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?
The first part of your argumentation is fine: $X_t$ is Gaussian with mean $0$. In order to compute the variance just note that by Fubini's theorem
$$\mathbb{E}(X_t^2) = \mathbb{E} \left( \left[ \int_0^1 W_t \, dt \right] \cdot \left[ \int_0^1 W_s \, ds \right] \right) = \int_0^1 \!\! \int_0^1 \mathbb{E}(W_s W_t) \, dt \, ds.$$ Now use that $\mathbb{E}(W_t W_s) = \min\{s,t\}$ in order to calculate the integral on the right-hand side.