Expectation Preserves Inequality

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Theorem

Let $X$, $Y$ be random variables.

Let $\map \Pr {X \ge Y} = 1$.


Then:

$\expect X \ge \expect Y$


Proof

Note that we have:

$\map \Pr {X - Y \ge 0} = 1$

From Expectation of Non-Negative Random Variable is Non-Negative, we then have:

$\expect {X - Y} \ge 0$

From Sum of Expectations of Independent Trials, we have:

$\expect X + \expect {-Y} \ge 0$

From Expectation of Linear Transformation of Random Variable, we have:

$\expect X - \expect Y \ge 0$

That is:

$\expect X \ge \expect Y$

$\blacksquare$