4
$\begingroup$

I've got a question about 1976 Black Model and Bachelier model.

I know that a geometric brownian motion in the P measure $dS_{t}=\mu S_{t}dt+\sigma S_{t} dW_{t}^{P}$ for a stock price $S_{t}$ leads (after a change of measure) to the Black-Scholes formula for a Call:

$$C= S_{0} N(d_{1}) − Ke^{−rT} N(d_{2})$$.

Where $d_{1} = \frac{ln(\frac{S_{0}}{K})+(r+\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$

I actually don't know how's possible to get the famous black formula on a forward contract:

$$C= e^{−rT}(F N(d_{1}) − KN(d_{2}))$$.

where now $d_{1} = \frac{ln(\frac{F}{K})+\frac{1}{2}\sigma^{2}T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$

Should I simply insert $F(0,T)=S_{0}e^{rT}$ in the first BS formula to get the second one?

I'm asking this because I've tried to derive the BS formula using an arithmetic brownian motion like $dS_{t}=\mu dt+\sigma dW_{t}^{P}$, and I get:

$$C= S_{0} N(d) + e^{−rT}[v n(d)-K N(d)]$$.

where $d=\frac{S_{0}e^{rT}-K}{v}$ and $v=e^{rT}\sigma\sqrt{\frac{1-e^{−2rT}}{2r}}$ and remembering that $N(d)$ and $n(d)$ are the CDF and PDF.

but the previous substitution $F(0,T)=S_{0}e^{rT}$ doesn't seems to lead to the known result $C= e^{−rT}[(F-K)N(d)-\sigma\sqrt{T}n(d)]$

where now $d=\frac{F-K}{\sigma\sqrt{T}}$

I think I could reach the equations on forward both in the geometric brownian motion and arithmetic brownian motion using the equations

$dF=F\sigma dW_{t}^{Q}$ and $dF=\sigma dW_{t}^{Q}$ but I don't know how justify the use of them.

$\endgroup$
  • $\begingroup$ @Macro Welcome to Quant. S.E.! Do you want to price just forward contract or option on forward contract? $\endgroup$ – Neeraj Feb 28 '16 at 18:31
  • $\begingroup$ Hi Neeraj, thanks for your answer. I'd like to price an option on forward contract! $\endgroup$ – Marco Feb 28 '16 at 20:37
  • $\begingroup$ Just replace $S_0$ with $F e^{-rT}$ in your original BS formula or you can use risk neutral approach. Both will lead to the same valuation formula. $\endgroup$ – Neeraj Feb 29 '16 at 6:56
  • $\begingroup$ Ok, thanks. But can I do the same for the ABM? Because I can't get the result when I do this substitution. $\endgroup$ – Marco Feb 29 '16 at 8:02
1
$\begingroup$

European option on future

To price European Option on Future, you just need to replace $S_0$ with $Fe^{-rT}$ in your original BS formula or you can use risk neutral approach. Both will lead to same Valuation formula.

American option on future

Above procedure can not be used to price American option on future. In a paper, The valuation of options on future contracts by Ramaswamy, stated that

There are no known analytical solution to the valuation of American option on future contract.

Authors used implicit finite difference method to price American option on future contract.


Edit: Derivation of price of European option on future contract

Under risk neutral measure, future price, $F_t$ satisfy following SDE: $$dF_t = \sigma F_t dW_t$$ where, $W_t$ is a Wiener process. It can be easily shown that: $$F_T|F_t= F_t e^{-\frac{1}{2}\sigma^2 (T-t) + \sigma (W_T- W_t)} $$ $$F_T|F_t \sim logN \left( ln(F_t) - \frac{1}{2}\sigma^2 (T-t), \sigma^2(T-t)\right) $$

The price of option on future contract $(C_t)$ under risk neutral measure is: $$C_t = e^{-r(T-t)}E_\mathbb{Q} [(F_T - K)^+]$$

You can easily solve the above expression to get the price of option written on future. The distribution of $F_T$ is very similar to $S_T$ (see this answer). If you replace $$ln(F_t) =ln(S_t) + r(T-t) $$ then you will get the same distribution of $S_T$ as under risk neutral measure. This is the reason, to get the price of option on future, we replace $S_t$ with $F_t e^{-r(T-t)}$ in BS model of European call option price.

$\endgroup$
  • $\begingroup$ Hi Neeraj, actually I'd like to price an European option starting from an ABM. $\endgroup$ – Marco Feb 29 '16 at 16:16
  • $\begingroup$ @Marco please check edit answer. $\endgroup$ – Neeraj Feb 29 '16 at 18:13

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.