I'm working through problems I found on the net for which there are no answers given. Therefore I'm looking for someone to check my work.
Q: $P\left(\int_0^1W(t)dt>\frac{2}{\sqrt3}\right)$ where $W(t)$ is a Wiener process (Brownian motion).
So, let's denote $X_t = \int_0^TW(t)dt$
Then $X_t \sim N(0,\sigma(t)^2)$ since $W(t) \sim N(0,t)$ and the sum of Normal R.V.s is still Normal, correct?
Then, since Gaussians are parametarized by mean (which is zero) and variance, I just need to find (verify above) the variance $\mathbb{E}(X_t^2)$ and then I can find the probability from a Normal CDF. I started with something like:
$\begin{align*} d(tW_t) = W_t dt + t dW_t \end{align*} \to $
$\int_0^TW_tdt=X_t=tW_t-\int{tdW_t} \to $
$X_t^2=t^2W_t^2+\left(\int{tdW_t}\right)^2-2tW_t\int{tdW_t}$
Now by linearity of expectation and Ito's Isometry: $\mathbb{E}(X_t^2)=t^3+\frac{t^3}{3}-2t*COV(W_t, \int{tdW_t})$
The $COV(W_t, \int{tdW_t})$ part is where I'm stuck. A wild guess would be that I could write $W_t$ as $\int{dW_t}$ (is this OK to do with a stochastic process/stochastic calculus?), and then by Ito's Isometry and the fact that it "respects the inner product" (whatever that means) we can turn $COV(\int{dW_t}, \int{tdW_t}) \to \int{tdt}=\frac{t^2}{2}$
Then finally I get $\mathbb{E}(X_t^2)=\frac{t^3}{3}$....which seems like it could be legit from some other things I've found online, but not totally sure.
Can someone check my logic in all of this and verify?