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

$max(K-(F_1-F_2);0)=max(0-(F_1-F_2);0)=max(F_2-F_1;0)$

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:

$p=exp(-rT)*(-F_1N(-d_1)+F_2N(-d_2))$

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?

Thanks

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  • $\begingroup$ The exercise probability is given by $P(F_2 \ge F_1)$. What is it? $\endgroup$ – Gordon Jun 5 '18 at 13:03
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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:

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

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  • $\begingroup$ 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 '18 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 '18 at 13:46
  • $\begingroup$ By the way I added the final expression to my answer. $\endgroup$ – byouness Jun 5 '18 at 13:53
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    $\begingroup$ 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 '18 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 '18 at 21:08

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