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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.

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  • $\begingroup$ 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, 2019 at 18:02
  • $\begingroup$ @siou0107 that is a typo. I deleted that part from my question. $\endgroup$
    – Idonknow
    Dec 6, 2019 at 18:05

1 Answer 1

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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.

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  • $\begingroup$ So the differential is not needed? $\endgroup$
    – Idonknow
    Dec 6, 2019 at 23:10
  • $\begingroup$ Right. It is a substitution of variables that redefines the horizontal axis of the graph. $\endgroup$
    – Alex C
    Dec 7, 2019 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, 2019 at 13:00

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