= Solution
{c}
Every $X_k$ is a <Borel measurable function> of $X$, by the formula involving the <floor function>. Thus
$$
\sigma(X_n,X_{n+1},\ldots)\subseteq\sigma(X).
$$
Conversely, part (i) gives the everywhere identity $X=\lim_{k\to\infty}X_k$. For fixed $n$, this is the <pointwise limit> of <random variables> <measurable> with respect to $\sigma(X_n,X_{n+1},\ldots)$. Such a <limit of a sequence> is <measurable>, so $\sigma(X)\subseteq\sigma(X_n,X_{n+1},\ldots)$. Consequently
$$
\boxed{\sigma(X_n,X_{n+1},\ldots)=\sigma(X)\quad(n\geq1).}
$$
These are increasing-precision observations even when the first finitely many are discarded; they do not behave like the <independent random variables> whose <tail sigma-algebra> appears in Question 1(b).
Back to article page
