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 Apr25 comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges. And now, duplicates, near-duplicates, functional duplicates of the question abound. Apr25 comment Random variables set representation in the sample space Apr25 comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges. @Golbez The mystery is solved, please see my previous comment. Apr25 comment Prove $\{s_n\}$ converges if $\{a_n = s_n + 2s_{n+1}\}$ converges. @Umakant You know what? The conditions $a_n = s_n + 2s_{n\color{red}{-1}}$ and $a_n = s_n + 2s_{n\color{red}{+1}}$ are not equivalent. Apr25 comment Random variables set representation in the sample space Obviously. To steal a quote from a famous philosopher of the late 20th century... what else? Apr25 comment Conditional expectation, sigma algebra ?? Each $A_i$ is in $\Sigma_Y$ hence each $\mathbf 1_{A_i}$ is $\Sigma_Y$-measurable hence $Z=E(X\mid Y)$ is $\Sigma_Y$-measurable, an assertion which is strictly equivalent to the assertion that $\Sigma_Z\subseteq\Sigma_Y$. Apr25 comment Random variables set representation in the sample space Well, $P(Y=f(X))=P(A)$ where $A=\{\omega\in\Omega\mid Y(\omega)=f(X(\omega))\}$. Is this your question? Apr25 comment How do I show that \$0