# Covariance functional of a generalized real-valued Brownian motion

Let $(B_t)_{t\ge 0}$ be a real-valued Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$, $\lambda$ be the Lebesgue measure on $[0,\infty)$ and $$\langle W,\phi\rangle:=\int\phi(t)B_t\;{\rm d}\lambda\;\;\;\text{for }\phi\in\mathcal D:=C_c^\infty([0,\infty))\;.$$

We can prove that the expectation $$\operatorname E[W](\phi):=\operatorname E\left[\langle W,\phi\rangle\right]\;\;\;\text{for }\phi\in\mathcal D$$ is $0$. Now I want to prove that the covariance $$\rho[W](\phi,\psi):=\operatorname E\left[\langle W,\phi\rangle\langle W,\psi\rangle\right]=\int\int\min(s,t)\phi(s)\psi(t)\;{\rm d}\lambda(s)\;{\rm d}\lambda(t)\;\;\;\text{for all }\phi,\psi\in\mathcal D\;.$$

Obviously, we somehow need to use that $$\operatorname{Cov}[B_s,B_t]=\min(s,t)\;\;\;\text{for all }s,t\ge 0\;.$$ How can we do that?

[As a secondary issue: How can we generalize this result to Hilbert space valued Brownian motions and cylindrical Brownian motions?]

• Well, just apply Fubini's theorem... and you are done? – saz Jan 31 '16 at 12:46
• @saz How do you want to apply Fubini's theorem? We're taking the expectation of a product of two Lebesgue integrals. – 0xbadf00d Jan 31 '16 at 12:49
• Yeah, so what? That's exactly what's Fubini's theorem for. By Fubini, $$\langle W,\phi \rangle \langle W,\psi \rangle = \iint B_s B_t \phi(s) \psi(t) \, d\lambda(s) \, d\lambda(t)$$ and therefore, again by Fubini, $$\mathbb{E}(\langle W,\phi \rangle \langle W,\psi \rangle ) = \iint \mathbb{E}(B_s B_t \phi(s) \psi(t)) \lambda(ds) \, \lambda(dt).$$ (Obvioiusly, we have to check that Fubini is applicable, but since $\psi$ and $\phi$ have compact support, that's hardly a problem.) – saz Jan 31 '16 at 12:52
• @saz That's not the Fubini that I know: Given $\sigma$-finite measure spaces $(\Omega_i,\mathcal A_i,\mu_i)$ and $f:\Omega_1\times\Omega_2\to\mathbb R$ measurable with respect to $\mathcal A_1\otimes\mathcal A_2$ and nonnegative or $(\mu_1\otimes\mu_2)$-integrable, Fubini tells us that $$\int f\;{\rm d}(\mu_1\otimes\mu_2)=\int\int f\;{\rm d}\mu_1{\rm d}\mu_2=\int\int f\;{\rm d}\mu_2{\rm d}\mu_1\;.$$ So, I don't understand how you derive your equation from that. – 0xbadf00d Jan 31 '16 at 12:58
• That's not any different from the version I know. Apply it with $\mu_1 = \mathbb{P}$ and $\mu_2 = (\lambda \otimes \lambda) = \lambda^2$ (i.e. two-dim. Lebesgue measure) – saz Jan 31 '16 at 13:04