Brownian Correlation and Measure Changes in the BGM Forward Rate Model

Summary

This note raises a consistency question about the BGM, or Libor Market, forward-rate model. It compares a forward rate written under its own pricing measure with a representation under another measure, where the drift changes but the diffusion is expressed using a different Brownian motion. The author takes quadratic covariations and concludes that the stated equations would require every pairwise Brownian correlation to equal one.

The document does not include an answer or a derivation resolving the apparent contradiction. Its learning value is therefore as a prompt to distinguish Brownian motions under different measures and to track which covariance relation applies to each representation. It also highlights that finite-variation drift terms do not contribute to quadratic covariation, while the correlation of the relevant Brownian increments does. The question alone does not establish that the model imposes unit correlations; additional clarification of notation and measure-specific Brownian motions is needed.

Key ideas

• The note compares forward-rate dynamics under different pricing measures.
• It questions whether a change in drift can create an inconsistency in Brownian correlations.
• Quadratic covariation depends on diffusion terms rather than finite-variation drift terms.
• The document offers no resolution, so its correlation conclusion should be treated as an open question.

Tags

# Change of measure for BGM (LMM) Model


# Change of measure for BGM (LMM) Model












I've been checking the demos for BGM (LFM) forward rate model. Here's a short reminder to help you follow:

Now, take the following

$$\frac{dL_j(t)}{L_j(t)} = \sigma_j. dW^j(t) = \mu_{ij} dt + \sigma_j. dW^i(t) $$

if we consider the same brownians as in the definition of the BGM model, where we particularly have that $$\langle dW^i(t), dW^j(t) = \rho_{ij} dt $$. We would get an inconsistency such that:

$$ \langle\frac{dL_j(t)}{L_j(t)}\rangle = \langle\sigma_j. dW^j(t)\rangle = \sigma_j^2 dt $$ Whereas, on the other hand, $$\langle\frac{dL_j(t)}{L_j(t)}\rangle = \langle\sigma_j. dW^j(t), \mu_{ij} dt + \sigma_j. dW^i(t)\rangle = \sigma_j^2. \rho_{ij} dt $$

This means that $ \rho_{ij} = 1 $ for all $i$'s and $j$'s!

My question: Is my reasoning false and why plz?

Perhaps the brownians that we define by a change of measure (from measure $Q^i$ to $Q^j$) are not the ones considered in the definition of BGM.

Thanks

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