Expectation of Bounded Random Variable

Theorem

Let $X$ be a random variable.

Let $a$ and $b$ be real numbers with $b \ge a$.

Let:

$\map \Pr {a \le X \le b} = 1$

Then:

$a \le \expect X \le b$

where $\expect X$ denotes the expectation of $X$.

Proof

From:

$\map \Pr {a \le X \le b} = 1$

it follows that:

$\map \Pr {X \ge a} = 1$

That is:

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

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

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

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

$\expect X - a \ge 0$

That is:

$\expect X \ge a$

Note that we also have:

$\map \Pr {X \le b} = 1$

That is:

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

Again applying Expectation of Non-Negative Random Variable is Non-Negative, we have:

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

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

$b - \expect X \ge 0$

giving:

$\expect X \le b$

Putting these inequalities together, we have:

$a \le \expect X \le b$

$\blacksquare$