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).

Answer 3

Edit: the below answer contains some mistakes and needs some further work, as per the comments underneath.

Intro: I think intuition is super important for this one, so my answer below focuses on the intuition here.

Short answer

The volatility parameter is meant to describe the behaviour of $S(t)$, whilst the drift only describes the probabilities of the instrument $S(t)$ moving up or down: changing the drift (and therefore the probabilities of "up" and "down") allows us to come up with equivalent probability-measures, which in turn allows us to price derivatives (for example).

If we, however, changed the volatility parameter, the equation for $S(t)$ would no longer describe the instrument $S(t)$.

Long answer

I will give a "hand-waving" answer, at an intuitive level.

It is well known that a multiplicative binomial tree model converges to the continuous GBM model that you have included in your question (see for example here). Therefore, we can take the binomial tree as an intuitive example without loss of generality.

Suppose we draw a binomial tree for the process $S(t)$ under the measure $\mathbb{P}$: rates are zero and the probabilities of an up-move and down-move are both at 50%. The tree is as follows:

This tree produces a "drift" of 2.5% per period. Also, at an intuitive level, the size of the "up" and "down" move describe the volatility of this instrument $S$: the expected value under $\mathbb{P}$ after the first period is 102.5, and the variance in that first period is $0.5*(110-102.5)^2 +0.5* (95-102.5)^2=56.25$ (variance ~ volatility).

(Suppose another instrument with an initial price of 100 had a different tree with prices 107 and 98 after the first period: it would also have a drift of 2.5%, but the variance would be - of course - smaller.)

Now the point is that under this measure $\mathbb{P}$ (for argument's sake suppose it is the real-world measure), you cannot price derivatives on the instrument $S$, because this real world measure $\mathbb{P}$ is not arbitrage free. So you want to come up with an equivalent measure $\mathbb{Q}$ that will allow you to price derivatives on $S$ without arbitrage: but you want to preserve the original tree which depicts the price map of $S$.

In other words, you don't want to change the behaviour of the instrument $S$, you just want to play around with the probabilities of the "up" and "down" moves; the idea is that the tree of prices describes the "real" behaviour of the instrument $S$, so changing it would defeat the purpose of pricing derivatives on $S$ (and as alluded to above, changing the tree is equivalent to changing the variance, i.e. the "volatility", i.e. $\sigma$).

Denoting the probability of an "up-move" as $p$, the measure $\mathbb{Q}$ which is arbitrage free solves the following equation (remember that rates are set to zero in our toy example):

$$110*p^{\mathbb{Q}}+95(1-p^{\mathbb{Q}})=100$$

Which gives $p^{\mathbb{Q}}=\frac{1}{3}$.

Under $\mathbb{Q}$, the drift is now zero ($\frac{1}{3}*110 + \frac{2}{3}*95=100$): so we have changed the drift, but we have not changed the volatility, which in effect allowed us to preserve the original tree depicting the behaviour of $S$.

PS: rates

Under the Libor market model (see here), the forward rates must be martingales under the T-forward measure: so the process for rates under the T-forward measure must always be driftless (otherwise the process would not describe a martingale).