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Let $B_H$ be a fractional Brownian motion, $f\in C^1[0,1]$ and $a>0$. Define the process $$X(t)=f(t)+aB_H(t),\qquad t\in[0,1].$$

I would like to show, that $\mathbb{P}\big(X(0)=X(1)\big)=0$ holds. Is it right, that I also can write $E[X(0)]\neq E[X(1)]$, which means $f(0)\neq f(1)$?

Thank you.

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    $\begingroup$ Show that $X(1)-X(0)$ is Gaussian with nonzero variance, hence equals 0 with zero probability. It doesn't follow that $E[X(0)] \ne E[X(1)]$; nothing in the information given prevents you from having $f(0)=f(1)$. $\endgroup$ – Nate Eldredge Oct 16 '13 at 18:15
  • $\begingroup$ Can you please tell me, why the argumentation of the nonzero variance leads to the assertion? $\endgroup$ – Ichigo Oct 16 '13 at 19:37
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    $\begingroup$ A Gaussian random variable $Y$ with a nonzero variance has a continuous distribution; in particular, for any number $a$, we have $P(Y=a)=0$. $\endgroup$ – Nate Eldredge Oct 16 '13 at 23:02
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From the comments by Nate Eldredge: introduce random variable $Y=X(1)−X(0)$. Recall that the increments of (fractional) Brownian motion follow normal distribution with nonzero variance. Hence, the probability of $Y=a$ is zero for every $a$, in particular for $a=f(0)-f(1)$. This establishes $P(X(0)=X(1))=0$.

However, it doesn't follow that $E[X(0)]\ne E[X(1)]$. Also, nothing in the information given prevents you from having $f(0)=f(1)$.

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