# Is a stopped Ito-integral integrable if the Ito integrand is only square-integrable on an open interval?

Assume a filtered probability space $$(\Omega,\{\mathcal F_t\}_{t\in[0;T)}, \mathbb P)$$ with an $$\mathbb R^n$$-valued Brownian motion $$\{W_t\}_{t\in[0;T)}$$ and the filtration $$\{\mathcal F_t\}_{t\in[0;T)}$$ being the filtration generated by the Brownian motion, augmented by the nullsets.

Assume an $$\mathbb R^n$$-valued, progressively measurable stochastic process $$\{Z_t\}_{t\in[0;T)}$$ with the property $$\mathbb E\bigg( \int_0^t \Vert Z_t \Vert^2 \mathrm dt \bigg) < \infty$$ for all $$t \in [0;T)$$, but possibly $$\mathbb E\bigg( \int_0^T \Vert Z_t \Vert^2 \mathrm dt \bigg) = \infty$$

Assume a stopping time $$\tau \colon \Omega \to [0;T)$$.

I want to prove (but I don't know whether it's true...) $$\mathbb E\bigg( \int_0^\tau Z_t \mathrm dW_t \bigg) = 0.$$ I think this can be reduced to a proof of the statement $$\mathbb E\bigg( \Big| \int_0^\tau Z_t \mathrm dW_t \Big| \bigg) < \infty$$ because then I could use the dominated convergence theorem with the sequence $$1_{\{\tau \le T - 1/n\}} \cdot \int_0^\tau Z_t \mathrm dW_t$$.

But how to obtain this integrability property? Or is it even wrong? Thanks!

## 1 Answer

A counterexample should be just the deterministic $$Z_t = \frac{1}{\sqrt{T-t}}$$ with $$\tau := \inf{\biggl\{s>0 : \int_0^s Z_u \, dW_u =1\biggr\}}$$. You have $$\tau < T$$ a.s.and thus $$\mathbb{E}\biggl[ \int_0^\tau Z_u \, dW_u\biggr] =1$$.

• Thanks. Intuitively, I see why $\mathbb P[\tau < T] = 1$. But is there a way to briefly argue for this mathematically? – Kolodez Feb 15 at 7:59
• Consider a time change $u = Log(T-t)$. Then this is equivalent to that Brownian motion hits 1 in finite time. – Stephan Sturm Feb 15 at 8:14
• You might also try to look up “suicide strategy” in the financial math literature. It is intimately tied to this problem as well as the classic (D. Bernoulli in the 18th century) St. Petersburg paradox. – Stephan Sturm Feb 15 at 8:18