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?

  • 1
    $\begingroup$ I suspect it has to do with equivalent probability measures. See also the accepted answer here: mathoverflow.net/questions/51090/… $\endgroup$
    – user34971
    Commented Jan 24, 2022 at 15:30
  • 1
    $\begingroup$ 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. $\endgroup$
    – shalop
    Commented Jan 24, 2022 at 21:43
  • $\begingroup$ 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? $\endgroup$
    – dm63
    Commented Jan 25, 2022 at 13:35
  • $\begingroup$ @dm63: agree. something not quite right with what I said, I need to rethink this stmt. Sorry. $\endgroup$
    – nbbo2
    Commented Jan 25, 2022 at 14:19
  • $\begingroup$ No prob. It is a subtle thing for sure, can’t say I fully understand it. $\endgroup$
    – dm63
    Commented Jan 25, 2022 at 21:55

2 Answers 2


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

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


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

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.
  • 1
    $\begingroup$ Just what I was looking for, thanks! $\endgroup$ 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.