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Let $X$ be a Feller process on $mathbb{R}$ with generator $Gf=frac{1}{2}f''-f'$ on $C_c^2$. Let $tau_b$ be the first time that $X$ hits $binmathbb{R}$. Show that for $x > 0$, $bP_x(tau_b < tau_0)to 0$ as $btoinfty$ and compute $E_x[tau_0]$.

The way I have solved problems similar to the second part is to use Doobs stopping theorem on some appropriate martingale. Indeed we even have a variety of martingales at our disposal thanks to the fact that $M_t=f(X_t)-int_0^t Gf(X_s),ds$ is a martingale for $fin C_c^2$. But I don't see an appropriate $f$ that would help me out here, neither for the first nor the second part. I would appreciate any hints.

Hints for Part 1: Set $$tau :=tau_{0,b} := min{tau_0,tau_b}.$$

1. Set $g(x) := tfrac{1}{2} exp(2x)$. Use Dynkin's formula to show that $$mathbb{E}^x(g(X_{t wedge tau}))-x = mathbb{E}^x left( int_0^{t wedge tau} (Gg)(X_s) , ds right).$$
   


2. As $Gg(x)=0$ it follows from Step 1 that $$mathbb{E}^x g(X_{t wedge tau}) = g(x).$$ Apply the dominated convergence theorem to conclude that $$mathbb{E}^x g(X_{tau})=g(x).$$
   


3. Show that $$mathbb{E}^x g(X_{tau}) = frac{1}{2} left( e^{2b} mathbb{P}^x(tau_b<tau_0)+ mathbb{P}^x(tau_0<tau_b) right)$$ and deduce from Step 2 that $$limsup_{b to infty} e^{2b} mathbb{P}^x(tau_b<tau_0)< infty;$$ hence $$lim_{b to infty} b mathbb{P}^x(tau_b<tau_0)=0.$$
   

Hints for Part 2: Set $$tau :=tau_{0,b} := min{tau_0,tau_b}.$$

1. Set $f(x) := x$. Use Dynkin's formula to show that $$mathbb{E}^x(f(X_{t wedge tau}))-x = mathbb{E}^x left( int_0^{t wedge tau} (Gf)(X_s) , ds right).$$
   


2. Since $Gf(x)=-1$ it follows that $$mathbb{E}^x f(X_{t wedge tau})-x = - mathbb{E}^x (t wedge tau).$$
   


3. Use $|f(X_{t wedge tau})| leq b$ and the monotone convergence theorem to prove that $mathbb{E}^x(tau)<infty$.


4. Conclude from Step 2 and 3 that $$b mathbb{P}^x(tau_b<tau_0)-x = mathbb{E}^x f(X_{tau})-x = - mathbb{E}^x(tau). tag{1}$$
   


5. Use Part I and the monotone convergence theorem to let $b to infty$ in $(1)$. Conclude that $mathbb{E}^x(tau_0)=x$.

Remark: The process $X$ is actually a Brownian motion with drift, $$X_t = B_t-t; $$ this follows from the particular form of the generator.