# Is the stochastic integral invariant under equivalent change of probability?

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?

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.

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.

• You appear to be confusing indistinguishability of processes with their distributional properties. The same random variable will have different distributions under different equivalent measures. Does not change the fact that it is the same r.v. Jan 13 '20 at 10:35
• @encore Thanks for the comment. Upon re-reading the question and my answer over two years later, I think I indeed misread 'the same' as 'have the same distribution'. Your answer makes sense and fully clarifies the situation. Jan 15 '20 at 9:22
• I agree, it very much depends on one's mindset. I was working on this particular aspect of the theory when I bumped into the question. Jan 15 '20 at 19:15