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In elementary discussions on change of measure for geometric Brownian motion, one often find statements like "change of measure = change of drift". Given a general continuous diffusion process of the form

$$ dX_t = \mu(X_t,t,\theta)dt + \sigma(X_t,t,\theta)dW_t $$

is it still true that the change of measure is always only given by the change of drift $\mu(X_t,t,\theta)$? Furthermore, is the Radon–Nikodym derivative for this change of measure always unique and always given by the Doléans-Dade exponential?

If I think of the change of measure as a change of variable, then there are really no restrictions on how one may transform, say, a normal variable $X$: one can just shift it (i.e. $Y = a + X$, which would correspond to the change of drift) or one can scale it and shift it (i.e. $Y = a + bX$, which would also change the standard deviation or "volatility"), or apply any other transformation and still have a valid probability density $f_Y$ for the transformed variable $Y$ with the corresponding "change of measure" given by $\frac{f_Y}{f_X}$. So, what is stopping us doing the same in the case of the diffusion process? Why do we only seem to talk about change of drift?

Confounded
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  • We change measures when we change the numéraire, yet we require numéraire-rebased assets to be martingales under the numéraire's measure. A martingale process has no drift thus the measure changes performed in derivatives pricing are designed so as to suppress drifts and are not related to the diffusion coefficient. – Daneel Olivaw Apr 11 '19 at 13:37
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    @DaneelOlivaw Thank you for your comment. Yes, that the aim of the transformation is to obtain a drift-less process for a "numeraire-discounted" asset is clear, but it doesn't mean that the diffusion coefficient has to remain the same, does it? I mean, we can still have a martingale under a different measure but with a different diffusion, no? – Confounded Apr 11 '19 at 15:07
  • I'm having some of the same questions with learning this material. Have you found and answer, and would you want to share it? – Alex Lapanowski Dec 07 '21 at 02:24
  • @AlexLapanowski I haven't got much further with this, unfortunately, but the answer below from dm63 seems to offer an explanation. – Confounded Dec 07 '21 at 14:02

3 Answers3

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I have read that for diffusion processes, indeed the volatility must be preserved under a change of measure. This old question appears to be relevant :

Version of Girsanov theorem with changing volatility

In particular, I quote from the above answer : a probability measure assigns relative likelihood to different trajectories of the Brownian motion. Variance of the Ito process can be recovered from the shape of a single trajectory (quadratic variation), so it does not depend on the relative likelihood of the trajectories, hence, does not depend on the choice of the probability measure.

In other words , changing measure is a process of assigning different probabilities than before to the same set of possible outcomes. When you change the diffusion coefficient , you change the set of possible outcomes. Hence not allowed.

dm63
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Change of measure and change of variable are two separate things. In measure change, you keep the same variable and redistribute the probability. Keeping the variable the same is the key to the concept. This induces a change in drift. Which is a massive help because once you can manipulate the drift then everything becomes easy. For example, one can then bring in the well developed martingale theory to analyse the processes.

There is another transformation, called Lamperti transformation not commonly referred to by this name, that can be used to change the diffusion coefficient, though I have seen it used in 1 dimension only.

Magic is in the chain
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  • But we do change $W_t$ to some other $W_{t}^{} = \mu^ t + W_t$, which I see as a change of variable from a zero-mean normal to a non-zero-mean normal. Can we not also, use, say, $W_{t}^* = \mu^* t +\sigma^* W_t$ to obtain another continuous measure? – Confounded Apr 23 '19 at 11:36
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    In case of simple random variables it is easy to see how probability changes when you change the variables but things get quite complicated when one moves to a process view. The example you provided , $\sigma W_t$ is discussed in Klebaner’s introduction to stochastic calculus, where it is shown that the measures induced by this change is singular. – Magic is in the chain Apr 23 '19 at 21:05
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Although this is a quite old thread, but I actually had the same confusion as you pointed out in the question. I just found a relevant question that gives a reasonable explanation to the question:

https://math.stackexchange.com/questions/79875/laws-of-b-t-t-in-0-t-and-2b-t-t-in-0-t-singular

In short, in the change of measure, the considered measures are required to have the same diffusion coefficient, because otherwise we can always find a collection of realisations of processes that has probabilities greater than zero in one measure, while having zero probability in the other measure. In such cases, the Radon-Nikodym derivative is not well-defined and therefore the change of measure does not apply.

Milk_Tutu
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