Expected time in half-line for random walk For a one-dimensional random walk (starting at $0$) for which we move $1$ unit to the right with probability $p$ and $1$ unit to the left with probability $q=1-p>p$, what is the expected time spent in the interval $(0,\infty)$ (equivalently, $[1,\infty)$)?
It is not too hard to come up with an expression for this involving infinite sums and binomial coefficients, but there does not seem to be a straightforward way to put it in closed form. Alternatively, I'm looking for some sort of (asymptotic) estimate if possible.
 A: There are several combinatorial shortcuts which make the calculation not too difficult, and it is worth knowing that $\frac{x}{1-x}=x(1-x)+2x^2(1-x)+3x^3(1-x)+\cdots$ for $|x|<1$
Assuming $0 < p < \frac12$, the probability of moving right initially and then returning to $0$ is $p$.  By symmetry, the probability of moving left initially and then returning to $0$ is also $p$, so the probability of moving left initially but never returning to $0$ is $1-2p$.
So the expected number of returns to $0$ is $$1\times 2p(1-2p) + 2\times (2p)^2(1-2p)+3\times (2p)^3(1-2p)+ \cdots = \frac{2p}{1-2p}$$ and half of these are expected to start by moving right into positive territory, so $\dfrac{p}{1-2p}$.
Suppose the conditional expected time positive on a trip which starts at $0$, moves right, and then returns to $0$ is $T$.  Then the first step takes you to $+1$ and then

*

*with probability $1-p$ you return immediately to $0$ having spent time $1$ positive,

*with probability $p(1-p)$ you take a trip to the left of expected time $T+1$ returning to $+1$ and then immediately to $0$ having spent and expected time $T+2$ positive,

*with probability $p^2(1-p)$ you take two trips to the left each of expected time $T+1$ returning to $+1$ and then immediately to $0$ having spent and expected time $2T+3$ positive,

and so on.  This means  $$T= 1\times (1-p) + (T+2)\times p(1-p)+(2T+3)\times p^2(1-p)  + \cdots \\= 1+ (T+1)\frac{p}{1-p}$$
which gives $T=\dfrac{1}{1-2p}$.
Multiplying by the overall expected number of trips into positive territory then gives the overall expected time in positive territory of $$\frac{p}{(1-2p)^2}.$$
For $p=0.4$ this would give an expected time in positive territory of $10$, though with a widely dispersed distribution of possible values.  To check, here is a simulation in R:
above0 <- function(p, top){
  steps <- sample(c(+1,-1), ceiling(top/(1-2*p)), replace=TRUE, prob=c(p,1-p))
  cumsteps <- cumsum(steps)
  sum(cumsteps > 0)
}

set.seed(2021)
simabove0 <- replicate(10^5, above0(0.4, 1000))
max(simabove0)
# 350
mean(simabove0)
# 10.03401

