Error in calculation of hitting time of -1 in a biased random walk on the integers We have a random walk on the integers ${...,-1,0,...}$ which starts at $0$. It moves left with probability $p$ and right with probability $1-p$. I wanted to calculate the expected amount of time it would take this walk to reach $-1$ for the first time. The formula I arrived at is:
$\mathbb{E}{[T]} = \sum\limits_{n = 0}^{\infty}(2n+1)p^{n+1}(1-p)^{n}C_n$, where $C_n$ is an $n^{th}$ Catalan number. This is because: 


*

*You cannot reach $-1$ in an even number of steps.

*It takes $n+1$ left steps and $n$ right steps to reach $-1$ in $2n+1$ steps.

*The number of ways to order these steps is equal to the number of ways you can organize parentheses $(( () ))$ in a legal way = the $n^{th}$ Catalan number.
What I noticed (using the estimation $C_n \sim 4^n / (n^{3/2} \sqrt{\pi})$ is that this formula actually converges for any $p$ other than $p = 1/2$, including $p > 1/2$. However this does not make sense; seemingly it should converge for only values of $p$ smaller than $1/2$.
Where is my error? My thinking is that maybe the correct calculation should include also the probability that $T = \infty$, i.e. $P(T = \infty) \cdot \infty$ should be formally taken into account in the summation. But I do not know if this is correct.
 A: If $H$ is the event that $-1$ is ever hit then $\mathbb P(H)=\sum\limits_{n = 0}^{\infty}p^{n+1}(1-p)^{n}C_n$ 
This will be $1$ when $\frac12 \lt p \lt 1$  but less than $1$ when $0 \lt p \lt \frac12$
Your expression actually calculates
$$\mathbb E[T\mid H] \,\mathbb P(H) = \sum\limits_{n = 0}^{\infty}(2n+1)p^{n+1}(1-p)^{n}C_n$$ and it is quite reasonable that this converges when $p \not = \frac12$
You can turn this into the conditional expectation
$$\mathbb E[T\mid H] = \dfrac{\sum\limits_{n = 0}^{\infty}(2n+1)p^{n+1}(1-p)^{n}C_n}{\sum\limits_{n = 0}^{\infty}p^{n+1}(1-p)^{n}C_n}$$ which is $\frac{1}{2p-1}$ for $\frac12 \lt p \lt 1$, but is $\frac{1}{1-2p}$ for $0 \lt p \lt \frac12$. You could say  $$\mathbb E[T\mid H]=\left|\dfrac{1}{2p-1}\right|$$ 
So for example with $p=0.9$ or $p=0.1$ you get $\mathbb E[T\mid H] =1.25$, while with $p=0.6$ or $p=0.4$ you get $\mathbb E[T\mid H] =5$.  This makes sense intuitively since with $p$ close to $1$ you are likely to hit $-1$ early, while with $p$ close to $0$ you are unlikely to ever hit $-1$ unless you hit it in the first attempt   
