# Are Stochastic Differential Equation diffusion terms always invariant under a change of measure?

I'm struggling with learning change of numeraire, and stochastic differential equations. I'm reading the beginning of Brigo and Mercurio's Interest Rate Models- Theory and Practice, and I'm on the Change of Numeraire Toolkit section. They present a formula for how the drift term of the SDE for a process $$X_t$$ changes when we change from one numeraire to another. The setup is as follows.

Let $$N$$ be a numeraire with corresponding martingale measure $$\mathbb{P}^{N}$$. Let $$W^{N}_t=(W_{1}^{N}(t),\dots, W_{n}^{N}(t))$$ be an $$n$$-dimensional brownian motion under $$\mathbb{P}^{N}$$. Let's suppose that we have a process $$X_t$$ governed by the SDE under $$\mathbb{P}^{N}$$: $$dX_t = \mu^{N}(t, X_t)\,dt + \sigma(t, X_t)CdW^{N}_{t},$$ where $$C\in \mathbb{R}^{n}$$, and $$\mu^{N}$$ and $$\sigma^{N}$$ are drift and diffusion functions (the superscript on the drift but not the diffusion will be important).

My goal is to understand how to take this formula and adapt it to a new numeraire $$S$$. In the text, they start with a similar equation for the process under the new martingale measure $$\mathbb{P}^{S}$$: $$dX_t = \mu^{S}(t, X_t)\,dt + \sigma(t, X_t)CdW^{S}_{t}.$$ Notice how the drift term has changed, but the diffusion term is the exact same as before.

My question is simple: why is this the case? Is this always the case under change of numeraire? Is it always the case under equivalent measures?

There are multiple threads which have asked similar questions, but I'm still struggling with a proper understanding. Here are some threads:

Are all changes of measures for continuous diffusion processes given by the change of drift?

Version of Girsanov theorem with changing volatility

Now for my solution. The diffusion is related to quadratic variation. According to Shreve, Vol II., expressions of the form $$(\text{Something})\,dW_t$$ are merely shorthand for expressing quadratic variations. So, the quadratic variation of $$X_t$$ over the interval $$[0,T]$$ is $$\int_{0}^{T} dX_t\,dX_t = \int_0^T \sigma(t, X_t)^2 \|C\|^{2}\,dt.$$ The right hand doesn't have any martingale measure in it. The quadratic variation of a sample path is independent of which measure is placed on the probability space. All the measure does is give probabilities of the sample paths. Since quadratic variation is independent of the measure, and because quadratic variation and diffusion are intertwined, the diffusion term in a SDE is independent of which measure we use.

This a heuristic argument, and it would seem to suggest that the diffusion term in any SDE is invariant under all measures. But this can't be right, can it?

There's also the line of think which goes that Girsanov's Theorem either assumes that the martingale measures are equivalent, or otherwise specifies when they are equivalent. And under equivalence the Theorem gives the formula for the Radon-Nikodym derivative. If the process had different diffusions under the martingales $$\mathbb{P}^N$$ and $$\mathbb{P}^{S}$$, the set $$\{\omega\in \Omega: [X_t, X_t](\omega)^{S}\neq [X_{t}, X_t](\omega)^{N}\}$$ would be a measurable set on which one of the two martingale measures would be singular with respect to the other (or both would be singular with respect to each other on subsets). That is, there are some sample paths in this set that one martingale measure gives positive probability to while the other gives zero probability to. Hence, the martingale measures would not be equivalent, and Girsanov's theorem doesn't hold. However, change of numeraires always have a Radon-Nikodym derivative given by the ratio of Numeraires (normalized by their initial values). Thus, a contradiction, and the diffusion term under both martingale measures must be the same.

Is my thinking correct? Are there more formal arguments to justify the invariance of the diffusion term under a change of measure?

• Maybe this question helps: quant.stackexchange.com/questions/14787/…. The key is that, for continuous processes (i.e. pure diffusions), there is a one-to-one mapping between the fact of (1) defining an equivalent probability measure (2) specifying the Radon-Nikodym derivative of the target vs. current measure as a stochastic exponential. When you do that, Girsanov tells you that a Brownian under the target measure, is a Brownian under the current measure minus a quadractic variation term (...) Dec 7, 2021 at 10:48
• (...) Because the latter term has finite variation it does not alter the instantaneous volatility of the Brownian process (be it stochastic or not): it is the same under the current and target measures. Note that the terminal volatility (or variance) will however be affected in general. Dec 7, 2021 at 10:49
• What does $C$ represent? Why is the constant $C$ "outside" of the function $\sigma(.,.)$? Dec 7, 2021 at 16:06