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I am having a slight brain meltdown because I do not seem to be able to understand the following basic thing.

Consider a BS economy, and two assets $X$ and $Y$ $$ dX = \sigma X dW $$ $$ dY = \nu Y dZ $$ $$ dWdZ = \rho dt $$

I would like to price a Margrabe option $(X_T - Y_T)_+$.

The first and most straightforward method is a change of numeraire approach. In other words $$ E_t(X_T - Y_T)_+ = Y_t E^{Q_Y}( X_T/Y_T -1 )_+ $$ where $Q_Y$ is the measure with $Y$ as numeraire. Now if you evaluate the above expression under this measure you get a relatively simple option price expression, and where the correlation $\rho$ will appear in the formula. Agree?

The second approach is to use conditioning. Does everyone agree that I can also price the options as follows: $$ E_t(X_T - Y_T)_+ = \int_0^\infty C(X_t, y) q(y) dy $$ where $C(X_t, y)$ is the single asset BS option price with strike $y$, and $q(y)$ is the lognormal distribution of $Y$.

I can always calculate using the numerical integration above right? If so, here is where I am confused: how does the correlation parameter $\rho$ appear in the numerical integration? I cannot see it, but it must somehow play a role.

Help!

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2 Answers 2

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Might you be using the tower law in a wrong way? I have the impression you derive your second equation by conditioning by the $\sigma$-algebra generated by $(Y_t)_{t\geq0}$, however note that: $$\mathscr{F}_t\nsubseteq\sigma(Y_t)_{0\leq t\leq T}$$ Hence: $$E\left((X_T-Y_T)_+|\mathscr{F}_t\right)\ \not= \ E\left(E(X_T-Y_T)_+|Y_T)|\mathscr{F}_t\right) \ = \ E(C(X_t,Y_T)|\mathscr{F}_t)$$

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    $\begingroup$ Yes I think you are right. What I have to do is split the Brownian motions $W$ and $Z$ into orthogonal components, and then I can use the tower law. So if I write $dZ = \rho dW + \sqrt{1-\rho^2} dW^\bot$, then I can condition on $W^\bot$, agree? $\endgroup$
    – user34971
    Commented Jan 23, 2020 at 10:27
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Agreeing to your first observation: After orthogonalization, with independent W and W’, and using self explanatory notation for the new diffusion coefficients, which obviously depend on $\rho$, we can show that, under $\mathbb{Q}_Y$, we have:

$$ dR = R[(\sigma_{XW} - \sigma_{YW})dW + (\sigma_{XW’} - \sigma_{YW’})dW’], $$

where $R=XY^{-1}$ (used only Ito calculations and the fact that R is a martingale under $\mathbb{Q}_Y$).

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  • $\begingroup$ Thank you, yes I understand now. Don't know where I was with my thoughts that I didn't see it (that I had to orthogonalize I mean). $\endgroup$
    – user34971
    Commented Jan 23, 2020 at 13:50

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