# Finding the expected value of of $\int_0^s \sqrt{t+B_t^2}dB_t$?

How can I find the expected value of $\int_0^s \sqrt{t+B_t^2}dB_t$? I know one condition is to show that if:

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

$$M_s := \int_0^s f(t) \, dB_t, \qquad s \geq 0,$$

is a martingale.

However, I don't know how to prove the above nor know of any good names for it. Is there a way to do it by Ito's formula for space and time or any other direct methods? Thanks.

• $E(\sqrt{t+B_t^2 }dB_t)$ are you asking about this Jan 13 '17 at 6:40
• I'm actually asking about $E(\int_0^s\sqrt{t+B_t^2 }dB_t)$ Jan 13 '17 at 7:22

About measurability. I assume that we are speaking about the filtration $\mathcal F_t$ that is generated by the given Brownian motion $B_t$. Then $f(t)=\sqrt{t+B_t^2}$ is progressively measurable with respect to $\mathcal F_t$. Indeed, $B_t$ itself is progressively measurable w.r.t. $\mathcal F_t$ (it is adapted to $\mathcal F_t$ by the definition of this filtration and has a.s. continuous paths). The function $g(t)=t$ is also progressively measurable w.r.t. $\mathcal F_t$ (it is non-random). Since $f(t)$ is a measurable function of $B_t$ and $g(t)$, it is progressively measurable as well.
About expectation. We have $$\mathbb E \left(\int_0^s|f(t)|^2 dt\right)=\mathbb E \left(\int_0^st+B_t^2 dt\right)=\mathbb E \left(\frac{s^2}{2}+\int_0^sB_t^2 dt\right)=\frac{s^2}{2}+\mathbb E \left(\int_0^sB_t^2 dt\right).$$ Now by Tonelli's theorem, we may interchange expectation and integration in the last term: $$\mathbb E \left(\int_0^sB_t^2 \ dt\right)=\int_0^s\mathbb E(B_t^2)\ dt=\int_0^s t \ dt=\frac{s^2}{2}.$$ We obtain that
$$\mathbb E \left(\int_0^s|f(t)|^2 dt\right)=s^2<\infty.$$ Now we conclude that your integral is a martingale, and its expected value is therefore zero.
in your case you have $f(t)= \sqrt{B_t^2+t}$, so you have to check if $$\Bbb E \biggl[ \int_0^s B_t^2+t \; dt \biggr] < \infty.$$ Note that $f$ is progressively measurable since $B$ is adapted. Interchanging both integrals we get that $$\Bbb E \biggl[ \int_0^s B_t^2+t \; dt \biggr]= \int_0^s \Bbb E \bigl[B_t^2+t \bigr] \; dt =\int_0^s 2t \; dt = s^2.$$ So the process $(\int_0^s \sqrt{B_t^2+t} \;dB_t)_{s \geq 0}$ is a martingale and hence has constant expectation. Therefore, $$\Bbb E \biggl[ \int_0^s \sqrt{B_t^2+t} \; dBt \biggr]=\Bbb E \biggl[ \int_0^0 \sqrt{B_t^2+t} \; dBt \biggr]=0.$$