Let $H\in$ $]0,1[$. A fractional Brownian motion $\left(B_H(t)\right)_{t\geq 0}$ can be represented as $${1\over C(H)}\int_\mathbb{R}\left((t-s)_+^{H-{1\over2}}-(-s)_+^{H-{1\over2}}\right)dB(s)$$ where $B(t)$ is the standard Brownian motion and $$C(H)=\left(\int_0^{\infty}\left((1+s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds+{1\over 2H}\right)^{1\over2}$$

I would like to understand the following equation:

Let $X(t)$ be the integral above, then: $$\mathbb{E}\left[X(t)^2\right]={1\over C(H)^2}\left[\int_{-\infty}^0\left((t-s)^{H-{1\over2}}-(-s)^{H-{1\over 2}}\right)^2ds+\int_0^t(t-s)^{2H-1}ds\right]={t^{2H}\over C(H)^2}\left[\int_{-\infty}^0\left((1-s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds+\int_0^1(1-s)^{2H-1}ds\right]=t^{2H}$$

My problem is what's happening in the brackets. The second summand is ok, but the first one, I don't know why it holds $$\int_{-\infty}^0\left((t-s)^{H-{1\over2}}-(-s)^{H-{1\over 2}}\right)^2ds=t^{2H}\int_{-\infty}^0\left((1-s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds$$

Anybody help me to understand this, please.

  • $\begingroup$ @ Ichiga: First factorize $t$ out of the integrand (to get a factor $t^{2H-1}$) then try the change of variable $u=\frac{s}{t}$ you'll see it works fine. Best regards $\endgroup$ – TheBridge Dec 7 '12 at 16:46
  • $\begingroup$ Did you try to follow this hint? $\endgroup$ – Did Jan 20 '13 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.