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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?

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  • $\begingroup$ Thanks to @Kurt-G for the correction. $\endgroup$ Commented Jan 4, 2022 at 14:03

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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.

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    $\begingroup$ Thanks. Agree, quadratic variation's distribution changes with changing the measure. In that case what does the statement "quadratic variation has not changes" mean? For instance, P. Protter in "Stochastic Integration and Differential Equation", page 123, mentions: "quadratic variation remain invariant with a change to an equivalent probability measure" . What remains invariant in that case? @Kurt G: thanks for spotting the typo, I fixed an original post $\endgroup$ Commented Jan 4, 2022 at 15:10
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    $\begingroup$ You can view the stochastic process $Y=[X,X]$ as a map $\Omega\times\mathbb R\to\mathbb R\,.$ This map is it what does not depend on the measure $\mathbb P\,.$ I do not think that this is a very deep theorem. $\endgroup$
    – Kurt G.
    Commented Jan 4, 2022 at 17:19
  • $\begingroup$ I was thinking in the same vein: assume $X$ is a solution of SDE: $dX_t=a(X_t)dt+b(X_t)dW_t$. Given that $[X,X]_T$ is a limit (in probability) of the Riemann sums $S^{(n)}_T$ taken along subdivisions $P_n=\{0=t_0<\ldots<t_n=T\}$ as $\sup_n(t_{n+1}-t_n)\to 0$, it follows that the definition of $[X,X]$ is not affected by probability $\mathbb{P}$ as Riemann sums $S^{(n)}_T$ must have same limit under equivalent probability measure $\mathbb{P}'$: if $S^{(n)}_T\to [X,X]_T$ in $\mathbb{P}$ then $S^{(n)}_T\to [X,X]_T$ in $\mathbb{P}'$ when $\mathbb{P}\sim \mathbb{P}'$. $\endgroup$ Commented Jan 5, 2022 at 8:03

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