I am trying to derive a martingale pricing equation (closed form solution) for a best-of option. But I am getting stuck at a point.

There are 2 stocks $U(t)$ and $V(t)$ they both follow GBM with a correlation of $\rho$.

I want to price an European option whose final payoff is $\max \left\{ U(t), V(t) \right\}$. For simplicity I am assuming that interest rates are 0.

$$ V(0) = \mathbb{E} \left[ \max \left\{ U(t), V(t) \right\} \right] $$

By theory of total expectation

$$ V(0) = \mathbb{E} \left[ U(t) \right] \mathbb{P} \left\{ U(t) > V(t) \right\} + \mathbb{E} \left[ V(t) \right] \left( 1 - \mathbb{P} \left\{ U(t) > V(t) \right\} \right) $$

But I guess this is not the correct approach.

Any help is highly appreciated.

  • $\begingroup$ Your approach is not correct. Ignoring discounting, you want to compute $\mathbb{E} \left[ \max \left\{ U_T, V_T \right\} \right] = \mathbb{E} \left[ U_T \mathrm{1} \left\{ U_T \geq V_T \right\} \right] + \mathbb{E} \left[ V_T \mathrm{1} \left\{ U_T < V_T \right\} \right]$. However, the payoff and the indicator in each of these expectations are not independent, so you can't simplify it to be the product of the expectations! This is just like the value of the plain vanillal option is not $\mathbb{E} \left[ S_T \right] \mathrm{1} \left\{ S_T > K \right\}$. $\endgroup$ Jul 27, 2017 at 6:36
  • $\begingroup$ The last sentence in my above comment should read "This is just like the value of the plain vanilla option is not $\mathbb{E} \left[ S_T \right] \mathbb{P} \left\{ S_T > K \right\}$." $\endgroup$ Jul 27, 2017 at 9:51

2 Answers 2


For best-of options, the usual approach is to use Magrabe's formula.

Starting from the following relationship:

$$\max\{U(t),V(t)\} = U(t) + \max\{V(t) - U(t), 0\} ,$$

you end up with the sum of one of the underlying asset, $U(t)$, and an exchange option paying $\max\{V(t) - U(t), 0\}$.

While the present value of the first term is easy to find, the second requires Magrabe's results stating that the price of the exchange option is given by:

$$V(t) N(d_U) e^{-q_V \tau} - U(t) N(d_V) e^{-q_U \tau}$$


$$d_V = \frac{\ln\left(V(t) / U(t)\right) + \left( q_U - q_V - \sigma^2 / 2\right)\tau}{\sigma \sqrt{\tau}}$$

$$d_U = d_V + \sigma \sqrt{\tau}$$


$$\sigma^2 = \sigma^2_U + \sigma^2_V - 2 \rho \sigma_U \sigma_V$$

with $\tau$ being the time-to-expiration, $q_x$ and $\sigma_x$ being respectively the dividend yield and the volatility of asset $x$, $x \in \{U,V \}$

Therefore, the price of your best-of option becomes:

$$U(t) e^{-q_U \tau} + V(t) N(d_U) e^{-q_V \tau} - U(t) N(d_V) e^{-q_U \tau}$$

which simplifies to

$$V(t) N(d_U) e^{-q_V \tau} + U(t) N(-d_V) e^{-q_U \tau}$$

  • 1
    $\begingroup$ So i was typing up an answer much like this - but i wanted to mention the issue with it. What volatility do you select? $\endgroup$
    – will
    Jul 27, 2017 at 9:31
  • $\begingroup$ Good question, there is no correct answer unfortunately. The most natural choice is to select the at-the-money implied volatility for both assets, but you could also use: from random import randint sigma = randint(0,100)/100... :) Ok maybe the latter is a bit extreme. What is worth mentioning is that the price of such derivatives is mainly driven by the underlying prices. Taking the ATM implied volatility would yield to more 'consistent' price. It would be interesting to plot the vega wrt one of the underlying's volatility to quantitatively assess its influence on the final price. $\endgroup$ Jul 27, 2017 at 9:48
  • 1
    $\begingroup$ Personally, i'd be much more comfortable sampling the terminal distributions, $\endgroup$
    – will
    Jul 27, 2017 at 10:00
  • $\begingroup$ May I ask you to develop what you mean by 'sampling' and which 'distributions' you are referring to? thanks $\endgroup$ Jul 27, 2017 at 10:07
  • $\begingroup$ Sample the smile implied distributions of S1 and S2. $\endgroup$
    – will
    Jul 27, 2017 at 10:08

Assuming GBM dynamics, I work through the (tedious) details of a worst-of call option in the more general case of a nonzero strike.


tab = Finance, subsection = Option models, file = Worst Of Call Option - Three Ways

In a real world application, you would use Monte Carlo with some sort of market calibrated local vol or stochastic vol model.

  • 1
    $\begingroup$ It's fine to link to other sites, but please also reproduce the content here, in case of link-rot. $\endgroup$ Jul 28, 2017 at 6:01
  • $\begingroup$ Thanks for the feedback. However, I am not quite sure what to do. The calculation is 20 pages of math, and would be of interest only to specialists or those who asked the question. Sorry, but I have not heard the term link-rot. If you have time, please look at the ref I provided and suggest a response appropriate for this forum. $\endgroup$
    – Tom Gladd
    Jul 29, 2017 at 6:49

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