A simple random walk $S\_n = X\_1 +\cdots +X\_n$, where $P(X\_i = 1) = p \not = 0.5$ and $P(X\_i=-1)= q \triangleq 1-p$, admits the following probability 
$$P(S\_n \textrm{ reaches } a \textrm{ before} -b) = \left(\frac{1-\left(\frac{q}{p}\right)^{b}}{1-\left(\frac{q}{p}\right)^{a+b}}\right),$$ 
for $a$ and $b$ positive integers. The expected number of steps to hit either of these boundaries is 
$$\frac{b}{q - p} - \left(\frac{a+b}{q-p}\right)\left(\frac{1-\left(\frac{q}{p}\right)^{b}}{1-\left(\frac{q}{p}\right)^{a+b}}\right).$$


My question is whether there are analogues to these simple expressions for positive $a$ and $b$ (not necessarily integer) when the $X\_i$ are continuous i.i.d. random variables on $[-1,1]$ with a given density $f$? I'm particularly interested in a case where the $X\_i$ have a finite positive mean $\mu\_X > 0$. 



Random walk with continuously distributed steps on [-1,1]
As already pointed out by Anthony in a comment, you can't really expect explicit formulae. Let's write $p(x)$ for the probability that the RW starting at $x\in (a,b)$ hits $a$ before it hits $b$. Then, by the same argument as in the discrete case (condition on what happens on the very next step),
$p(x) = \int p(x+t)f(t)\, dt$; here, we must interpret $p(s)=1$ for $s\le a$ and $p(s)=0$ for $s\ge b$ on the right-hand side. So by splitting off the parts where $x+t\notin (a,b)$, we can rewrite the integral equation as
$$
p(x) = F(a-x) + \int\_a^b f(t-x)p(t)\, dt ,\quad a<x<b .
$$
This is a Fredholm equation of the second type.
It won't be easy to come up with an $f$ for which this has an explicit solution. You can obtain a series representation of the solution by iterating, see [here](http://mathworld.wolfram.com/IntegralEquationNeumannSeries.html) for example (notice that the $\phi\_n$ increase pointwise and stay $<1$, so convergence is guaranteed).


