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If an expected value $E(X_n)$ with $X_n \geq 0$ is monotone in $n$, does this allow one to conclude that $X_n$ is monotone in $n$? Does the reverse hold, i.e. if $X_n \geq 0$ is monotone in $n$ does the same hold for $E(X_n)$?

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To the first question the answer is no. Consider e.g. $[0,1]$ with the Lebesgue measure, $X_{1}=$id and $X\equiv \frac{1}{2}$. Then $E(X_{1})=\frac{1}{2}$ and $E(X_{2})=\frac{1}{2}$ but neither $X_{1}\geq X_{2}$ nor $X_{1}\leq X_{2}$. So you can't really say anything about the behaviour of the sequence $X_{n}$ based on only their integrals.

Second question is yes e.g. due to integral's linearity. If $X_{1}\leq X_{2}$, then $X_{2}-X_{1}\geq 0$. So we have $E(X_{2})-E(X_{1})=E(X_{2}-X_{1})\geq 0$, whence $E(X_{2})\geq E(X_{1})$.

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No to the first (think of cases where $E(X_n)$ is constant), yes to the second ($X - Y \ge 0$ implies $E(X-Y) \ge 0$).

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