Expectation of Real-Valued Discrete Random Variable

Theorem
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be a discrete real-valued random variable.

Then $X$ is $\Pr$-integrable :


 * $\ds \sum_{x \in \Img X} \size x \map \Pr {X = x} < \infty$

Further:


 * $\ds \expect X = \sum_{x \in \Img X} x \map \Pr {X = x}$

Proof
From Characterization of Integrable Functions, we have:


 * $X$ is $\Pr$-integrable $\size X$ is $\Pr$-integrable.

Lemma
From the Lemma, we have:


 * $\ds \int \size X \rd \Pr = \sum_{x \in \Img {\size X} } x \map \Pr {\size X = x}$

We aim to show that:


 * $\ds \int \size X \rd \Pr = \sum_{x \in \Img X} \size x \map \Pr {X = x}$

Note that if $x = 0$, we have:


 * $\set {\size X = 0} = \set {X = 0}$

and if $x \ne 0$, we have:


 * $\set {\size X = x} = \set {X = x} \cup \set {X = -x}$

From Probability of Union of Disjoint Events is Sum of Individual Probabilities, we then have:


 * $\map \Pr {\size X = x} = \map \Pr {X = x} + \map \Pr {X = -x}$

We can therefore write:

Note that if $x \in \Img {\size X}$ but $x \not \in \Img X$, then:


 * there exists no $\omega \in \Omega$ such that $\map X \Omega = x$.

That is, for these $x$:


 * $\set {X = x} = \O$

So that, from Empty Set is Null Set:


 * $\map \Pr {X = x} = 0$

So, we have:


 * $\ds \sum_{x \in \Img {\size X} } x \map \Pr {X = x} = \sum_{x \in \Img {\size X} \cap \Img X} x \map \Pr {X = x}$

Note that if $x \in \Img {\size X} \cap \Img X$ then:


 * there exists $\omega \in \Omega$ such that $\map {\size X} \omega = x$

so $x \ge 0$, so $x \in \Img X$ and $x \ge 0$.

Conversely, note that if $x \in \Img X$ has $x \ge 0$, then:


 * there exists $\omega \in \Omega$ such that $\map X \omega = x$

Since we have $x \ge 0$, we also get:


 * $\map {\size X} \omega = x$

so:


 * $x \in \Img X \cap \Img {\size x}$

So:


 * $x \in \Img {\size X} \cap \Img X$ $x \in \Img X$ with $x \ge 0$.

So we have:

We transform the third term similarly.

Note that if $x \in \Img {\size X}$ but $-x \not \in \Img X$, then:


 * there exists no $\omega \in \Omega$ such that $\map X \Omega = -x$.

That is, for these $x$:


 * $\set {X = -x} = \O$

So that, from Empty Set is Null Set, we have:


 * $\map \Pr {\set {X = -x} } = 0$

So:


 * $\ds \sum_{x \in \Img {\size X} } x \map \Pr {X = -x} = \sum_{x \in \Img {\size X}, \, -x \in \Img X} x \map \Pr {X = -x}$