I have a question about spread options. I'm pricing a put option on two assets, with a strike value of 0:


I know this kind of options could be priced using Kirk approximation, or better in this case Margrabe formula, so the correct price of this put should be:


since this is a 0 strike option the delta should simply be: $\Delta_1=-N(-d_1)$ and $\Delta_2=N(-d_2)$

What I don't understand is: I know that for a vanilla option the delta value $exp(-rT)*N(d_1)$ is often used as a rough approximation of the exercise probability. What about a spread option like this one? How can I get a "Exercise probability" from the delta values?


  • $\begingroup$ The exercise probability is given by $P(F_2 \ge F_1)$. What is it? $\endgroup$
    – Gordon
    Jun 5, 2018 at 13:03

1 Answer 1


Seeing that your question is about the how, here is the idea of the derivation.

The exercise probability is simply $\mathbb{P}(F_{2,T} > F_{1,T})$, you assumed that both are lognormal: $$\begin{aligned} F_{1,T} & = F_{1,0} e^{rT - \frac{\sigma_1^2}{2}+\sigma_1\sqrt{T}Z_1} \\ F_{2,T} & = F_{2,0} e^{rT - \frac{\sigma_2^2}{2}+\sigma_1\sqrt{T}Z_2} \end{aligned}$$

where $Z_1$ and $Z_2$ are two standard gaussians, that are correlated.

Replacing in the probability, we get: $$\begin{aligned} \mathbb{P}(F_{2,T} & > F_{1,T}) \\ & = \mathbb{P}(\log(F_{2,0})- \frac{\sigma_2^2}{2}+\sigma_2\sqrt{T}Z_2 > \log(F_{1,0}) - \frac{\sigma_1^2}{2}+\sigma_1\sqrt{T}Z_1) \\ & = \mathbb{P}\left( \frac{1}{\sqrt{T}} \left[ \log\left(\frac{F_{2,0}}{F_{1,0}} \right)- \frac{\sigma_2^2 - \sigma_1^2}{2} \right] > \sigma_1 Z_1 - \sigma_2 Z_2\right) \end{aligned}$$

You know $Z_1$ and $Z_2$ are standard gaussian with a given correlation $\rho$, so you know that $(\sigma_1Z_1 - \sigma_2Z_2)$ is gaussian with mean zero and standard deviation: $$\sigma = \sqrt{\sigma_1^2 + \sigma_2 ^2 - 2\rho\sigma_1\sigma_2}$$

Writing $\sigma_1Z_1 - \sigma_2Z_2 = \sigma Z$, and replacing in the probability expression above will then give you the result you are looking for, using the gaussian cumulative distribution:

$$\mathbb{P}(F_{2,T} > F_{1,T}) = \mathcal{N}\left(\frac{1}{\sigma\sqrt{T}} \left( \log\left(\frac{F_{2,0}}{F_{1,0}}\right) - \frac{\sigma_2^2 - \sigma_1^2}{2} \right) \right)$$

  • $\begingroup$ Thanks for your reply. I'm looking for something more quick and dirty, a sort of analogy with vanilla options using N(d1) and N(d2) $\endgroup$
    – Marco
    Jun 5, 2018 at 13:41
  • $\begingroup$ You will get a similar expression if you carry out the derivation until the end, but it will involve the correlation as well. $\endgroup$
    – byouness
    Jun 5, 2018 at 13:46
  • $\begingroup$ By the way I added the final expression to my answer. $\endgroup$
    – byouness
    Jun 5, 2018 at 13:53
  • 1
    $\begingroup$ Thanks again. It's clear. How can i link this calculation to a more practical interpretation? I work with people used to look at N(d1) as the probability of an option to be ITM $\endgroup$
    – Marco
    Jun 5, 2018 at 20:37
  • $\begingroup$ The probability that $F_2$ finishes up above $F_1$ depends on: (1) the ratio of their respective starting points $\frac{F_{2,0}}{F_{1,0}}$, (2) the difference between their variances $\sigma_2^2 - \sigma_1^2$ and (3) their correlation $\rho$. The interest rate $r$ is not in the equation because it is the drift of both assets under the risk neutral measure ( $\neq$ vanilla option case where the strike is constant but the underlying has a drift $r$). I hope it's clearer now. $\endgroup$
    – byouness
    Jun 5, 2018 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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