Random process with Cauchy distribution

Question

The problem is as follows.

Let $X(t)$ be a stochastic process such that $X(t) = V + 2t, t \ge 0$, and $V$ has the Cauchy distribution $x_0 = 0, \gamma = 1$. Find the probability that $X(t) = 0$ for at least one $t \in (0, \frac{1}{2}]$.

The obvious trick is to find the probability $a$ that $\forall t \in (0, \frac{1}{2}]\ X(t) \not = 0$, and then to calculate $1 - a$. However, is not $a$ just equal to $1$?..

Answer 1

$P(\exists t\in[0,1/2): X(t)=0)= P(\exists t\in[0,1/2): V=2t)=P(1>V\geq 0)$

You can do the rest yourself using the CDF of a Cauchy distribution, given on Wikipedia.