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If the steps are iid uniform as in the title, is the return probability known? Is it positive? Answers, comments, references welcome. Clearly each of these steps is not equivalent to $d$ steps of type $\pm e_i$, especially for large $d$.

Its distribution after $n$ steps is that of $d$ independent uniform $\pm 1$ sums isn't it?

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These random walks are recurrent when $d\le 2$ and transient when $d \ge 3$. That behavior happens for a wide variety of random walks.

The expected number of returns to the origin is $$\sum_{n=1}^\infty \left(\frac{1}{2^{2n}}{2n \choose n}\right)^d.$$

If the expected number of returns is $x \lt \infty$, then the probability of returning is $1-\frac{1}{x+1}$.

Mathematica evaluates the probability of returning for $d=3$ as

$$1-\frac{\Gamma(3/4)^4}{\pi}$$

but for greater $d$ the methods I tried in Mathematica left it as a particular value of a hypergeometric function which restates the series.

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  • $\begingroup$ thanks. Bear with me, I am a bit confused, and didn't expect this to be so simple. If $A_n$ is the event that the walk returns to the origin at or after step $n$ isn't the expected number of returns then $\sum_{n=1}^{\infty} P(A_n)$? However your expression seems to be simply the sum of probabilities that the walk is at zero at each step. $\endgroup$
    – kodlu
    Commented Jan 14, 2016 at 4:15
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    $\begingroup$ @kodlu: If $A_n$ is the event that the walk first returns to the origin at or after step $n$, then $\sum_{n=0}^\infty P(A_n)$ is the expected number of steps before the first return. Note that there might not be a first return. That quantity is not the same as the count of the times that the walk returns to the origin (before wandering off to infinity). $\endgroup$
    – Douglas Zare
    Commented Jan 14, 2016 at 4:28

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