Finding an Equivalent Martingale Measure for Multivariate GBM
Summary
The document asks how to extend the one-dimensional Girsanov change of measure for geometric Brownian motion to multiple assets driven by a multidimensional Brownian motion. It gives each asset a drift above the risk-free rate and a vector of Brownian shocks, then proposes dividing each asset’s excess drift across its volatility coefficients to construct a market price of risk. The author is unsure whether this yields a single measure under which the whole price vector is a martingale.
The post also considers finding a separate measure for each asset and combining them as a product measure. It offers no accepted solution or evidence that this construction works. The key mathematical issue is that a common change of measure must solve a system relating all assets’ excess drifts to their volatility vectors; existence and uniqueness depend on that system, including the relationship between asset and Brownian dimensions. The suggested componentwise division and product-measure idea are therefore questions, not established results.
Key ideas
- A multivariate asset model may have more assets than Brownian risk factors, or the reverse.
- A common change of measure must adjust all asset drifts through one market-price-of-risk vector.
- The drift and volatility parameters form a system of equations whose solvability matters.
- Constructing separate measures for individual assets does not by itself establish a shared martingale measure.
- The document poses the extension problem without resolving it.
Tags
Full text
# multivariate geometric brownian motion equivalent martingale measure
# multivariate geometric brownian motion equivalent martingale measure
Suppose $W$ is a $\mathbb{P}$-Brownian motion and the process $S$ follows a geometric $\mathbb{P}$-Brownian motion model with respect to $W$. $S$ is given by \begin{equation} dS(t) = S(t)\big((\mu - r)dt + \sigma dW(t)\big) \end{equation} where $\mu, r, \sigma \in \mathbb{R}$. Thanks to Girsanov theorem, we know that $S$ is a $\mathbb{Q}$-martingale for $\mathbb{Q}$ defined by \begin{equation} \frac{d\mathbb{Q}}{d\mathbb{P}} = \mathcal{E}\Big(\int \lambda dW\Big) \end{equation} where $\lambda = \frac{\mu - r}{\sigma}$.
This is the case where the Brownian motion and $S$ are one dimensional. What can be said about multivariate case ?
I tried something : When $W$ is $d$-dimensional and $S$ is $n$-dimensional, with $n, d \geq 1$, $\forall i \in \{1,...,n\}$, $S$ is given by \begin{equation} dS_i(t) = S_i(t)\big((\mu_i -r)dt + \sum_{i = 1}^d \sigma_{ij}dW_j(t)\big) \end{equation} If we define $\lambda$ as \begin{equation} \lambda_{ij} = \frac{\mu_i - r}{\sigma_{ij}d} \end{equation} Maybe somehow we could use Girsanov theorem in a similar way as before to find a probability measure such that $S$ is a martingale ? I cannot find a way to do this properly. The problem is that with this approach, for each $i \leq n$, we only find a measure under which $S_i$ is a martingale, but I am not sure how to deduce a result for $S$.
edit : Is it enough to take each $i \leq n$ individually, find a measure $\mathbb{Q}_i$ under which $S_i$ is a martingale by using the approach I mentioned before, and then say that $S$ is a martingale under the product measure $\otimes_{i \leq n} \mathbb{Q}_i$ ? According to this, a sufficient condition for a multivariate process to be a martingale is if each component separately is a martingale.
Thank you in advance for your help.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.