Expectation of Non-Negative Random Variable is Non-Negative/Discrete

Theorem
Let $X$ be a discrete random variable.

Let $\map \Pr {X \ge 0} = 1$.

Then $\expect X \ge 0$, where $\expect X$ denotes the expectation of $X$.

Proof
Let $\map {\operatorname {supp} } X$ be the support of $X$.

Note that since $X$ is discrete, its sample space and hence support is countable.

Therefore, there exists some sequence $\sequence {x_i}_{i \in I}$ such that:


 * $\map {\operatorname {supp} } X = \set {x_i \mid i \in I}$

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 $x \not\in \map {\operatorname {supp} } X$.

We therefore have that any elements of $\map {\operatorname {supp} } X$ 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:


 * $\displaystyle \sum_{i \in I} x_i \map \Pr {X = x_i} \ge 0$

Hence, by the definition of expectation:


 * $\expect X \ge 0$