1

Fix $n ≥ 4$. Suppose there is a particle that moves randomly on the number line, but never leaves the set $\{1, 2, . . . , n\}$. The initial probability distribution of the particle is $π$ i.e., the probability that particle is in location $i$ is given by $π(i)$. In the first step, if the particle is at position $i$, it moves to one of the positions in $\{1, 2, . . . , i\}$ with uniform distribution; in the second step, if the particle is in location $j$, then it moves to one of the locations in $\{j, j + 1, . . . , n\}$ with uniform distribution. Suppose after two steps, the final distribution of the particle is uniform. What is the initial distribution $π$?

The answer given is $π(1) = 1$ and $π(i) = 0$ for $i=\{2, 3,...,n \}$. I've no idea how to derive this, please help me with a hint.

Tapi
  • 1,688
  • Hi, what have you tried ? – EDX Oct 17 '22 at 11:29
  • @EDX I don't understand how to approach this. I feel like there's no information regarding the initial distribution- need help with getting started. – Tapi Oct 17 '22 at 11:32
  • What means that after two steps the final distribution of the particule is uniform ? Because uniform distribution is here a dynamic idea, for switching from one step to another – EDX Oct 17 '22 at 11:35
  • What means that after two steps the final distribution of the particule is uniform ? Because uniform distribution is here a dynamic idea, for switching from one step to another. Actually if we're talking about the potential uniform distribution over ${1,2,..,i}$ for step three, this becomes very simple : indeed if is said "uniform" after step two which means will take values in ${1,2,..,n}$, that means that it needs to have $\pi(1)=1$ for the first step to choose in ${j=1,2,..n}$ for the second step. Noting that the only what to get the former set for step two is to have $\pi(1)=1$. – EDX Oct 17 '22 at 11:42

0 Answers0