Compute the probability $\mathbb{P}\big(\int_0^1 W(t) dt > \frac{2}{\sqrt 3} \big)$
Clearly, as I am not sure about where to start I am confused about how the $W(t)$ relates to an integral and how this number could relate to a probability.
1) Should I evaluate the integral first? Not sure how to do this.
2) Should I square both sides (risky)?
First noticed that $\int_{0}^{t}{W_sds}$ is Gaussian : $W$ is continuous, and can be written as the limit of a Riemann sum of Brownian increments.
$$\int_{0}^{t}{W_sds}=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}{W_{\frac{(k-1)t}{n}}\left(\frac{kt}{n}-\frac{(k-1)t}{n}\right)}=\lim_{n \rightarrow \infty}\sum_{k=1}^{n}{\left(W_{\frac{kt}{n}}-W_{\frac{(k-1)t}{n}}\right)\left(t-\frac{kt}{n}\right)}$$
The mean of the integral is $0$ and the variance is $\frac{t^3}{3}$.
Introduce the standard normal variable $Y$, and we have $$\int_{0}^{1}{W_sds}=\sqrt{\frac{1}{3}}Y$$
Therefore, $$P\left(\int_{0}^{1}{W_sds} >\frac{2}{\sqrt{3}}\right)=P\left(\sqrt{\frac{1}{3}}Y>\frac{2}{\sqrt{3}}\right)=P\left(Y>2\right)=1-P(Y \leq2)=1-\Phi(2)=\Phi(-2)$$
where $\Phi$ is the cdf of $Y$.