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Consider a market composed by two stocks whose prices $X$ and $Y$ are given by B&S diffusion

$$dX_t= \mu X_t dt+ \sigma X_tdW_t$$

$$dY_t= \mu Y_t dt+ \sigma Y_tdB_t$$ Supposing the market is complete, how to evaluate the fair price of an option whose payoff is $ \phi(X_T,Y_T)=(X_T-Y_T)_+$ ?

My idea was to apply a change of numeraire technique and so obtain price as a function of the B&S formula. However, I was not able to find it.

Any advice would be appreciated.

share|improve this question
So what work have you done so far (re answering this question)? Btw this site is intended for practitioners in the quant industry your question looks awfully like homework. – Matt Wolf Jun 7 '13 at 1:54
Look up Margrabe option. Essentially you fix the numeraire to be a unit of one the stocks, and price the other in those units. The pricing formula works out neatly for the payoff. – Veeken Jun 11 '13 at 5:09
See solution at quant.stackexchange.com/questions/21212/…. – Gordon Jan 26 at 19:09

That's a great question and it is what I always wanted to try to do.

I guess I found a solution using PDE approach. Change of numeraire would be more intuitive indeed, but I am not very good in stochastic calculus.

The idea is as follows:

1) Let's consider portfolio $\Pi = V(X,Y,t) - \Delta_X X - \Delta_Y Y$. I will found $\Delta_X$ and $\Delta_Y$ such that portfolio $\Pi$ would be riskless and earn risk-free rate of return $r$: $d\Pi = r\Pi dt$.

Assumption: $dX = \mu_X X dt + \sigma_X X dW^X$, $dY = \mu_Y Y dt + \sigma_Y Y dW^Y$ and $dW^X dW^Y = \rho dt$.

Hence, applying Ito's lemma I obtain: $d\Pi = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial X} dX + \frac{\partial V}{\partial Y} dY + \frac{1}{2} \sigma_X^2 X^2 \frac{\partial^2 V}{\partial X^2} dt+ \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 V}{\partial Y^2} dt+ \rho \sigma_X\sigma_Y XY \frac{\partial^2 V}{\partial X\partial Y} dt - \Delta_X dX - \Delta_Y dY =$

$\left( \frac{\partial V}{\partial t} + \frac{1}{2} \sigma_X^2 X^2 \frac{\partial^2 V}{\partial X^2}+ \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 V}{\partial Y^2} + \rho \sigma_X\sigma_Y XY \frac{\partial^2 V}{\partial X\partial Y} \right)dt + \left(\frac{\partial V}{\partial X} - \Delta_X \right) dX + \left(\frac{\partial V}{\partial Y} - \Delta_Y \right) dY$.

And all this is equal to $d\Pi = r\Pi dt = r\left(V - \Delta_X X - \Delta_Y Y\right)dt$

Now, set $\frac{\partial V}{\partial Y} = \Delta_Y$ and $\frac{\partial V}{\partial X} = \Delta_X$.

Left-hand side becomes $\left(\frac{\partial V}{\partial t} + \frac{1}{2} \sigma_X^2 X^2 \frac{\partial^2 V}{\partial X^2}+ \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 V}{\partial Y^2} + \rho \sigma_X\sigma_Y XY \frac{\partial^2 V}{\partial X\partial Y}\right) dt$

Right-hand side is now $r\left(V - \frac{\partial V}{\partial X} X - \frac{\partial V}{\partial Y} Y\right)dt$

The PDE is now $\frac{\partial V}{\partial t} + \frac{1}{2} \sigma_X^2 X^2 \frac{\partial^2 V}{\partial X^2}+ \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 V}{\partial Y^2} + \rho \sigma_X\sigma_Y XY \frac{\partial^2 V}{\partial X\partial Y} = r\left(V - \frac{\partial V}{\partial X} X - \frac{\partial V}{\partial Y} Y\right)$, or

$\frac{\partial V}{\partial t} + \frac{1}{2} \sigma_X^2 X^2 \frac{\partial^2 V}{\partial X^2}+ \frac{1}{2} \sigma_Y^2 Y^2 \frac{\partial^2 V}{\partial Y^2} + \rho \sigma_X\sigma_Y XY \frac{\partial^2 V}{\partial X\partial Y} + r\frac{\partial V}{\partial X} X + r \frac{\partial V}{\partial Y} Y = rV$

I forgot: the boundary condition is $V(X, Y, T) = (X - Y)^+$

2) Now, in order to solve this crazy PDE i will use substitution: $Z = \frac{X}{Y}$ and $V(X,Y,t) = G(Z, t)$.

