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.

That is, $X$ is $\Pr$-integrable :


 * $\ds \int \size X \rd \Pr < \infty$

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:


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

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:


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

We have, from the Lemma, 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 Case, we therefore have:


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