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Combining Correlated Geometric Brownian Motions for Margrabe Pricing

Article Quant Q&A · Author: SRKX

Summary

The document considers pricing a call-like payoff on the difference between the product of two correlated geometric Brownian motions and a scaled version of one process. It explains how to express the product process as a single geometric Brownian motion, which is needed to apply Margrabe’s exchange option formula.

Using Itô’s product rule, the product’s instantaneous variance is the sum of both variance contributions and their correlation term. The answer constructs a Brownian motion for the combined diffusion and identifies an additional drift from the cross variation. It then computes the covariance between this diffusion and the scaled underlying process, a quantity needed for the formula. The derivation offers a framework for the stochastic-process step, but does not complete the option price or discuss the formula’s market assumptions, such as whether the assets are tradable and the relevant discounting setup.

Key ideas

  • Applying Itô’s product rule to two correlated GBMs produces a cross-variation drift term.
  • The product process can be represented using a combined Brownian motion with volatility determined by both volatilities and their correlation.
  • The product process’s drift includes the correlation contribution as well as the usual volatility adjustment in its exponential solution.
  • Margrabe pricing requires the covariance of the two diffusion terms, which can be derived from the Brownian correlation structure.
  • The document gives process calculations but no completed option valuation.

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Full text
# What is the correlation between these two functions of GBMs?


# What is the correlation between these two functions of GBMs?












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?

## Answer by Richi Wa (score 2, accepted)

https://quant.stackexchange.com/a/21328

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: $$ \begin{eqnarray} Cov(\bar{\sigma} Z_t, \lambda \sigma^A W^A_t) = Cov(\sigma^A W_t^A + \sigma^R W_t^R, \lambda \sigma^A W^A_t) &= \\ Cov(\sigma^A W_t^A,\lambda \sigma^A W^A_t) + Cov(\sigma^R W_t^R,\lambda \sigma^A W^A_t) &= \\ \sigma^A \lambda Cov(W_t^A,W_t^A) + \sigma^R \lambda \sigma^A Cov(W_t^R,W_t^A) &= \\ \sigma^A \lambda t + \sigma^R \lambda \sigma^A \rho t.& \end{eqnarray} $$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.