0
$\begingroup$

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 $C$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.