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 integrable :
:$\ds \sum_{x \in \Img X} \size x \map \Pr {X = x} < \infty$
in which case: 
:$\ds \expect X = \sum_{x \in \Img X} x \map \Pr {X = x}$
== Proof ==
From Characterization of Integrable Functions, we have: 
:$X$ is $\Pr$-integrable $\size X$ is $\Pr$-integrable.
That is, $X$ is integrable :
:$\ds \int \size X \rd \Pr < \infty$ 
=== Lemma ===
From Absolute Value of Real-Valued Random Variable is Real-Valued Random Variable:
:$\size X$ is a random variable. 
Applying 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:
:$\ds \sum_{x \in \Img {\size X} } x \map \Pr {X = -x} = \sum_{x \in \Img {\size X}, \, -x \in \Img X} x \map \Pr {X = -x}$
Suppose that $x \in \Img {\size X}$ and $-x \in \Img X$, we then have: 
:$x \ge 0$ and $-x \in \Img X$ 
as before.
Conversely, suppose that: 
:$x \ge 0$ and $-x \in \Img X$
Then: 
:there exists $\omega \in \Omega$ such that $\map X \omega = -x$.
We then have: 
:$\map {\size X} \omega = \size {-x} = x$ 
So:
:$x \in \Img {\size X}$ 
So:
:$x \in \Img {\size X}$ with $-x \in \Img X$ $-x \in \Img X$ with $x \ge 0$. 
Hence: 
:$\ds \sum_{x \in \Img {\size X} } x \map \Pr {X = -x} = \sum_{-x \in \Img X, \, x \ge 0} x \map \Pr {X = -x}$
We can rewrite this: 
Putting it all together, we get: 
So we have:
:$X$ is $\Pr$-integrable :
::$\ds \sum_{x \in \Img X} \size x \map \Pr {X = x} < \infty$ 
We now show that if: 
:$\ds \int \size X \rd \Pr < \infty$
then:
:$\ds \int X \rd \Pr = \sum_{x \in \Img X} x \map \Pr {X = x}$
From Positive Part of Real-Valued Random Variable is Real-Valued Random Variable and Negative Part of Real-Valued Random Variable is Real-Valued Random Variable, we have: 
:$X^+$ and $X^-$ are real-valued random variables.
We can also see that $X^+ \ge 0$ and $X^- \ge 0$ by the definition of the positive part and negative part.
Applying the lemma, we have that: 
:$\ds \int X^+ \rd \Pr = \sum_{x \in \Img {X^+} } x \map \Pr {X^+ = x}$
and:
:$\ds \int X^- \rd \Pr = \sum_{x \in \Img {X^-} } x \map \Pr {X^- = x}$
Note that $x \ge 0$ has: 
:$x \in \Img {X^+}$
 there exists $\omega \in \Omega$ such that: 
:$\map {X^+} \omega = x$
That is, from the definition of positive part, we have: 
:$\max \set {0, \map X \omega} = x$ 
Since $x \ge 0$, this is equivalent to:
:$\map X \omega = x$ for some $\omega \in \Omega$.
This is equivalent to: 
:$x \in \Img X$
We also have, for $x \ge 0$: 
so:
:$\map \Pr {X^+ = x} = \map \Pr {X = x}$ 
giving: 
:$\ds \sum_{x \in \Img {X^+} } x \map \Pr {X^+ = x} = \sum_{x \in \Img X, \, x \ge 0} x \map \Pr {X = x}$
Similarly, $x \ge 0$ has: 
:$x \in \Img {X^-}$
 there exists $\omega \in \Omega$ such that: 
:$\map {X^-} \omega = x$
That is, from the definition of negative part, we have: 
:$-\min \set {0, \map X \omega} = x$
Since $x \ge 0$, this is equivalent to:
:$\map X \omega = -x$ for some $\omega \in \Omega$.
This is equivalent to: 
:$-x \in \Img X$
We also have, for $x \ge 0$: 
so:
:$\map \Pr {X^- = x} = \map \Pr {X = -x}$ 
giving: 
Then: 
So:
:$\ds \int X \rd \Pr = \sum_{x \in \Img X} x \map \Pr {X = x}$
From the definition of Expectation: General Definition, we therefore have: 
:$\ds \expect X = \sum_{x \in \Img X} x \map \Pr {X = x}$
== Sources ==
* : $10$: Probability: $10.1$: Basics
Category:Discrete Random Variables
Category:Expectation
Category:Expectation of Discrete Random Variable