Summing 1d discrete random walks in opposite directions Fix some time $T$.  Suppose we have $A$ with $A_0 = 0$, and $A_t = A_{t-1} + X_{t - 1}$ for $T \geq t > 0$ where $X_t$ takes value either $+1/2$ or $-1/2$ with equal probability; and that we have $B$ with $B_T = 0$, and $B_{t-1} = B_t + X'_t$, with $X'_t$ defined as $X_t$.
That is to say, a random walk $A$, starting at time $0$, running to time $T$; and $B$ starting at $T$ running backwards to time $0$.
I'm interested in $C_t = A_t + B_t$.  How much time does it spend with $|C| > v$ for some value $v$, and how can I calculate $P(C_T > a | C_t = b)$
I suspect there aren't nice answers with $X$ so defined, so if there are nicer results when $X \sim N(0,1)$, then let that be my question instead.
 A: Writing down the combined step and the corresponding probability we get a random walk $C = A - B$ with
$$C_t = C_{t-1} + Y$$
where 
$$Y=\left(
\begin{array}{cc}
 +1 & p=\frac{1}{4} \\
 0 & p=\frac{1}{2} \\
 -1 & p=\frac{1}{4} \\
\end{array}
\right)$$
This gives a generating function
$$g(z,t) = \left(\frac{1}{4}\left(z+2+\frac{1}{z}\right)\right)^t$$
and the final probability for the distance between the two Walkers to be $x$ at the time $t$ is 
$$p(t,x) =\frac{1}{4^t}\binom{2 t}{t+x}$$
The first few values in the format $(t,p(t,x)$ with $( x=-t..+t))$ are
$$p(t,x) = \left(
\begin{array}{cc}
 1 & \left\{\frac{1}{4},\frac{1}{2},\frac{1}{4}\right\} \\
 2 & \left\{\frac{1}{16},\frac{1}{4},\frac{3}{8},\frac{1}{4},\frac{1}{16}\right\} \\
 3 & \left\{\frac{1}{64},\frac{3}{32},\frac{15}{64},\frac{5}{16},\frac{15}{64},\frac{3}{32},\frac{1}{64}\right\} \\
 4 & \left\{\frac{1}{256},\frac{1}{32},\frac{7}{64},\frac{7}{32},\frac{35}{128},\frac{7}{32},\frac{7}{64},\frac{1}{32},\frac{1}{256}\right\} \\
 5 & \left\{\frac{1}{1024},\frac{5}{512},\frac{45}{1024},\frac{15}{128},\frac{105}{512},\frac{63}{256},\frac{105}{512},\frac{15}{128},\frac{45}{1024},\frac{5}{512},\frac{1}{1024}\right\} \\
\end{array}
\right)$$
