How to price an exchange option using B&S framework?

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.

• 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. Jun 7, 2013 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. Jun 11, 2013 at 5:09

Measure change is still the most natural approach for such problems. We assume that, under the measure $P$, \begin{align*} dX_t &= \mu X_t dt + \sigma X_t dW_t^1,\\ dY_t &= \mu Y_t dt + \sigma Y_t \left(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2 \right), \end{align*} based on the Cholesky decomposition, where $\{W_t^1, t \ge 0\}$ and $\{W_t^2, t \ge 0\}$ are two independent standard Brownian motions, and $\rho$ ($|\rho|<1$) is the correlation. We define the measure $Q$ such that \begin{align*} \frac{dQ}{dP}\big|_t = \exp\left(-\frac{1}{2}\sigma^2 t + \sigma\left(\rho W_t^1 + \sqrt{1-\rho^2} W_t^2 \right) \right). \end{align*} Then, $\{\widehat{W}_t^1, t \ge 0\}$ and $\{\widehat{W}_t^2, t \ge 0\}$ are two independent standard Brownian motions under $Q$, where \begin{align*} \widehat{W}_t^1 &= W_t^1 - \rho\sigma t,\\ \widehat{W}_t^2 &= W_t^2 - \sqrt{1-\rho^2}\sigma t. \end{align*} Moreover, \begin{align*} dX_t &= (\mu +\rho\sigma^2) X_t dt + \sigma X_t d\widehat{W}_t^1,\\ dY_t &= (\mu + \sigma^2) Y_t dt + \sigma Y_t \left(\rho d\widehat{W}_t^1 + \sqrt{1-\rho^2} d\widehat{W}_t^2 \right), \end{align*} and then \begin{align*} Y_T &= Y_0 \exp\left(\Big(\mu + \frac{1}{2}\sigma^2 \Big)T + \sigma\left(\rho \widehat{W}_T^1 + \sqrt{1-\rho^2} \widehat{W}_T^2 \right) \right),\\ \frac{X_T}{Y_T} &= \frac{X_0}{Y_0}\exp\left((\rho-1)\sigma^2T +\sigma(1-\rho)\widehat{W}_T^1-\sigma\sqrt{1-\rho^2} \widehat{W}_T^2 \right)\\ &=\frac{X_0}{Y_0}\exp\left((\rho-1)\sigma^2T +\sqrt{2(1-\rho)}\sigma \frac{\sigma(1-\rho)\widehat{W}_T^1-\sigma\sqrt{1-\rho^2} \widehat{W}_T^2}{\sqrt{2(1-\rho)}\sigma} \right)\\ &=\frac{X_0}{Y_0}\exp\left(-\frac{1}{2}\hat{\sigma}^2T +\hat{\sigma} W_T \right), \end{align*} where $\hat{\sigma} = \sqrt{2(1-\rho)}\sigma$, and \begin{align*} W_t =\frac{\sigma(1-\rho)\widehat{W}_t^1-\sigma\sqrt{1-\rho^2} \widehat{W}_t^2}{\sqrt{2(1-\rho)}\sigma} \end{align*} is a standard Brownina motion, by Levy's characterization. Therefore, \begin{align*} E_P\left( (X_T-Y_T)^+\right) &= E_P\left(Y_T \left(\frac{X_T}{Y_T}-1\right)^+\right)\\ &=E_Q\left( \left( \frac{dQ}{dP}\big|_T\right)^{-1}Y_T \left(\frac{X_T}{Y_T}-1\right)^+\right)\\ &=Y_0e^{(u+\sigma^2)T}E_Q\left(\left(\frac{X_T}{Y_T}-1\right)^+ \right)\\ &= Y_0e^{(u+\sigma^2)T} \left[\frac{X_0}{Y_0}\Phi(d_1) - \Phi(d_2) \right]\\ &= e^{(u+\sigma^2)T} \Big[X_0\Phi(d_1) - Y_0 \Phi(d_2) \Big]. \end{align*} where \begin{align*} d_1 &= \frac{\ln \frac{X_0}{Y_0} + \frac{1}{2}\hat{\sigma}^2T}{\hat{\sigma}\sqrt{T}}\\ &=\frac{\ln \frac{X_0}{Y_0} + (1-\rho)\sigma^2T}{\sqrt{2(1-\rho)}\sigma\sqrt{T}},\\ d_2 &= d_1 - \hat{\sigma}\sqrt{T}\\ &= d_1 -\sqrt{2(1-\rho)}\sigma\sqrt{T}. \end{align*}

I think there are 2 ways to get the answer. First way is what Gordon said. But when I first saw his answer, I didn't know why he defined Radon–Nikodym like that, so I thought about it for a long time, trying to give my understanding here which was inspired by this answer.

