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 $


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?


$\int_0^1 W_s ds$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.