Covariance of increments of fractional Gaussian noise

Question

Let $B^H(t)$ be a fractional Brownian motion with Hurst parameter $H\in (0,1)$. We define fractional Gaussian noise as $X(t)=B^H(t+1)-B^H(t)$. We know the fBm has covariance $R(s,t)=E(B^H(t)B^H(s))=\frac{1}{2}(t^{2H}+s^{2H}-|t-s|^{2H})$. We can build up the covariance of increments of fBm from this and then even further we can ask about:

$$E((X(t)-X(s))(X(v)-X(u))$$

Is there any non horrible way of writing down what this covariance should be? I can write it in terms of a bunch of $R$s but nothing better.

Answer 1

Hint: $\mathbb{E}\left[\left(B_{t}^{H}-B_{s}^{H}\right)\left(B_{u}^{H}-B_{v}^{H}\right)\right]=\frac{1}{2} \mathbb{E}\left[\left(B_{t}^{H}-B_{v}^{H}\right)^{2}+\left(B_{s}^{H}-B_{u}^{H}\right)^{2}-\left(B_{t}^{H}-B_{u}^{H}\right)^{2}-\left(B_{s}^{H}-B_{v}^{H}\right)^{2}\right]=\frac{1}{2}\left(|t-v|^{2 H}+|s-u|^{2 H}-|t-u|^{2 H}-|s-v|^{2 H}\right)$

Answer 2

By expansion and direct substitutions, the expression

$$E\left[ ( X(t)-X(s) )( X(v)-X(u) ) \right]$$

yields

$$\frac{1}{2} \left(-\left| s-u\right| ^{2 H}+\left| s-v\right| ^{2 H}+\left| t-u\right|^{2 H}-\left| t-v\right| ^{2 H}\right).$$

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If you are actually asking about the covariance of a process defined as the increments of fractional Gaussian noise (fGn), called the differentiated fractional Gaussian noise (DfGn), $Y(t) = X(t+1) - X(t)$, where $X(t)$ is fGn, then, assuming it has $\mu = 0$ and $\sigma^2 = 1$ for simplicity, employing the covariance of fGn (which can be calculated in the same manner):

$$E\left[ X(t)X(s) \right] = \frac{1}{2}\left( |t-s-1|^{2H} + |t-s+1|^{2H} - 2 |t-s|^{2H} \right),$$

one can expand

$$E\left[ Y(s)Y(t) \right] = E\left[ (X(s+1)-X(s))(X(t+1)-X(t)) \right]$$

to obtain

$$2\left( |t-s-1|^{2H} + |t-s+1|^{2H} \right) -3|t-s|^{2H} - \frac{1}{2}\left( |t-s-2|^{2H} + |t-s+2|^{2H} \right).$$

The variance then is obtained by setting $s=t$, and yields

$$4-4^H,$$

which depends only on $H$, and decreases monotonically from 3 at $H=0$ to 0 at $H=1$. For comparison, the variance of fBm is $t^{2H}$, and fGn's is $1$.