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Consider the following sort of random walk. The position of the walker at time $t$ is represented by the random variable $r(t)$, with $r(0) = 0$. The variable satisfies the following equation, $$ r(t+1) = r(t) + \xi(\, r(t), t \, ), $$ where $\xi(\, r(t), t \, )$ is an increment with two possible outcomes, $0$ and $k$, where $k$ is some positive integer. If $r(t) > t$, the increments are distributed according to, $$ \mathbb{P}(\xi = k ) = \frac{C}{r(t) - t}, $$ and $\mathbb{P}(\xi = 0 ) = 1 - \mathbb{P}(\xi = k )$, where $C$ is a positive constant less than 1. If for some $t$ it happens that $r(t) = t$, then take $r(t+1) = r(t) + k$ with probability $1$. The case $r(t) < t$ cannot occur given the previous definitions.

How does the expectation $\mathbb{E}[r(t)]$ grow with $t$? In particular I would like to know if $\mathbb{E}[r(t) - t]$ grows to infinity with $t$ for some values of $k$ and $C$.

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Just define $a(t) = r(t) - t$ and define the new process, $$ a(t+1) = a(t) + \xi - 1 = a(t)+ \xi^{\prime},$$ where $\xi^{\prime} = k-1$ with probability $C\, /\, a(t)$ or $-1$ otherwise. Observe that the expected increment is $0$ for a constant value $a^{\star}$. This means that $a(t)$ will not go to infinite with positive probability.

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