Quadratic variation and measure change

Let $W_t$ be a Brownian motion defined on probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and assume $X_t$ is a process given by SDE $$ dX_t=W_tdW_t, W_0=0 $$ i.e. $X_t=\int_0^tW_sdW_s$. With respect to the probability measure $\mathbb{P}$ we can calculate the quadratic variation $$ [X,X]_t=\int_0^t(W_s)^2ds, \qquad (1) $$ Let us change from the measure $\mathbb{P}$ to an equivalent $\mathbb{P}'$ (using Girsanov theorem) where $W_t'=W_t-at$ is Brownian motion and $a\neq 0$ under new measure $\mathbb{P}'$.

In that case $$ X_t=\int_0^t(W_s'+as)d(W_s'+as)=\int_0^t(W_s'+as)dW_s'+\int_0^ta(W_s'+as)ds $$ therefore under $\mathbb{P}'$ quadratic variation becomes $$ [X,X]_t=\int_0^t(W_s'+as)^2ds, \qquad (2) $$

Although I am aware that quadratic variation should not change under equivalent measure changes, quadratic variations (1) and (2) look completely different unless $a=0$.

Am I missing something here?

Solution 1:

I think you wanted to replace $W_s$ in (1) by $W'_s+at$ where $W'$ is the $\mathbb P'$-Brownian motion. This directly leads to $$ [X,X]_t=\int_0^t(W_s)^2\,ds=\int_0^t(W'_s+as)^2\,ds\,. $$ The quadratic variation has in fact not changed. The two different ways of writing it only reflect the fact that $[X,X]$ has a different distribution under $\mathbb P$ than under $\mathbb P'\,.$ This is a property which quadratic variation shares with every other stochastic process.