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

BGM Definition

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 $$ <dW^i(t), dW^j(t) = \rho_{ij} dt $$. We would get an inconsistency such that:

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

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.


  • $\begingroup$ In the first equation you write there is an indexing error. The left hand side has no index $i$, whereas the right hand side does, and hence this makes no sense. This further means there is no need for an $i$ index in the final bracket process. Ultimately I think the reasoning is false/erroneous because the indices you've been using are erroneous. Being more rigorous with the indices and the bracket process acting on scalars/vectors should help you tidy up your problem. $\endgroup$
    – oliversm
    Sep 5 '19 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.