Expectation of Non-Negative Random Variable is Non-Negative
Theorem
Let $X$ be a random variable.
Let $\map \Pr {X \ge 0} = 1$.
Then $\expect X \ge 0$, where $\expect X$ denotes the expectation of $X$.
Proof
Discrete Random Variable
Let $\map \supp X$ be the support of $X$.
Note that since $X$ is discrete, its sample space and hence support is countable.
Therefore, there exists some sequence $\sequence {x_i}_{i \mathop \in I}$ such that:
- $\map \supp X = \set {x_i \mid i \in I}$
for some $I \subseteq \N$.
By the definition of a sample space, we have:
- $\map \Pr {X = x_i} \ge 0$
for all $i \in I$.
Note that since $\map \Pr {X \ge 0} = 1$, we have $\map \Pr {X < 0} = 0$.
So, for any $x < 0$ we necessarily have $\map \Pr {X = x} = 0$, meaning that $x \notin \map \supp X$.
We therefore have that any elements of $\map \supp X$ are non-negative.
That is:
- $x_i \ge 0$
for all $i \in I$.
Therefore:
- $x_i \map \Pr {X = x_i} \ge 0$
for all $i \in I$.
Summing over all $i \in I$, we have:
- $\ds \sum_{i \mathop \in I} x_i \map \Pr {X = x_i} \ge 0$
Hence, by the definition of expectation:
- $\expect X \ge 0$
$\blacksquare$
Continuous Random Variable
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