Poisson process and an independent Wiener process Let $\{W_t\}_{t\geq 0}$ be a standard one-dimensional Wiener process and $\{N_t\}_{t\geq 0}$ an independent rate-1 Poisson process. Define $T$ to be the first time (if ever) $t \geq 0$ such that $W_t \geq N_t + 1$. I want to find $P\{T < \infty\}$. It is not difficult to find an expression involving expectations of some functions of hitting times of SBM. But I heard from some source that it would be the solution to a transcendental algebraic equation. Any suggestions?
 A: Assume that $N$ is a Poisson process with intensity $a$. 
For every nonnegative $x$, let $T_x=\inf\{t\geqslant0\mid W_t-N_t\geqslant x\}$ and $t(x)=P[T_x\lt\infty]$. When $s\to0$, during the time interval $(0,s)$, $N$ jumps from $0$ to $1$ with probability $as+o(s)$ hence the (weak) Markov property of the process $W-N$ at time $s$ yields 
$$
t(x)=ast(x+1)+(1-as)E[t(x-W_s)]+o(s),
$$
where $t(x)=1$ for every $x\leqslant0$.
The distribution of $W_s$ is centered normal with variance $s$ hence 
$$
E[t(x-W_s)]=t(x)+\tfrac12st''(x)+o(s),
$$ 
and
$$
t''(x)=2a(t(x)-t(x+1)).
$$
The process $W-N$ has no positive jump and has stationary increments hence the (strong) Markov property of $W-N$ at time $T_x$ yields $t(x+y)=t(x)t(y)$ for every nonnegative $x$ and $y$. Since $t(\ )$ is nonincreasing, 
$$
t(x)=\mathrm e^{-\tau(a) x},
$$ 
for every nonnegative $x$, for some $\tau(a)$. The delayed differential equation above shows that, for each positive $a$, $\tau(a)$ is the unique positive solution of the identity
$$
\tau(a)^2=2a(1-\mathrm e^{-\tau(a)}).
$$
Finally, $P[T\lt\infty]=t(1)=\mathrm e^{-\tau(a)}$ solves
$$
(\log P[T\lt\infty])^2=2a(1-P[T\lt\infty]).
$$