We want to define a new measure $$\mathbb{Q}^{1} \sim \mathbb{Q}$$ which uses the $$Y_{T}$$ to be the numeraire. Then we define,

$$\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}}=\frac{Y_T}{Y_0} \frac{B_0}{B_T}=\frac{Y_T}{Y_0} e^{-r T} .$$

The reason why we define this Radon–Nikodym are

1. $$E^{\mathbb{Q}}[\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}}] = 1$$(new measure)

2.$$\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}} > 0$$ under BS model can make sure $$\mathbb{Q}^1 \sim \mathbb{Q}$$ and use ratio of 2 numeraire can make sure $$\mathbb{Q}^1$$ an equivalent martingale measure.($$Y_{T}$$/numeraire is a martingale under corresponding measure)

$$\frac{Y_{0}}{B_{0}} = E^{\mathbb{Q}}[\frac{Y_{T}}{B_{T}}] = E^{\mathbb{Q}^1}[\frac{\mathrm{~d} \mathbb{Q}}{\mathrm{d} \mathbb{Q}^1}\frac{Y_{T}}{B_{T}}] = E^{\mathbb{Q}^1}[\frac{Y_{T}}{Y_{T}}\frac{Y_{0}}{B_{0}}]$$ Thus we have $$E^{\mathbb{Q}^1}[\frac{Y_{T}}{Y_{T}}] = \frac{Y_{0}}{Y_{0}}$$

Back to the question, under measure $$\mathbb{Q}$$, $$\begin{gathered} d X_t=r X_t d t+\sigma X_t d W^{1}_t \\ d Y_t= r Y_t d t+\sigma Y_t d W^{2}_t \end{gathered}$$ where $$W^{1}_{t}$$ and $$W^{2}_t$$ are 2 Brownian Motion under $$\mathbb{Q}$$ and $$dW^{1}_{t}dW^{2}_t = \rho t$$.

We know that $$Y_{T} = Y_{0}e^{(r - \frac{1}{2}\sigma^2)T - \sigma W^{2}_{T}}$$. If we just substitude this $$Y_{T}$$ formula into $$\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}}$$, this Radon–Nikodym only contatins one Brownian Motion $$W^{2}_{T}$$, we can not use multi-dimension Girsanov. Thus here is the trick, we use cholesky decompostion to obtain that \begin{aligned} d X_t&=r X_t d t+\sigma X_t d W^{'}_t \\ d Y_t&= r Y_t d t+\sigma Y_t d (\rho W^{'}_t + \sqrt{1 - \rho^2} W^{''}_t) \end{aligned} where $$W^{'}_t$$ and $$W^{''}_t$$ are 2 independent Brownian Motion.Thus we have $$Y_{T} = Y_{0}e^{(r - \frac{1}{2}\sigma^2)T - \sigma(\rho W^{'}_T + \sqrt{1 - \rho^2} W^{''}_T)}$$. Now we substitude $$Y_{T}$$ into $$\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}}$$, then $$\frac{\mathrm{d} \mathbb{Q}^1}{\mathrm{~d} \mathbb{Q}}=\frac{Y_T}{Y_0} e^{-r T} = \frac{Y_{0}e^{(r - \frac{1}{2}\sigma^2)T - \sigma(\rho W^{'}_T + \sqrt{1 - \rho^2} W^{''}_T)}}{Y_0} e^{-r T} = e^{-\frac{1}{2}\sigma^2T - \sigma(\rho W^{'}_T + \sqrt{1 - \rho^2} W^{''}_T)}$$ which is coninside with how Gordon defined.

