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Divide the interval $[0,1]$ in $n$ subintervals with length $\frac{1}{n}$. The $n$ subintervals are numerated from $1$ to $n$. We have a particle that, after an exponential time of parameter $1$, chooses a site $y\in[0, 1]$ according to the kernel $J:[0,1]\times[0,1]\to\mathbb R^+$, in other words, if the particle is situated in $x$, the probability to choose a site in a region $A\subset[0, 1]$ is given by $\int_AJ(x, y)dy$. Once that the particle chooses the site it jumps on it but with the following restriction: jumps from even subintervals to odd subintervals are not allowed, all the other jumps are allowed.

Let $X_n(t)$ be the position of the particle at time t. I would like to understand which is the distribution of the position of the particle when the parameter $n\to +\infty$.

Has someone any idea to compute such limit distribution?

Thank you

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For a fixed $t>0$, the limit does not need to exist. For example, take $J \equiv 1$, $t = 1$, and the starting position $X(0) = a$ to be in the set of simply normal numbers in base $2$. Then as $n\to \infty$, the probability that a jump occurred by time $t = 1$ is going to oscillate between $e ^{-1}$ and $e^{-\frac 12 t}$. Thus, depending on whether the $n$-th digit of $a$ is $0$ or $1$, the distribution of $X_n(1)$ is going to be either $$ e ^{-1} \delta _a + (1 - e ^{-1}) u_{[0,1]} $$ or $$ e ^{-\frac 12 } \delta _a + (1- e^{-\frac 12 }) u_{[0,1]} $$ where $u_{[0,1]}$ is the uniform distribution on $[0,1]$.

Note also that the set of simply normal numbers is of Lebesgue measure one.

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