Standard Deviation of Fractional Brownian Motion
Let 
![E[B_{H}(t)B_{H}(s)]=\frac{1}{2}(|t|^{2H}+|s|^{2H}-|t-s|^{2H})](http://s0.wp.com/latex.php?latex=E%5BB_%7BH%7D%28t%29B_%7BH%7D%28s%29%5D%3D%5Cfrac%7B1%7D%7B2%7D%28%7Ct%7C%5E%7B2H%7D%2B%7Cs%7C%5E%7B2H%7D-%7Ct-s%7C%5E%7B2H%7D%29&bg=e6e6e6&fg=333333&s=0 "E[B_{H}(t)B_{H}(s)]=\frac{1}{2}(|t|^{2H}+|s|^{2H}-|t-s|^{2H})")
Then(by taking s=t in (1))
![Var(B_{H}(t))=E[(B_{H}(t)-E(B_{H}(t)))^2]=E[B_{H}^{2}(t)]=|t|^{2H}](http://s0.wp.com/latex.php?latex=Var%28B_%7BH%7D%28t%29%29%3DE%5B%28B_%7BH%7D%28t%29-E%28B_%7BH%7D%28t%29%29%29%5E2%5D%3DE%5BB_%7BH%7D%5E%7B2%7D%28t%29%5D%3D%7Ct%7C%5E%7B2H%7D&bg=e6e6e6&fg=333333&s=0 "Var(B_{H}(t))=E[(B_{H}(t)-E(B_{H}(t)))^2]=E[B_{H}^{2}(t)]=|t|^{2H}")
Which gives the standard deviation as 
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