Let $W_{t}$ be a Brownian motion independent of $B_{t} .$ Find the probability $P\left(\int_{0}^{1} e^{t} d B_{t}>\int_{0}^{1 / 2} t d W_{t}\right)$.

Question

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!

Answer 1

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 $P(\sqrt {\frac {e^{2}-1} 2} Z >\sqrt {\frac 1 {24}} W)=\frac 1 2$.

Answer 2

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 $0.5$.