The Behavior of a Jump for Processes with Independent Increments

Let $\xi(t), t \geq 0$ be a process with independent increments, $\xi(0)=0$, $\tau_y=\inf\{t:\xi(t)\ge y\}$, $\Gamma_y=\xi(\tau_y)-y$.
We prove that:
1) $\mathbf{P}\{\Gamma_y>0\}=0$ for all $y>0$ if and only if
$$ \ln\mathbf{M}e^{i\lambda\xi(t)}=t\biggl\{ia\lambda-\frac{\sigma^2\lambda^2}{2}+\int_{-\infty}^0 \biggl(e^{i\lambda x}-1-\frac{i\lambda x}{1+x^2}\biggr)ds(x)\biggr\}; $$

2) $\mathbf{P}\{\Gamma_y>0\}>0,\ y>0,\ y\ne kh,\ k=1,2,\dots,h>0$, and $\mathbf{P}\{\Gamma_{kh}>0\}=0$, $k=1,2,\dots$ if and only if
$$ \ln\mathbf{M}e^{i\lambda\xi(t)}=t\biggl\{p_1(e^{i\lambda h-1})+\sum_{k=-\infty}^{-1}p_k(e^{i\lambda kh}-1)\biggr\}, $$

$$ p_k\ge 0,\quad k=1,-1,-2,\dots,\quad \sum_{k=-\infty}^{-1}p_k<\infty,\quad p_1>0; $$

3) in all other cases $\mathbf{P}\{\Gamma_y>0\}>0$ for all $y>0$.