Discrete random walk with uniformly distributed transition p, set initially

Question

I've been working on a discrete version of the "unreliable friend" distribution. It would seem that what I've come up with is equivalent to the following random walk:

Choose $p$ from $U(0,1)$

Start at location $x_0 = a, a \in \mathbb{N} $

Move to $x_{i+1} = x_i - 1$ with probability $p$ and $x_{i+1} = x_i + 1$ with probability $1-p$

What is the distribution on your first arrival at $0$?

Is this a known/explored distribution? Information on r.w. distributions with known $p$ is abundant, but so far I haven't found anything for $p \sim U(0,1)$.

UPDATE: After more research it's clear the distribution is related to the Hitting Time Theorem (alluded to by Jon Peterson below). I've yet to find the discussion of the case where $p$ itself is set by sampling from the uniform. Still looking...

Answer 1

I'm assuming you're asking about the distribution of the hitting time 0 by this random walk. That is, $\tau = \inf\{n\geq 0: X_n = 0\}$. One way to describe the distribution is through the moment generating function $E[e^{\lambda \tau}]$. To calculate this in your model, let $E_{p,a}$ denote expectations for the random walk which steps to the left with probability $p$ and which starts at $X_0 = a \in \mathbb{N}$. As you noted, much is known about these simple random walks. In particular, if I'm remembering correctly, it is known that $E_{p,a}[ e^{\lambda \tau} ] = \left( \frac{1-\sqrt{1-4p(1-p)e^{2\lambda}}}{2(1-p)e^\lambda} \right)^a$ for $\lambda \leq 0$ (the moment generating function is finite for small enough $\lambda>0$ if $p>1/2$ but is infinite for any $\lambda > 0$ if $p \leq 1/2$). Therefore, the moment generating function for $\tau$ in your model is obtained by just averaging this formula over the distribution on $p$ (the uniform distribution on $(0,1)$). Thus, $$E[e^{\lambda \tau}] = \int_0^1 \left( \frac{1-\sqrt{1-4p(1-p)e^{2\lambda}}}{2(1-p)e^\lambda} \right)^a \, dp, \qquad \lambda \leq 0. $$ I'm not sure if this integral can be explicitly calculated or not for $\lambda <0$.