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Math 6010 • Linear Models

2004 • Question 10

Define \(X_0=1\). Then iteratively define \(X_n\) so that for all Borel sets \(A\subseteq \mathbb {R}\), \[ \mathbb {P}\left (X_n\in A\mid X_0,\ldots ,X_{n-1}\right ) = \frac {\text {The lebesgue measure of }A\cap [0,X_{n-1}]}{X_{n-1}}. \] Prove that \(\lim _{n\to \infty }2^nX_n=0\) a.s. You may use–without proof–the following version of the Borel–Cantelli lemma [due to Paul Lévy]: If \(\{\mathcal {F}_n\}_{n=1}^{\infty }\) is a filtration and \(A_n\)’s are events such that \(A_n\in \mathcal {F}_n\) and \(\sum _{n=1}^{\infty }\mathbb {P}(A_n\mid \mathcal {F}_{n-1})=\infty \) a.s., then \(\mathbb {P}(A_n\text { infinitely often})=1\).

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Define \(X_0=1\). Then iteratively define \(X_n\) so that for all Borel
sets \(A\subseteq\mathbb{R}\),
\[
  \Prob\left(X_n\in A\mid X_0,\ldots,X_{n-1}\right)
  =
  \frac{\text{The lebesgue measure of }A\cap[0,X_{n-1}]}{X_{n-1}}.
\]
Prove that \(\lim_{n\to\infty}2^nX_n=0\) a.s. You may use--without proof--the
following version of the Borel--Cantelli lemma [due to Paul L\'evy]: If
\(\{\mathcal{F}_n\}_{n=1}^{\infty}\) is a filtration and \(A_n\)'s are events such that
\(A_n\in\mathcal{F}_n\) and
\(\sum_{n=1}^{\infty}\Prob(A_n\mid\mathcal{F}_{n-1})=\infty\) a.s., then
\(\Prob(A_n\text{ infinitely often})=1\).