Brownian Motion and Its Square

Let $B_t$ be a standard Brownian motion, so $B_t$ is normal with mean $0$ and variance $t$ . What is the correlation between $B_t$ and $B_t^2$ ?

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#The covariance vanishes

The covariance is the mixed moment minus the product of means,

\Cov(B_t, B_t^2) = \E[B_t \cdot B_t^2] - \E[B_t]\,\E[B_t^2] = \E[B_t^3] - 0 = 0, \tag{1}

since every odd moment of a mean-zero normal vanishes by symmetry.

#Dependent yet uncorrelated

The correlation is therefore $0$ , even though $B_t^2$ is a deterministic function of $B_t$ . Knowing $B_t$ fixes $B_t^2$ exactly, but the link is the symmetric parabola, which carries no linear trend, and correlation measures only the linear part of a relationship.

Knowing B_t pins its square exactly, so the two are perfectly dependent. Yet the relation is the symmetric parabola, which has no linear trend, and correlation sees only the linear part, so it reads zero.

#Read it off

\Cor(B_t, B_t^2) = 0. \tag{2}

It is a clean reminder that zero correlation is far weaker than independence. Here the two variables are as dependent as possible, one determining the other, yet their correlation is exactly zero.