Let \(X_{t}=b t+W_{t}+Z_{t}\) where \(W\) is a Brownian motion and \(Z_{t}=\sum_{k=1}^{N_{t}} Y_{k}\) a \((\lambda, F)\)-compound Poisson process independent of \(W\). The first passage time above the level \(x\) is

\[T_{x}=\inf \left\{t: X_{t} \geq x\right\}\]

and the overshoot is \(O_{x}=X_{T_{x}}-x\). Let \(\Phi_{x}\) be the Laplace transform of the pair \(\left(T_{x}, O_{x}\right)\), i.e.,

\[\Phi_{x}(\theta, \mu, x)=\mathbb{E}\left(e^{-\theta T_{x}-\mu O_{x}} \mathbb{1}_{\left\{T_{x}<\infty\right\}}\right)\]

Let \(\tau_{1}\) be the first jump time of \(N\). We wish to establish an integral equation for \(\Phi_{x}\) using the following computation:

(1) Prove that \[\mathbb{E}\left(e^{-\theta T_{x}-\mu O_{x}} \mathbb{1}_{\left\{T_{x}<\tau_{1}\right\}}\right)=e^{(b-\alpha) x}\]
where \(\alpha=\sqrt{b^{2}+2(\theta+\lambda)}\).
(2) Prove that \[\begin{aligned}
\mathbb{E}\left(e^{-\theta T_{x}-\mu O_{x}} \mathbb{1}_{\left\{T_{x}=\tau_{1}\right\}}\right) & =\frac{e^{(b-\alpha) x}}{\alpha(\mu-b+\alpha)} \int_{[0, x[}\left(e^{(\alpha-b) y}-e^{-\mu y}\right) F(d y) \\
& +\frac{1}{\alpha(\mu-b-\alpha)} \int_{[x, \infty[}\left(e^{(b-\alpha)(y-x)}-e^{-\mu(y-x)}\right) F(d y) \\
& +\frac{e^{\mu x}-e^{(b-\alpha) x}}{\alpha(\mu-b+\alpha)} \int_{[x, \infty[} e^{-\mu y} F(d y) \\
& +\frac{e^{(b-\alpha) x}}{\alpha(\mu-b-\alpha)} \int_{[0, \infty[}\left(e^{-(\alpha+b) y}-e^{-\mu y}\right) F(d y) .
\end{aligned}\]
(3) Prove that \[\begin{gathered}
\mathbb{E}\left(e^{-\theta T_{x}-\mu O_{x}} \mathbb{1}_{\left\{\tau_{1}=\frac{1}{\alpha} \int_{\mathbb{R}} F(d y) \int_{-\infty}^{(x-y) \wedge x} e^{b z}\left(e^{-\alpha|z|}-e^{(2 x-z) \alpha}\right) \Phi_{x-z-y}(\theta, \mu) d z \end{gathered}\]
(4) Deduce an equation for \(\Phi\) by adding the three components above.