Let $\{X_n\}$ be a sequence of independent and identically distributed random variables, $\mathbb{E}X_n=a=\mathrm{const}$.
How to prove that $$ \mathbb{E}\Big(\sum_{k=1}^{n}(X_{k+1}-X_k)\,\mathrm{sign}(a-X_k)\Big)>0 $$ for sufficiently large $n$ $(n\geq N)\,$?
Thank you very much in advance!
As I see, if $X_k<a$ then one "expects" that $X_{k+1}>X_k$,
and if $X_k>a$ then one "expects" that $X_{k+1}<X_k$
(since $(X_{k+1}-X_k)\,\mathrm{sign}(a-X_k)>0$ is equivalent to $(X_{k+1}-X_k)(a-X_k)>0$).