Expression for discrete Brownian motion with reflection

Question

For a discrete Brownian motion with an absorbing state we can express the distribution of the position as a linear sum of two binimial distributions as described here when the odds for +1 and -1 steps are 1:1, and here for the more general case when the odds are not equal.

How is the situation for a discrete Brownian motion with a reflecting wall?

Say we describe the position at time $t$ as $X(t)$ with starting position $X(0) = 1$. With probabability $p$ we make a step forward $X(t+1) = X(t) + 1$ and with probabability $1-p$ we make a step backwards $X(t+1) = X(t) - 1$, except when $X(t) = 0$ in which case we always make a step forward.

Answer 1

If $p=q= 0.5$ then we can also represent it as a mixture sum of two binomial distributions as the R-code below demonstrates. But for different $p \neq q$ this doesn't work and a more complicated sum with multiple terms might be necessary.

t = 20
t2 = t/2
p = 1/2
q = 1-p

###
### Compute with Markov chain
###


## start position 
x = c(0,1)

### create a simulation of t steps by taking t/2 times 2 steps 
for(i in 1:t2) { 
   ### computing the evolution after 2 steps
   ### p^2 probability for +2
   ### 2pq probability for +0
   ### q^2 probability for -2
   x_new = c(0,x*p^2)+
           c(  2*x*q*p,  0)+
           c(  x[-1]*q^2,0,0)

   ### correction for the reflection
   ### steps to position [1] are reflected to position [2]
   x_new[2] = x_new[2] + x_new[1] 
   x_new[1] = 0  

   ### new positions after 2 extra steps
   x = x_new
}

plot(0:t2, x[-1], 
     xlab = "position (X-1)/2", 
     ylab = "probabability",  
     main = "example of probabability mass distribution \n for position after 20 steps")



###
### computing as sum of two binomials
###

z = 0:t2
y1 = t2+z
y2 = t2+z+1

f = dbinom(y1,t,p) 
f = f + dbinom(y2,t,p)

lines(z,f) ### this computation with two binomials matches the simulation