Compute $P\left(\int_0^1W(t)dt>\frac{2}{\sqrt3}\right)$ where $W(t)$ is a Wiener process

Question

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?

Answer 1

$\int_0^1 W_s ds$ is indeed gaussian with mean 0, but to arrive to this conclusion and get its variance the easiest way is to write it as a Wiener integral, i.e: $\int_0^1 ... dW_s$.

As a matter of fact, we can write it as:

$$ \int_0^t W_s ds = \int_0^t (t-s)dW_s$$ (see for example proof here: https://quant.stackexchange.com/a/29506/26242)

We know this Wiener integral is gaussian with mean $0$ and variance: $$Var\left(\int_0^t W_s ds\right) = \int_0^t(t-s)^2 ds =\frac{1}{3}t^3.$$

For $t = 1$, the mean is zero and variance is $\frac{1}{3}$.

Now you can compute your probability using the standard normal CDF $\Phi$: $$\begin{aligned} & P\left(\int_0^1 W_s ds > \frac{2}{\sqrt{3}}\right)\\ & = P\left(Z \frac{1}{\sqrt{3}}> \frac{2}{\sqrt{3}}\right)\\ & = P(Z>2)\\ & = 1 - P(Z \leq 2) \\ & = 1 - \Phi(2) \end{aligned}$$

In the above, $Z$ is a standard gaussian random variable.