$f(t,S)=S(t)-Ke^{-r(T-t)}-S(t)e^{\delta (T-t)}$$ or something similar? @Fab – user2069136 Apr 22 '14 at 12:25
• Hi, @user2069136: At time T, we want the value to be $S(T) - K$. Thus, need to buy $e^{-\delta T}$ shares at $t=0$, so that the number of shares grows to one at expiry. Value at time t then is: $S(t)e^{\delta(t-T)} - K e^{-r(T-t)}$. Thus, to make that 0 at $t=0$, choose $K = S(0)e^{(r-\delta)T}$ ("rates bring the fwd up, divs bring the forward down"). Plug it in to get the value of a forward contract on a stock with continuous div yield $\delta$ (expiring at T, entered at 0) at time t: $$S(t)e^{\delta(t-T)} - S(0)e^{rt-\delta T}$$ (Note that the delta of that contract <1.) – Fab Aug 11 '14 at 20:23