The second way is we investigate the dynamic of $$\frac{X_{T}}{Y_{T}}$$. Use ito formula to $$f:= \frac{x}{y}$$ we can derive that \begin{aligned} df &= \left(\sigma^2-\rho \sigma^2\right) f d t-\sqrt{2\sigma^2-2 \rho \sigma^2} f d W_{t} \end{aligned} Thus to make $$\frac{X_{T}}{Y_{T}}$$ a martingale, we need to have $$\bar{W}_{t}:= W_{t} - \frac{\sigma^2-\rho \sigma^2}{\sqrt{2\sigma^2-2 \rho \sigma^2} } t$$. Thus by girsanov, we define the similar Radon–Nikodym.(Actually I didn't use cholesky in this method, If we use cholesky first and do the ito formula, I think it will result in the same Radon-Nikodym by the reason maybe the uniqueness of Martingale Measure.)

By the way, I think in Gordon answer of \begin{aligned} E_P\left(\left(X_T-Y_T\right)^{+}\right) & =E_P\left(Y_T\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =E_Q\left(\left(\left.\frac{d Q}{d P}\right|_T\right)^{-1} Y_T\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =Y_0 e^{\left(u+\sigma^2\right) T} E_Q\left(\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =Y_0 e^{\left(u+\sigma^2\right) T}\left[\frac{X_0}{Y_0} \Phi\left(d_1\right)-\Phi\left(d_2\right)\right] \\ & =e^{\left(u+\sigma^2\right) T}\left[X_0 \Phi\left(d_1\right)-Y_0 \Phi\left(d_2\right)\right] . \end{aligned} should be \begin{aligned} E_P\left(\left(X_T-Y_T\right)^{+}\right) & =E_P\left(Y_T\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =E_Q\left(\left(\left.\frac{d Q}{d P}\right|_T\right)^{-1} Y_T\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =Y_0 e^{\mu T} E_Q\left(\left(\frac{X_T}{Y_T}-1\right)^{+}\right) \\ & =Y_0 e^{\mu T}\left[\frac{X_0}{Y_0} \Phi\left(d_1\right)-\Phi\left(d_2\right)\right] \\ & =e^{\mu T}\left[X_0 \Phi\left(d_1\right)-Y_0 \Phi\left(d_2\right)\right] . \end{aligned}

• Could you give me some upvotes please if anyone find that helpful! (since I don't have enough reputation to comment. hahah) Sep 13 at 16:44

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 $G(Z,T) = Y(Z - 1)^+$

UPDATE: PREVIOUS VERSION WAS NOT COMPLETELY CORRECT

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.

Relatively quick Solution

If $U$ and $V$ be normally distributed with means $\mu_u\,,\,\mu_v$, variances $\sigma^2_u\,,\,\sigma^2_v$ and correlation $\rho$ then we can show ( by definition of expectation and apply joint density function ) $$\mathbb{E}\left[\left(e^U-e^V\right)^+\right]={\large{e^{\mu_u+\frac{1}{2}\sigma_u^2}}}\Phi\left(d_1\right)-{\large{e^{\mu_v+\frac{1}{2}\sigma_v^2}}}\Phi\left(d_2\right)\qquad (1)$$ where $$d_1=\frac{\mu_u-\mu_v+\sigma_u^2-\rho\sigma_u\sigma_v}{\sqrt{\sigma_u^2-2\rho\sigma_u\sigma_v+\sigma_v^2}}$$ and $$d_2=\frac{\mu_v-\mu_u+\sigma_v^2-\rho\sigma_v\sigma_u}{\sqrt{\sigma_v^2-2\rho\sigma_v\sigma_u+\sigma_u^2}}$$ By application of Ito's lemma and Girsanov theorem we have $$\ln X_T=\ln X_t+\left(r-\frac{1}{2}{\sigma_x}^2\right)(T-t)+{\sigma_x}({W_T}^{\mathbb{Q}}-{W_t}^{\mathbb{Q}})$$ $$\ln Y_T=\ln Y_t+\left(r-\frac{1}{2}{\sigma_y}^2\right)(T-t)+{\sigma_y}({B_T}^{\mathbb{Q}}-{B_t}^{\mathbb{Q}})$$ let $U=\ln X_T$ and $V=\ln Y_T$ apply $(1)$

