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I'm trying to show that for an event A with positive probability there is some n bounded by 1/P(A) such that $P(A \cap$ T$^{-n}A) > 0$, where T is a probability-preserving transformation. I'm getting stuck because I'm not seeing what's keeping all the $T^{-k}A$ from clustering around a particular portion of the sample space. Is there anything we can say about $T^{-j}A \cap T^{-k}A$?

I get where n comes from: since $0 < P(A) < 1$, there's some $n$ such that $P(A) < 1/n$. This flows nicely with the relations $1 \ge nP(A) = \sum_{k=1}^n P(A) = \sum_{k=1}^n P(T^{-k}A) \ge P(\bigcup_{k=1}^n T^{-k} A)$. My idea is to assume for each $k \le n, P(A \cap T^{-k} A) = 0$ and derive some contradiction, but if $P(T^{-j}A \cap T^{-k}A)$ is big enough I can't be sure that the orbits cover a sufficient portion of the sample space.

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Choose $n$ such that $P(A)>1/(n+1)$.

Suppose $P(A\cap T^{-m}A)=0$ for $0<m\leq n$. Then for all $j<k$ with $k-j\leq n$, $$P\big(T^{-j}A\cap T^{-k}A\big)=P\big(T^{-j+j}A\cap T^{-k+j}A\big)=P\big(A\cap T^{-(k-j)}A\big)=0,$$ since $T$ is measure preserving. Now $$P\bigg(\bigcup_{m=0}^n T^{-m}A\bigg)=\sum_{m=0}^n P \big(T^{-m}A\big)=(n+1)P(A)>1,$$ since the sets $T^{-m}A$ with $m=0,\ldots,n$ can be made disjoint by changing them on null sets.

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