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


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.


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.