Expectation of Real-Valued Discrete Random Variable

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then $X$ is integrable if and only if:

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

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 if and only if $\size X$ is $\Pr$-integrable.

That is, $X$ is integrable if and only if:

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

Lemma

Let $X$ be a discrete real-valued random variable such that:

$\map X \omega \ge 0$ for all $\omega \in \Omega$.


Then:

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

$\Box$


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:

\(\ds \map \Pr {X = 0} + \sum_{x \in \Img {\size X} } x \map \Pr {\size X = x}\) \(=\) \(\ds 0 \times \map \Pr {X = 0} + \sum_{x \in \Img {\size X} } x \paren {\map \Pr {X = x} + \map \Pr {X = -x} }\)
\(\ds \) \(=\) \(\ds 0 \times \map \Pr {X = 0} + \sum_{x \in \Img {\size X} } x \map \Pr {X = x} + \sum_{x \in \Img {\size X} } x \map \Pr {X = -x}\)

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:

$x \in \Img {\size X} \cap \Img X$ if and only if $x \in \Img X$ with $x \ge 0$.

So we have:

\(\ds \sum_{x \in \Img {\size X} \cap \Img X} x \map \Pr {X = x}\) \(=\) \(\ds \sum_{x \in \Img X, \, x \ge 0} x \map \Pr {X = x}\)
\(\ds \) \(=\) \(\ds \sum_{x \in \Img X, \, x > 0} \size x \map \Pr {X = x}\)


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 $-x \in \Img X$ if and only if $-x \in \Img X$ 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:

\(\ds \sum_{-x \in \Img X, \, x \ge 0} x \map \Pr {X = -x}\) \(=\) \(\ds \sum_{y \in \Img X, \, y \le 0} \paren {-y} \map \Pr {X = y}\)
\(\ds \) \(=\) \(\ds \sum_{y \in \Img X, \, y < 0} \size y \map \Pr {X = y}\)


Putting it all together, we get:

\(\ds \int \size X \rd \Pr\) \(=\) \(\ds 0 \times \map \Pr {X = 0} + \sum_{x \in \Img X, \, x > 0} \size x \map \Pr {X = x} + \sum_{x \in \Img X, \, x < 0} \size x \map \Pr {X = x}\)
\(\ds \) \(=\) \(\ds \sum_{x \in \Img X} \size x \map \Pr {X = x}\)

So we have:

$X$ is $\Pr$-integrable if and only if:
$\ds \sum_{x \in \Img X} \size x \map \Pr {X = x} < \infty$

$\Box$


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:

$x \in \Img {X^+}$

if and only if 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 $x \ge 0$:

\(\ds \set {\omega \in \Omega : \map {X^+} \omega = x}\) \(=\) \(\ds \set {\omega \in \Omega : \max \set {0, \map X \omega} = x}\) Definition of Positive Part
\(\ds \) \(=\) \(\ds \set {\omega \in \Omega : \map X \omega = x}\)

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:

$x \in \Img {X^-}$

if and only if 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 $x \ge 0$:

\(\ds \set {\omega \in \Omega : \map {X^-} \omega = x}\) \(=\) \(\ds \set {\omega \in \Omega : -\min \set {0, \map X \omega} = x}\) Definition of Negative Part
\(\ds \) \(=\) \(\ds \set {\omega \in \Omega : \map X \omega = -x}\)

so:

$\map \Pr {X^- = x} = \map \Pr {X = -x}$

giving:

\(\ds \sum_{x \in \Img {X^-} } x \map \Pr {X^- = x}\) \(=\) \(\ds \sum_{-x \in \Img X, \, x \ge 0} x \map \Pr {X = -x}\)
\(\ds \) \(=\) \(\ds -\sum_{-x \in \Img X, \, x \ge 0} \paren {-x} \map \Pr {X = -x}\)
\(\ds \) \(=\) \(\ds -\sum_{y \in \Img X, \, y \le 0} y \map \Pr {Y = y}\)

Then:

\(\ds \int X \rd \Pr\) \(=\) \(\ds \int X^+ \rd \Pr - \int X^- \rd \Pr\) Definition of Integral of Measure-Integrable Function
\(\ds \) \(=\) \(\ds \sum_{x \in \Img X, \, x \ge 0} x \map \Pr {X = x} + \sum_{x \in \Img X, \, x \le 0} x \map \Pr {X = x}\)
\(\ds \) \(=\) \(\ds \sum_{x \in \Img X, \, x \ge 0} x \map \Pr {X = x} + \sum_{x \in \Img X, \, x < 0} x \map \Pr {X = x}\) since $0 \times \map \Pr {X = 0} = 0$
\(\ds \) \(=\) \(\ds \sum_{x \in \Img X} x \map \Pr {X = x}\)

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:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Expectation_of_Real-Valued_Discrete_Random_Variable&oldid=744959"