Expected value and variance of Z = $\int_0^1 (B_t + t) dB_t$ where B is the standard Brownian motion (also known as the Wiener process)
Expected Value Steps:
Let $Y_t = \frac{1}{2} B^2_t + tB_t$
$dY_t = (t + B_t)dB_t + B_tdt + \frac{1}{2}dt$
$\int_0^1 (B_t + t) dB_t = \frac{1}{2} B^2_1 + tB_1 - \int_0^1 B_tdt - \int_0^1 \frac{1}{2}dt$
$E[Z] = E[\int_0^1 (B_t + t) dB_t] = \frac{1}{2}*1 + 0 - 0 - \int_0^1 \frac{1}{2}dt = \frac{1}{2} - \frac{1}{2} = 0$
Variance Steps:
$Var[Z] = E[Z^2] - (E[Z])^2$
Z^2 = \int_0^1 (B_t + t)^2 dB_t = \int_0^1 (B^2_t + 2tB_t + t^2) dB_t$
Let $X_t = \frac{1}{3}B^3_t + tB^2_t + t^2B_t$
$dX_t = (B^2_t + 2tB_t + t^2)dB_t + (B^2_t +2tB_t)dt + (B_t + t)dt$
$Z^2 = \int_0^1 (B_t + t)^2 dB_t = \frac{1}{3}B^3_1 + 1*B^2_1 + 1^2*B_1 - \int_0^1 (B^2_t + 2tB_t) dt - \int_0^1 (B_t + t)dt$
$E[Z^2] = 0 + 1 + 0 - \int_0^1 E[(B^2_t + 2tB_t)] dt - \int_0^1 E[(B_t + t)]dt = 1 - \int_0^1 E[(t + 0)] dt - \int_0^1 E[(0 + t)]dt = 1 - \frac{1}{2} - \frac{1}{2} = 0$
$Var[Z] = 0 - 0^2 = 0$
I feel like I messed up somewhere in calculating the Expected Value and Variance but I can't pinpoint it.
Because $B_t+t$ is square integrable $\mathbb{E}[Z]=0$ by the martingale property of the Ito integral. The by Ito Isometry:
$$Var(Z)=\mathbb{E}\int_0^1 (B_t + t)^2 dt$$
$$=\int_0^1 \mathbb{E}(B_t^2 +2B_tt+ t^2) dt=$$
$$\int_0^1 (t +t^2) dt=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}.$$