Is the stochastic integral invariant under equivalent change of probability?

Question

Let $(\Omega,\mathcal F, \mathbb F,\mathbb P)$ be a filtered probability space under the usual conditions and suppose $\mathbb Q\sim\mathbb P$ is an equivalent probability measure. Let $X$ be a $\mathbb P$-semimartingale and $H$ a predictable process $\mathbb P$-integrable with respect to $X$.

By Girsanov-Meyer theorem we know that $X$ is also a $\mathbb Q$ semimartingale. So we can ask:

1) Is $H$ $Q$-integrable with respect to $X$?

2) If so, is the stochastic integral $\int H dX$ the same under any of the two probabilities?

Answer 1

The measure-invariance of stochastic integrals with (locally) bounded integrands is shown in Meyer (1976, VI.26) or Dellacherie & Meyer (1978, VIII.12). Analogous statement is available for general integrands, e.g. Protter (2004, Theorem II.14).

Meyer, P. A., Un cours sur les integrales stochastiques, Semin. Probab. X, Univ. Strasbourg 1974/75, Lect. Notes Math. 511, 245-400 (1976). ZBL0374.60070.

Dellacherie, Claude; Meyer, Paul-Andre, Probabilities and potential. Transl. from the French, North-Holland Mathematics Studies, 29. Amsterdam - New York -Oxford: North-Holland Publishing Company. VIII, 189 p. (1978). ZBL0494.60001.

Protter, Philip E., Stochastic integration and differential equations, Applications of Mathematics 21. Berlin: Springer (ISBN 3-540-00313-4/hbk; 978-3-642-05560-7/pbk). xiii, 415 p. (2004). ZBL1041.60005.

Answer 2

1) A predictable process $H$ is integrable with respect to a semimartingale if it is locally bounded. So we need to show that if $H$ is locally bounded under $\mathbb{P}$, it is so under $\mathbb{Q}$.

Let $\tau_n$ be a localizing sequence of stopping times for the local boundedness under $\mathbb{P}$. For each $n$ there is a constant $K_n>0$ such that $$\mathbb{P}\left[\sup_{t\geq 0}\mathbf{1}_{\tau_n\geq 0}|H^{\tau_n}_t|<K_n\right]=1.$$ Therefore, $$\mathbb{P}\left[\sup_{t\geq 0}\mathbf{1}_{\tau_n\geq 0}|H^{\tau_n}_t|\geq K_n\right]=0.$$ Since $\mathbb{Q\ll P}$, we also have $$\mathbb{Q}\left[\sup_{t\geq 0}\mathbf{1}_{\tau_n\geq 0}|H^{\tau_n}_t|\geq K_n\right]=0 \quad\Leftrightarrow\quad \mathbb{Q}\left[\sup_{t\geq 0}\mathbf{1}_{\tau_n\geq 0}|H^{\tau_n}_t|< K_n\right]=1,$$ which shows that $H$ is $\mathbb{Q}$-locally bounded.

2) The answer to this is no in general. For a simple counterexample we choose $H_t\equiv 1$ and $X_t=W_t$ a standard Wiener process under $\mathbb{P}$ with $W_0=0$. So, $$\int_0^tH_sdX_s=W_t.$$ Let $\mathbb{Q}=Z_t\cdot\mathbb{P}$ on $\mathcal{F}_t$, where $Z_t$ is given by $$Z_t=\exp\left[\int_0^th_s dW_s-\frac{1}{2}\int_0^th_s^2ds\right].$$ By Girsanov's theorem, $$\tilde W_t\doteq W_t-\int_0^th_sds$$ is a $\mathbb{Q}$-Wiener process, which means that in general $W$ is not a $\mathbb{Q}$-Wiener process. So the stochastic integral $\int HdX$ in this case has different distributions under $\mathbb{P}$ and $\mathbb{Q}$ whenever $h$ is non-zero.