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

Question

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?

Answer 1

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):

About changing the measure

(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.

Answer 2

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).