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I want to study the following market:

$$S_1(t)=S_1(t)(\mu_1dt + \sigma_1dW_1(t))$$


for $t\in [0,T]$, constants $\mu_i,\sigma_i$, initial values $S_i(0)>0$, $i\in\{1,2\}$ and $W_1,W_2$ two Brownian Motion with the following bracket:

$$\langle W_1,W_2\rangle =\rho t$$ with $\rho\in\mathbb{R} $. We look at the following payoff:


and I want to determine the price at $0$ of an option with the above payoff. It was hinted (on an old exercise from last year) to find a measure $Q$ under which the quotient $S_1/S_2$ is a martingale and calculate the price at $0$ of a European call option with strike $K=1$. Since no bank account was mentioned in the exercise text, I'm a little bit confused. Supposing there is a bank account $B(t)$, and assuming absence of arbitrage, we know there is a EMM $Q^1$ under which the discounted assets are martingales, i.e. $S_i(t)/B(t)$ are martingales. In this case I would define the density (of $Q$ w.r.t. to $Q^1$) on $\mathcal{F}_T$ as follows:


and $Z_t:=E[Z_T|\mathcal{G}_t]=\frac{1}{B(t)}\frac{S_2(t)}{S_2(0)}$. Then using Bayes formula I'm able to prove that $\frac{S_1}{S_2}$ is a $Q$-martingale. Are you think my observations are correct so far?

Now I want to compute $$E_Q[(\frac{S_1(t)}{S_2(t)}-1)^+|\mathcal{F}_0]$$

I wrote down:


Using Bayes gives


Under $Q^1$, $\frac{S_i}{B}$ is given by


Where $W^*=W_t+\frac{\mu_i-r}{\sigma}t$ is $Q^1$ Brownian motion. Hence the above is under $Q^1$, $\exp{(Z)}$ distributed, where $Z$ is normal distributed with mean $-\frac{1}{2}\sigma_i^2t$ and Variance $\sigma^2t$. Is it true, that this conditional expectations reduce to expectations? If not, how would you calculate this?

However I am not at all sure about this derivation so far. Since the bracket of $W_1$ and $W_2$ is explicitly given, I thought somewhere I have to use it. It would be appreciated, if I did a error in reasoning, that someone could explain how to solve the problem, i.e. calculate the price. Thanks in advance


share|cite|improve this question
1. looks fine $\text{ }{}$ 2. you can look this up under option to exchange or better of ${}\text{ }$3. there are issues (dividends, r) but basically you factor $S_2(t)$ out and all that factor does is change the drift of $W_2$ – mike Nov 26 '12 at 14:08

Let $\beta(t) := e^{rt}$, where $r$ is the domestic risk-free rate. Let $Q$ be the risk-neutral measure.

The value is:

$v_t := E^Q[\frac{\beta(t)}{\beta{T}}S_1(T) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t] - E^Q[\frac{\beta(t)}{\beta{T}}S_2(T) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t]$

$=: I_1 + I_2$.

To evaluate $I_1$, do a change of probability measure to $S_1(T)$ as the numeraire asset:

$ E^{P_1}[S_1(t) 1_{\{S_1(T) > S_2(T)\}} | \mathscr{F}_t] = S_1(t)P_1\{S_1(T) > S_2(T) | S_1(t),S_2(t)\}$;

where the Radon-Nikodym derivative to go from $Q$ to $P_1$ was:

$\frac{dP_1}{dQ} = \frac{S_1(T)\beta(0)}{S_1(0)\beta(T)}$

meaning that, we can read off this Doleans exponential using Girsanov's theorem to get our new Wiener process:

$dW_{P_1}(t) = dW_1^Q - \sigma_1 dt$

$W_{P_1}(t) = W_1^Q - \sigma_1 t$

Now, $S_1(t)P_1\{S_1(T) > S_2(T) | S_1(t),S_2(t)\} = S_1(t)P_1\{\text{ln}\frac{S_1(T)}{S_2(T)} > 0\}$

by the monotone increasing property of $\text{ln}$. Under measure $P_1$ and $\mathscr{F}_t$,

$S_1(T) = S(t) e^{(r+\frac12 \sigma_1^2)(T-t) + \sigma_1 (W_1(T) - W_1(t))}$.

Under $Q$, we can decompose $S_2(T)$'s Wiener process ($W_2^Q$) into Wiener processes $W_1^Q(t)$ and $W_1^\perp(t)$, where $W_1^\perp(t)$ is a BM uncorrelated to $W_1^Q(t)$. Under $Q$:

$W_2^Q = \rho W_1^Q + \sqrt{1-\rho^2}W_1^\perp$

... You then use Levy's characterization to solve the rest of the problem.

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thanks for your answer. Just one question: Why can we decompose $W^Q_2=\rho W^Q_1+\sqrt{1-\rho^2}W^\perp_1$ ? Is this always possible, i.e. for a general Brownian motion? – math Dec 1 '12 at 9:24
@math Unfortunately this is a black box to me (I lifted it from lectures). All I know is that it very very easily allows one to compute the solution to options such as the Magrabe option (which is the one you're asking about). – Jase Dec 1 '12 at 10:55
Thanks for your quick response. I will check my book about stochastic calculus. There is another point, which is not completely clear to me: Why is this equality true: $P_1\{S_1(T)>S_2(T)|S_1(t),S_2(t)\} = P_1\{\log(\frac{S_1(T)}{S_2(T)}>0\}$? – math Dec 1 '12 at 10:59
@math I made it ambiguous by skipping a few steps and also dropping the conditional $P\{...|...\}$ notation. Here's what I did: (1) Divide both sides by $S_2(T)$ (doesn't change inequality, GBM is strictly positive), (2) log both sides; $\ln{1} = 0$ so the RHS just becomes 0. Inequality still holds because $ln$ is monotone increasing. – Jase Dec 1 '12 at 11:22
Actually the thing which was important for me is: "and also dropping the conditional P". The whole rest is clear to me :) thx again! – math Dec 1 '12 at 11:24

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