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.
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 $f(0)=f(1)$.