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Let's say that I have two correlated GBMs:

$$dA_t = A_t \sigma^A dW^A_t$$ $$dR_t = R_t \sigma^R dW^R_t$$ $$dW^R_t dW^A_t = \rho dt$$

I am trying to price a derivative which payoff at time $T$ is:

$$\text{Payoff}_T = (A_TR_T - A_T \lambda)^+ $$

My idea was to apply Margrabe's formula, but for this I need to formulate the two processes $X_t = A_t R_t$ and $Y_t = \lambda A_t$ as GBMs as well in order to find their respective volatilities and their correlation.

The first one is quite trivial:

$$dY_t = d(\lambda A_t) = \lambda dA_t + \frac{1}{2} 0 = \lambda A_t \sigma^A dW^A_t = dY_t \sigma^A dW^A_t$$

which is clearly a GBM and $\sigma_Y = \sigma^A$.

But I'm a struggling to express the second one, what I came up with so far is:

$$ \begin{align} d(A_t R_t) & = & A_t dR_t + R_t dA_t + dA_tdR_t \\ & = & A_t R_t \sigma^R dW^R_t + A_t R_t \sigma^A dW^A_t + A_t R_t \sigma^R \sigma^A \underbrace{dW^A_t dW^R_t}_{\rho dt} \\ & = & A_t R_t \left[ \sigma^R dW^R_t + \sigma^A dW^A_t + \sigma^R \sigma^A \rho dt \right] \end{align}$$

But this is where I'm stuck, I can't figure out how to express this as a "simple" GBM as it's quite clearly multivariate... Am I missing something?

Is there a way I can still use the Margrabe formula to price my option?

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    $\begingroup$ As $(A_TR_T - A_T \lambda)^+ = A_T(R_T - \lambda)^+$, you can use $A_T$ as the numeraire, and then price using Black's formula. $\endgroup$ – Gordon Oct 20 '15 at 13:39
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    $\begingroup$ I am not experience with approaches using numeraires. Does it mean that one prices $(R_T-\lambda)^+$ and how precisely can I use the numeraire? This really sound interesting! $\endgroup$ – Ric Oct 21 '15 at 14:50
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    $\begingroup$ @Richard agreed I'll ask another question about this. $\endgroup$ – SRKX Oct 22 '15 at 1:02
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Coming back to the line where you are stuck. If we define $$ Z_t = \sigma^A/\bar{\sigma} W_t^A + \sigma^R/\bar{\sigma} W_t^R, $$ with $\bar{\sigma}^2 = (\sigma^A)^2 + 2 \sigma^A \sigma^R \rho + (\sigma^R)^2$, then $Z_t$ is a Brownian motion in its own filtration and the first and second moment are correct.

Then we write your last line using $X_t = A_t R_t$ as $$ dX_t = X_t (\bar{\sigma} dZ_t + \sigma^A\sigma^R \rho dt), $$ with the solution $$ X_t = X_0 \exp((\sigma^A\sigma^R \rho - \bar{\sigma}^2/2) t + \bar{\sigma} Z_t), $$ which is a GBM with (!) drift.

In Magrabe's formula you need the covariance (vol times vol times correlation) of the two diffusion terms: $$ \begin{eqnarray} Cov(\bar{\sigma} Z_t, \lambda \sigma^A W^A_t) = Cov(\sigma^A W_t^A + \sigma^R W_t^R, \lambda \sigma^A W^A_t) &= \\ Cov(\sigma^A W_t^A,\lambda \sigma^A W^A_t) + Cov(\sigma^R W_t^R,\lambda \sigma^A W^A_t) &= \\ \sigma^A \lambda Cov(W_t^A,W_t^A) + \sigma^R \lambda \sigma^A Cov(W_t^R,W_t^A) &= \\ \sigma^A \lambda t + \sigma^R \lambda \sigma^A \rho t.& \end{eqnarray} $$

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  • $\begingroup$ And then $dZ_t dW^A_t = ( \frac{\sigma^A }{\bar{\sigma}} + \frac{ \sigma^R }{\bar{\sigma}} \rho ) dt$? So the correlation to be used withing Margrabe's formula is $\frac{\sigma^A }{\bar{\sigma}} + \frac{ \sigma^R }{\bar{\sigma}} \rho$? $\endgroup$ – SRKX Oct 20 '15 at 16:25
  • $\begingroup$ @SRKX: That is correct. $\endgroup$ – Gordon Oct 20 '15 at 20:47
  • $\begingroup$ See my thoughts above about this. $\endgroup$ – Ric Oct 21 '15 at 8:28

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