# What is the expectation of $\int_0^t \sqrt{s+B_s^2}dB_s$?

I am trying to find the expectation of $\int_0^t \sqrt{s+B_s^2}dB_s$, but am unable to use Ito's Formula because of the nasty integral. Is there another solution I am missing? Thanks!

• Hint : why should it be different form 0 ? I don't see any reason. Best regards – TheBridge Apr 29 '16 at 8:05

If $f: (0,\infty) \times \Omega \to \mathbb{R}$ is progressively measurable and $$\mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right)<\infty \quad \text{for all t \geq 0}$$ then

$$M_t := \int_0^t f(s) \, dB_s, \qquad t \geq 0,$$

is a martingale. This implies in particular

$$\mathbb{E}(M_t) = \mathbb{E}(M_0) = 0.$$

Since $f(s,\omega) := \sqrt{s+B_s^2(\omega)}$ is progressively measurable and

$$\mathbb{E} \left( \int_0^t f(s)^2 \, ds \right) = \mathbb{E} \left( \int_0^t (s+B_s^2) \, ds \right) = \int_0^t (s+s) \, ds < \infty,$$

we find that

$$\mathbb{E} \left( \int_0^t \sqrt{s+B_s^2} \, dB_s \right)=0$$

for all $t \geq 0$.

• Thank you! I am just wondering if there was a name for the above result that if $f$ is $L^2$ bounded, it is a martingale? – user136503 May 1 '16 at 3:04
• @user136503 I don't think so... – saz May 1 '16 at 6:14
• Is there a way to prove the above result? – user136503 May 1 '16 at 10:46
• @user136503 Well, sure... you'll find it in any (reasonable) book on this topic. – saz May 1 '16 at 10:50