compute the variance of the area under a brownian path

Question

Let $P(t)$ be the brownian path wrt t, for a random outcome $\omega \in \Omega$

Letting $A(\omega) = \int_{0}^{1}P(t)dt$, how to show the Var$[A] = \frac{1}{3} $?

It is clear that $E[A] = 0$ so Var$[A]=E[A^2] = E[\int_{0}^{1}P(s)ds\int_{0}^{1}P(t)dt]$

$=\int_{0}^{1}\int_{0}^{1}E[P(s)]E[P(t)-P(s)]dtds$

but $E[P(s)]=0$ right? So the variance is not turning into $\frac{1}{3}$ I'm not sure where I went wrong.

Answer 1

Let $W_t$ be a standard Wiener process (Brownian motion). We have

\begin{align*} \operatorname{Var}\left(\int_{0}^{t}W_sds\right)&=\mathbb{E}\left[\left(\int_{0}^{t}W_sds\right)^2\right]=\mathbb{E}\left[\int_{0}^{t}\int_{0}^{t}W_s\,W_u du\,ds\right]\\ &=\int_{0}^{t}\int_{0}^{t}\mathbb{E}[W_sW_u]duds=\int_{0}^{t}\int_{0}^{t}\min\{s,u\}duds\\ &=\int_{0}^{t}\int_{0}^{s}u\,duds+\int_{0}^{t}\int_{s}^{t}s\,duds=\frac 13 t^3 \end{align*} Now set $t=1$.