Let $W_{t}$ be a Brownian motion independent of $B_{t} .$ Find the probability $P\left(\int_{0}^{1} e^{t} d B_{t}>\int_{0}^{1 / 2} t d W_{t}\right)$. Note $B_0$ = 0.
My attempt
By Ito's zero mean property an Isometry property we know that:
$E(\int_{0}^{1} e^{t} d B_{t}) = 0 $ and
Variance of $(\int_{0}^{1} e^{t} d B_{t})$ = $\int_{0}^{1} E\left(e^{2 t}\right) d t=\frac{e^{2}-1}{2}$
Similarly:
$E(\int_{0}^{1 / 2} t d W_{t}) = 0 $.
Variance of $(\int_{0}^{1 / 2} t d W_{t})$ is $\left.(\int_{0}^{1 / 2} E(t^2)   dt\right)=\frac{1}{24}$
I know introduce standard normal variable Z and we have:
$\int_{0}^{1} E\left(e^{2 t}\right) d t= \sqrt{\frac{e^{2}-1}{2}} Z$
$\left.(\int_{0}^{1 / 2} E(t^2)   dt\right)=\sqrt{\frac{1}{24}}Z$
Now we have:
$P\left(\int_{0}^{1} e^{t} d B_{t}>\int_{0}^{1 / 2} t d W_{t}\right)$. = $P(\sqrt{\frac{e^{2}-1}{2}} Z > \sqrt{\frac{1}{24}}Z )$
I am bit stuck now on how to conclude and I am not sure if what I have done is right or wrong.  Any help is highly appreciated. Thank you!
 A: You cannot use  the same $Z$ for $(B_t)$ and $(W_t)$ because they are independent. So you have to evaluate $P(\sqrt {\frac {e^{2}-1} 2} Z >\sqrt {\frac 1 {24}} W)$ where $Z,W$ are i.i.d. $N(0,1)$. Since $\sqrt {\frac {e^{2}-1} 2} Z -\sqrt {\frac 1 {24}} W$ is normal with mean $0$ we see that $P(\sqrt {\frac {e^{2}-1} 2} Z >\sqrt {\frac 1 {24}} W)=\frac 1  2$.
A: The first process is described by the SDE
$$dX_t=e^tdB_t$$
The characteristic function solves the ODE
$$\frac{\partial}{\partial t}\hat{P}_X(\omega,t)=-\frac{\omega^2}{2}e^{2t}\hat{P}_X(\omega,t)$$
Therefore
$$\hat{P}_X(\omega,t)=\exp\left\{-\frac{\omega^2}{4}(e^{2t}-1)\right\}$$
which implies $X_1\sim \mathcal{N}(0,\frac{e^{2}-1}{2})$. The second process is described by the SDE
$$dY_t=tdW_t$$
The characteristic function solves the ODE
$$\frac{\partial}{\partial t}\hat{P}_Y(\xi,t)=-\frac{\xi^2}{2}t^2\hat{P}_Y(\xi,t)$$
And its solution is
$$\hat{P}_Y(\xi,t)=\exp\left\{-\frac{\xi^2}{6}t^3\right\}$$
which implies $Y_{\frac{1}{2}}\sim \mathcal{N}(0,\frac{1}{3\cdot 2^3}=\frac{1}{24})$. Finally
$$P(X_1>Y_{\frac{1}{2}})=P(V=X_1-Y_{\frac{1}{2}}>0)$$
Since $V \sim \mathcal{N}(0,\frac{e^{2}-1}{2}+\frac{1}{24})$ this probability is $0.5$.
