Let $U\_1,U\_2,\ldots$ be iid random variables distributed uniformly on $[0,1]$. I am interested in the random walk $X\_i = \sum\_{j \leq i} U\_j$. In particular,



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> What is the expected number of points appearing in an interval $[x,x+1]$?
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Experimentally, this seems to converge to $2$.


Here is an intuitive explanation. Suppose that $x+1-\delta$ is the first point within the interval $[x,x+1]$ which the random walk hits. One can calculate that the expected number of points within the interval is $e^\delta$. To estimate the distribution of $\delta$, consider the point $x-\epsilon$ just preceding $x+1-\delta$. Assuming that $\epsilon$ is uniformly distributed (which is close to the truth when $x$ is large), the density of $\delta$ at a point $\delta'$ is proportional to $\Pr[\epsilon \leq \delta'] = \delta'$, and so it is $2\delta'$. Therefore the expected number of points in $[x,x+1]$ is
$$
\int\_0^1 2\delta e^\delta \, \mathrm{d}\delta = 2.
$$



Random walk with positive uniformly distributed steps
Let $X\_t$ be the number of points in $[0,t]$. Then, $X\_t$ is a renewal process. Let $m(t) = E[X\_t]$. Then [renewal theorem](http://www.randomservices.org/random/renewal/LimitTheorems.html) says
$$
 m(t+h) -m(t) \stackrel{t \to \infty}{\longrightarrow} \frac{h}{\mu}
$$
where $\mu = E[U\_1]$ is the expected increment. 


With $\mu = 1/2$ and $h = 1$ in this case, one gets $2$ for the limit. 


