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)}$

This isn't graded homework or assignment. (It is ungraded homework)

  • $\begingroup$ 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. $\endgroup$
    – Jase
    Commented Nov 23, 2012 at 14:45
  • $\begingroup$ 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? $\endgroup$
    – Richi Wa
    Commented Nov 23, 2012 at 20:34

2 Answers 2


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... :-)

  • $\begingroup$ 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 $\endgroup$ Commented Apr 22, 2014 at 12:25
  • $\begingroup$ 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.) $\endgroup$
    – Fab
    Commented Aug 11, 2014 at 20:23

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.