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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.

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  • $\begingroup$ @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
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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.

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  • $\begingroup$ 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
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Here's a simple way to get the price of the call on the forwards price using risk neutral pricing.

Suppose we have a European call that pays at $t = T$, $(For(T,T^*) - K)^+$, where $T^* \geq T$. Further assume interest rates are constant and are represented by "$r$". Let $c^{For}(t, s)$ be the price of the call where $S(t) = s$.

Then if the stock pays no dividends:

$c^{For}(t,s) = \widetilde{\mathbb{E}}[e^{-r(T-t)}(For(T,T^*) - K)^+|S(t) = s]$, By replication it can be shown, $For(T,T^*) = S(T)e^{r(T* - T)}$, and
$c^{For}(t,s) = \widetilde{\mathbb{E}}[e^{-r(T-t)}(S(T)e^{r(T* - T)} - K)^+|S(t) = s]$

You should immediately notice since interest rates are constant, and thus deterministic, we can pull the "$e^{r(T^*-T)}$" term out of the expectation:

$c^{For}(t,s) = e^{r(T* - T)}\widetilde{\mathbb{E}}[e^{-r(T-t)}(S(T) - e^{-r(T* - T)}K)^+|S(t) = s]$

Thus this is now proportional to the Black Scholes call price with strike $X = e^{-r(T* - T)}K$

$c^{For}(t,s) = e^{r(T* - T)}c^{B.S.}(t,s | X = e^{r(T* - T)K}$) $c^{For}(t,s) = e^{r(T^* - T)}[SN(d_+) - e^{-r(T-t)}e^{-r(T* - T)}KN(d_-)]$ $c^{For}(t,s) = e^{r(T^* - T)}[SN(d_+) - e^{-r(T* - t)}KN(d_-)]$
$c^{For}(t,s) = e^{-r(T - t)}(FN(d_+) - KN(d_-))$, where $F = Se^{r(T^* - t)}$

also:
$d_{\pm} = \frac{1}{\sigma\sqrt{T-t}}[ln(\frac{S}{K}) + (r \pm \frac{1}{2}\sigma^2)(T-t)]$

This is the "famous black formula on a forward contract". I hope this helps!

Please note that that the forward price and the price of the forward contract are not the same. The price of the forward contract at time 0 is 0, but may change, the forward price is the price you agree to pay at delivery.

If you are curious what it would be if it were a call on the futures price instead of a call on the forward price, I claim if the asset price is not correlated with the interest rate, then they are the same otherwise there would be arbitrage (under assumptions of no counterparty risk, etc.). I encourage you to try and show this.

(P.S. To the previous commenters response about there being no formula for an American option on the forward price, this doesn't stop us from using monte carlo!)

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