Edit for SRKX

We know $${{f}_{U,V}}(u,v)=\frac{1}{2\pi {{\sigma }_{u}}{{\sigma }_{v}}\sqrt{1-{{\rho }^{2}}}}{{e}^{\large{-\frac{1}{2(1-{{\rho }^{2}})}\left[ {{\left( \frac{u-{{\mu }_{u}}}{{{\sigma }_{u}}} \right)}^{2}}-2\rho \left( \frac{u-{{\mu }_{u}}}{{{\sigma }_{u}}} \right)\left( \frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}} \right)+{{\left( \frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}} \right)}^{2}} \right]}}}$$ let $$I=\int_{-\infty }^{+\infty }{\int_{v }^{\infty }{{{e}^{u}}}}{{f}_{U,\,V}}(u,v)dudv$$ and define $$g(u,v)=\frac{1}{{{\sigma }_{u}}\sqrt{2\pi (1-{{\rho }^{2}})}}{{e}^{\large{-\frac{1}{2(1-{{\rho }^{2}})}{{\left[ u-\left( {{\mu }_{u}}+\rho \sigma_u \left( \frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}} \right) \right) \right]}^{2}}}}}$$ Indeed $g$ is probability density function normal distribution. we have $$I=\int_{-\infty }^{\infty }{\frac{1}{{{\sigma }_{v}}\sqrt{2\pi }}{{e}^{\large{-\frac{1}{2}{{\left( \frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}} \right)}^{2}}}}}}\left[ \int_{v }^{\infty }{{{e}^{u}}}g(u,v)du \right]dv$$ thus we can write $$I={{e}^{{{\mu }_{u}}+\frac{1}{2}(1-{{\rho }^{2}}){{\sigma }_{u}}^{2}}} \int_{-\infty }^{\infty }{\frac{1}{{{\sigma }_{v}}\sqrt{2\pi }}{{e}^{\large{-\frac{1}{2}{{\left[ \left(\frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}}\right)^2-2\rho\sigma_v\left(\frac{v-\mu_v}{\sigma_v}\right) \right]}}}}}}\Phi \left[ \Lambda \right]dv$$ where $$\Lambda =\frac{{{\mu }_{v}}+\rho {{\sigma }_{u}}\left( \frac{v-{{\mu }_{v}}}{{{\sigma }_{v}}} \right)+{{(1-\rho )}^{2}}{{\sigma }_{u}}^{2}-v}{{{\sigma }_{u}}\sqrt{1-{{\rho }^{2}}}}$$ let $y=\large \frac{v-{{\mu }_{v}}-\rho {{\sigma }_{u}}{{\sigma }_{v}}}{{{\sigma }_{v}}}$ by Change of variable and setting good variable we have $$I={{e}^{{{\mu }_{u}}+\frac{1}{2}\sigma _{u}^{2}}}\Phi ({{d}_{1}})$$ using similar steps for the case $$J=\int_{-\infty }^{+\infty }{\int_{v }^{\infty }{{{e}^{v}}}}{{f}_{U,\,V}}(u,v)dudv$$ as a result

$$J={{e}^{{{\mu }_{V}}+\frac{1}{2}\sigma _{V}^{2}}}\Phi ({{d}_{2}})$$ Now we should use $I$ and $J$ and calculate $$E\left[(e^U-e^V)^+\right]=\int_{-\infty}^{\infty}\int_{v}^{\infty}(e^u-e^v)f_{U,V}(u,v)dudv$$

• The whole trick here is to compute the expectation you label as (1), so it's not "really" a quick solution. Do you know somewhere where we could find the details of this derivation?
– SRKX
Jun 18, 2016 at 13:15
• Hi SRKXI , I use relatively quick Solution as label. However you are right , I saw equation (1) in exam of advance Statistics, although i can not solved it but i can show it now. Unfortunately I have not reference.
– user16651
Jun 18, 2016 at 13:30
• If you could add the demo at the end it would be be really good.
– SRKX
Jun 18, 2016 at 13:56
• Ok I will do it
– user16651
Jun 18, 2016 at 14:35
• It was edited..
– user16651
Jun 18, 2016 at 16:16

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

• 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, 2013 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. Jun 7, 2013 at 15:03
• Then use a comment.
– SRKX
Jun 7, 2013 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, 2013 at 17:03