Distribution of Increments of a fractional Brownian Motion

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According to my Textbook, a fBm Process $B_t$ has the same distribution as its increments:

$(B_{t+h}^H-B_t^H)_{t\geq 0} \stackrel{d}{=}(B_t)_{t\geq 0}$

To proof this, note that both sides are centered Gaussian Processes. Hence it is enough to show that they have the same covariance i.e.:

$Cov(B_{t+h}^H-B_t^H,B_{s+h}^H-B_s^H) = \frac{1}{2}(t^{2H}+s^{2H}-|t-s|^{2H})$

But I am not able to proof this. So far I have:

$Cov(B_{t+h}^H-B_t^H,B_{s+h}^H-B_s^H) \\ =E((B_{t+h}^H-B_t^H)(B_{s+h}^H-B_s^H)) \\ =\frac{1}{2}\left(|(t+h)-s|^{2H}+|t-(s+h)|^{2H}-|(t+h)-(s+h)|^{2H}-|t-s|^{2H}\right) \\ =\frac{1}{2}\left(|(t+h)-s|^{2H}+|t-(s+h)|^{2H}-2|t-s|^{2H}\right)$

How do I continue from this or did I already make a mistake on the way ?

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I messed up the indices. It should be:

$(B_{t+h}^H-B_{\mathbf{h}}^H)_{t\geq 0} \stackrel{d}{=}(B_t)_{t\geq 0}$

Now we can derive the desired result:

$Cov(B_{t+h}^H-B_h^H,B_{s+h}^H-B_h^H) \\ =E((B_{t+h}^H-B_h^H)(B_{s+h}^H-B_h^H)) \\ =\frac{1}{2}\left( |t+h|^{2H}+|s+h|^{2H}-|t-s|^{2H} \\ -|t+h|^{2H}-|h|^{2H}+|t|^{2H} \\ -|h|^{2H}-|s+h|^{2H}+|s|^{2H} \\ +|h|^{2H}+|h|^{2H}-|h-h|^{2H}\right) \\ = \frac{1}{2}\left(|t|^{2H}+|s|^{2H}-|t-s|^{2H} \right) $