Suppose I have an adapted process $X_t$. I have available option prices on $X_t$ for a range of strikes and maturites. In particular, I have

$$ C_0(K, T_1) = D(T_1)\mathbb{E}_Q[(X_{T_1} - K)_+], $$


$$ C_0(K, T_2) = D(T_2)\mathbb{E}_Q[(X_{T_2} - K)_+] $$ where $D(t)$ is the discounting factor (and I am assuming determinisitc IR). I want to synthesise an option with payoff $$ \bigg( \frac{X_{T_2}}{X_{T_1}} - K^\prime \bigg)_+ $$ at time $T_2$ (assuming $T_1$ < $T_2$). Is this possible using the available option prices?

Additional Info:

  • I am aware of Margrabe's formula for the pricing of an option on the ratio of two different assets, but I don't think this can be applied here?
  • Is there any literature on options like this? It is hard to search for something when you don't know the technical name.



This is a forward starting option. It's price cannot be determined by the vanillas you have because vanillas only determine the marginal distributions of the random variables - whereas to price this one, you need to somehow produce a joint distribution (correlate $X_{T1}$ and $X_{T2}$).

An equivalent way of doing the same thing is the following. You need to be able to have a good forecast of the forward volatility (i.e., what price would a vanilla option with maturity $T2$ trade on date $T1$?). The expected value of this volatility will determine the price of this option. Note that this is equivalent to specifying the joint distribution in paragraph one, because talking in terms of forward volatility is the same as talking in terms of (intertemporal) correlation. High correlation means low forward vol and vice versa.

Edit: In case you just want to spit out some price without caring too much about fair value, you may correlate these by a copula, and integrate the payoff.

  • $\begingroup$ could you please explain further the assertion "High correlation means low forward vol and vice versa" $\endgroup$ – Bogaso Aug 6 '20 at 20:08
  • $\begingroup$ Consider the degenerate case where correlation is perfect: one variable is a deterministic function of the other. Then forward vol is exactly 0 - the random variable is already known. $\endgroup$ – Arshdeep Singh Duggal Aug 7 '20 at 4:53

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.