Given a standard European call option on a non-dividend-paying stock. Draw the graph of call price at time $t$ versus the future price $F(t,T)$. The future price $F(t,T)$ is observed at time $t$, prior to maturity. The futures contract and the option both mature at the same date $T.$

Note that $F(t,T) = S(t)e^{r(T-t)}$ where $S(t)$ is the stock price at time $t$ and $r$ is interest rate.

Let $c$ be the call option value and $F$ be the future price. By Chain rule, we have $$\frac{\partial c}{\partial F} = \frac{\partial c}{\partial S} \cdot \frac{\partial S}{\partial F} = \Delta e^{-r(T-t)} = N(d_1) e^{-r(T-t)}.$$

Initially I thought that I can just solve the differential equation above and obtain $c$ in terms of $F.$ But it seems that it is not so straightforward.

Any hint is appreciated.

  • $\begingroup$ Your call value at time $t$ is wrong : here you just take intrinsic value + difference between value of the strike price at times $t$ and $T$ (why?) $\endgroup$
    – siou0107
    Dec 6 '19 at 18:02
  • $\begingroup$ @siou0107 that is a typo. I deleted that part from my question. $\endgroup$
    – Idonknow
    Dec 6 '19 at 18:05

Since you have that proportionality between the stock price $S$ and the futures price $F = Se^{rT}$, you just have to slightly shift your graph but the shape is the same.

  • $\begingroup$ So the differential is not needed? $\endgroup$
    – Idonknow
    Dec 6 '19 at 23:10
  • $\begingroup$ Right. It is a substitution of variables that redefines the horizontal axis of the graph. $\endgroup$
    – Alex C
    Dec 7 '19 at 3:52
  • $\begingroup$ Is it possible to sketch a graph of it? I still couldn't get my head of it. $\endgroup$
    – Idonknow
    Dec 9 '19 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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