Why does the diffusion term remain the same when we change pricing measure?

Consider some Itô process $$dS(t)=\mu(t)dt+\sigma(t)dW^{\mathbb P}_{t}$$ under the measure $$\mathbb P$$, where $$W^{\mathbb P}$$ is a $$\mathbb P$$-Brownian motion

In plenty of interest rate examples, I have seen that we attempt attempt to find a measure under which $$S(t)$$ is driftless, i.e. $$dS(t)= \sigma(t) dW_{t}^{\mathbb Q}$$, where $$W^{\mathbb Q}$$ is a $$\mathbb Q$$ Brownian motion.

My question is simple: Why does the $$\sigma(t)$$ term remain the same under the measure change? does it have something to do the fact the measures $$\mathbb Q$$ and $$\mathbb P$$ are equivalent?

• I suspect it has to do with equivalent probability measures. See also the accepted answer here: mathoverflow.net/questions/51090/…
– user34971
Commented Jan 24, 2022 at 15:30
• The simple mathematical answer is that if you change sigma then the law of the new process becomes singular with respect to the old one , i.e. the two processes are no longer equivalent. For a simple example you can consider drift zero and two different values of sigma and then a simple application of the law of the iterated logarithm will easily give you singularity of the two measures. A similar argument can be used in the general case of variable coefficients as well. Commented Jan 24, 2022 at 21:43
• Hey @noob2, is your last sentence true ? If you change the probabilities on a tree without changing the possible outcomes for the stock, I think you do change the variance don’t you?
– dm63
Commented Jan 25, 2022 at 13:35
• @dm63: agree. something not quite right with what I said, I need to rethink this stmt. Sorry. Commented Jan 25, 2022 at 14:19
• No prob. It is a subtle thing for sure, can’t say I fully understand it.
– dm63
Commented Jan 25, 2022 at 21:55

Extract from my answer about what the VIX measures (more details on the notation and the conventions I am using can be found in the preceding sections from that answer):

(This section is based on currently non-public lecture notes by Dylan Possamaï for a course on mathematical finance. I will update it with precise references if the lecture notes are published. For now, I will insert [RefN] where a precise reference is needed.)

TODO: Update this part for non-zero interest rate.

For simplicity I will assume that the interest rate is zero (it is actually not hard to incorporate a constant interest rate $$r$$.) We deal with only one security, which is an Itô process of the form $$S_t=S_0+\int_0^tb_s\,\mathrm ds+\int_0^t\mathfrak S_s\,\mathrm dW_s.$$

If there is no arbitrage [RefN], then there always exists an $$\mathbb F$$-predictable stochastic process $$\lambda$$ such that

$$\mathfrak S_s(\omega)\lambda_s(\omega)=b_s(\omega)$$

for $$\mathrm dt\otimes\mathsf P$$-almost all $$(s,\omega)\in[0,T]\times\Omega$$.

We will assume that

$$Z_t\overset{\text{Def.}}=\exp\left(-\int_0^t\lambda_s\,\mathrm dW_s-\frac12\int_0^t\lambda_s^2\,\mathrm ds\right), \quad t\in[0,T]$$

is well-defined and an $$(\mathbb F,\mathsf P)$$-martingale. In fact, if $$Z_t$$ is well-defined and an $$(\mathbb F,\mathsf P)$$-martingale, then it follows that there is no arbitrage in the market (up to time $$T$$).

In this case, we can prove that the measure $$\mathsf Q$$ given by $$\frac{\mathrm d\mathsf Q}{\mathrm d\mathsf P}=Z_T$$ is an equivalent (local) martingale measure for the financial market up to time horizon $$T$$:

By Girsanov's Theorem [RefN], the stochastic process $$(W^{\mathsf Q}_t)_{t\in[0,T]}$$ given by

$$W_t^{\mathsf Q}=W_t+\int_0^t\lambda_s\,\mathrm ds$$

is an $$(\mathbb F,\mathsf Q)$$-Brownian motion (up to to time $$T$$).

Therefore, for any progressively measurable stochastic process $$\mathfrak S=(\mathfrak S)_{s\in[0,T]}$$ with $$\mathsf E^{\mathsf P}\left(\int_0^T\mathfrak S_s^2\,\mathrm ds\right)<\infty$$ and $$t\in[0,T]$$, we have

$$\int_0^t \mathfrak S_s\,\mathrm dW_s=\int_0^t \mathfrak S_s\,\mathrm d\left(W_s^{\mathsf Q}-\int_0^s\lambda_\tau\,\mathrm d\tau\right)=\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}-\int_0^t\mathfrak S_s\lambda_s\,\mathrm ds,$$

where the associativity of the stochastic integral [RefN] was used in the last equality.

Therefore, if the price process $$(S_t)_{t\in [0,T]}$$ is an Itô process

$$S_t = S_0+\int_0^tb_s\,\mathrm ds+\int_0^t\mathfrak S_s\,\mathrm dW_s,$$

then

$$S_t=S_0+\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}-\int_0^t\mathfrak S_s\lambda_s\,\mathrm ds+\int_0^tb_s\,\mathrm ds=S_0+\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}.$$

Furthermore, by the regularity assumed on $$\mathfrak S_s$$, the stochastic integral $$\int_0^t\mathfrak S_s\,\mathrm dW_s^{\mathsf Q}$$ is an $$(\mathbb F,\mathsf Q)$$-martingale [RefN] [see also Footnote 6 in my VIX answer]. This shows two things:

1. $$\mathsf Q$$ is an equivalent martingale measure for the market consisting only of the security with price process $$S$$.
2. The volatility term of $$S_t$$ remains unchanged when we go from $$\mathsf P$$ to $$\mathsf Q$$. However, the drift disappears so that $$S$$ is now a martingale.
• Just what I was looking for, thanks! Commented Jan 25, 2022 at 16:36

It has to do with the Girsanov theorem that relates the equivalent measures $$\mathbb Q$$ and $$\mathbb P\,.$$ To make intuitively clear what happens I like to give the following "baby Girsanov" example: Let $$X$$ be standard normal having probability density $$p(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ under $$\mathbb P\,.$$ If the equivalent measure $$\mathbb Q$$ is related to $$\mathbb P$$ by the Radon-Nikodym density $$q(x)=e^{\mu x-\mu^2/2}$$ then it is straightforward to see that $$X$$ has under $$\mathbb Q$$ the density $$\frac{1}{\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2}}\,.$$ Clearly, under $$\mathbb Q$$, $$X$$ has the same variance (diffusion parameter) but a different mean (drift).