Thanks to Wolfram Alpha, I have:

$\frac{\partial V}{\partial X} = \frac{1}{Y} \frac{\partial G}{\partial Z}$

$\frac{\partial V}{\partial Y} = -\frac{X}{Y} \frac{\partial G}{\partial Z}$

$\frac{\partial^2 V}{\partial X^2} = -\frac{1}{Y^2} \frac{\partial^2 G}{\partial Z^2}$

$\frac{\partial^2 V}{\partial Y^2} = \frac{X\left(2Y\frac{\partial G}{\partial Z}+X\frac{\partial^2 G}{\partial Z^2}\right)}{Y^4} $

$\frac{\partial^2 V}{\partial XY} = -\frac{Y\frac{\partial G}{\partial Z}+X\frac{\partial^2 G}{\partial Z^2}}{Y^3} $

Substituting into previous equation and cancelling the terms out we obtain:

$\dot{G} + [\sigma_X^2-\rho \sigma_X \sigma_Y]ZG' + \frac{1}{2}[\sigma_X^2-2\rho \sigma_X \sigma_Y + \sigma_Y^2]Z^2G'' = rG$, or

$\dot{G} + \mu_GZG' + \frac{1}{2}\sigma_G^2 Z^2G'' = rG$, where

$\dot{G} = \frac{dG}{dt}$, $G' = \frac{dG}{dZ}$

$\mu_G = \sigma_X^2-\rho \sigma_X \sigma_Y$, $\sigma_G = \sqrt{\sigma_X^2-2\rho \sigma_X \sigma_Y + \sigma_Y^2}$

And the boundary condition is $G(Z,T) = Y(Z - 1)^+$


3) Now the question is what to do with that $Y$ in the equation above? I employ next change of variables: $G(Z) = YF(Z)$.

Thanks to paper and pencil, I have:

$G' = (YF)' = Y\left(F' - \frac{F}{Z}\right)$ and $G'' = \left((YF)'\right)' = \text{after some calculations} = YF''$

Plugging this into $Z$'s PDE we obtain:

$\dot{F} + \mu_G Z F' + \frac{1}{2} \sigma_G^2 Z^2 F'' = (r+ \mu_G)F$ with boundary condition $F(Z,T) = (Z-1)^+$

Now denote $r^* = r+ \mu_G$ and equation becomes: $\dot{F} + (r^* - r) Z F' + \frac{1}{2} \sigma_G^2 Z^2 F'' = r^* F$

4) Now $r^*$ works like new risk-free rate and $r$ is like $Z$'s dividend yield and we can apply well-known formula for option on asset with continiously paid dividends:

$F(Z, T) = e^{-r^*T} N(d_1) Z_0 - e^{-rT} N(d_2) $, where $d_{1,2} = \frac{1}{\sigma_G\sqrt{T}}\left[\ln\left(Z_0\right)+\left(r^* - r \pm\frac{\sigma_G^2}{2}\right)T\right] =\frac{1}{\sigma_G\sqrt{T}}\left[\ln\left(Z_0\right)+\left(\mu_G \pm\frac{\sigma_G^2}{2}\right)T\right] $.

5) Now $V = e^{-r^*T} N(d_1) X_0 - e^{-rT} N(d_2) Y_0$, where $d_{1,2} =\frac{1}{\sigma_G\sqrt{T}}\left[\ln\left(\frac{X_0}{Y_0}\right)+\left(\mu_G \pm\frac{\sigma_G^2}{2}\right)T\right] $, where

$r^* = r+ \mu_G$

$\mu_G = \sigma_X^2-\rho \sigma_X \sigma_Y$

$\sigma_G = \sqrt{\sigma_X^2-2\rho \sigma_X \sigma_Y + \sigma_Y^2}$

Hope I was correct.

I also hope somebody would be able to propose any better solution, maybe using martingale approach.

share|improve this answer

You should study the dynamic of $X_t-Y_t$, don't forget about correlation and that the brownian motion is not the same.

I am pretty sure that there is some big flaws in your model (as taking same interest rate). You should really take a look at this: John C. Hull Options, Futures and Other Derivatives

share|improve this answer
You're not really answering the question here! It is pretty obvious that the payoff function dynamics will be the focus of the problem... – SRKX Jun 7 '13 at 14:24
It seems a homework question. I don't want to give him the answer but just a hint. I will try to give a better hint. – Were_cat Jun 7 '13 at 15:03
Then use a comment. – SRKX Jun 7 '13 at 15:28
@Imorin: It's not a homework question, even if it's a basic question. I've just got stuck and so I'd like some help. This question interest me as a inspiration for an bigger problem. – Paul Jun 7 '13 at 17:03

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