properties of expected value

1) (normalization) Let $X$ be almost surely constant random variable, i.e. $\Pr\left\{X=c\right\}=1$ ; then $E\left[X\right]=c$ .

3) (monotonicity) Let $X$ , $Y$ be random variables such that $\Pr\left\{X\leq Y\right\}=1$ and $E\left[\left|X\right|\right]<\infty$ , $E\left[\left|Y\right|\right]<\infty$ ; then $E\left[X\right]\leq E\left[Y\right]$ .

Proof.

1) Let’s define

F=\left\{\omega\in\Omega:X\left(\omega\right)=c\right\};

Then by hypothesis

\Pr\left\{\Omega\backslash F\right\}=0

and

\Pr\left\{F\right\}=1.

We have:

$\displaystyle E[X]$	$\displaystyle=$	$\displaystyle\int_{\Omega}X\left(\omega\right)dP$
	$\displaystyle=$	$\displaystyle\int_{\Omega\backslash F}X\left(\omega\right)dP+\int_{F}X\left(% \omega\right)dP$
	$\displaystyle=$	$\displaystyle\int_{F}X\left(\omega\right)dP$
	$\displaystyle=$	$\displaystyle\int_{F}cdP$
	$\displaystyle=$	$\displaystyle c\Pr\left\{F\right\}=c.$

2) [to be done].

3) Let’s define

F=\left\{\omega\in\Omega:X\left(\omega\right)\leq Y\left(\omega\right)\right\};

Then by hypothesis

\Pr\left\{\Omega\backslash F\right\}=0

and

\Pr\left\{F\right\}=1.

We have, keeping in mind property 2),

$\displaystyle E[Y]-E[X]$	$\displaystyle=$	$\displaystyle E[Y-X]$
	$\displaystyle=$	$\displaystyle\int_{\Omega}\left[Y\left(\omega\right)-X\left(\omega\right)% \right]dP$
	$\displaystyle=$	$\displaystyle\int_{\Omega\backslash F}\left[Y\left(\omega\right)-X\left(\omega% \right)\right]dP+\int_{F}\left[Y\left(\omega\right)-X\left(\omega\right)\right% ]dP$
	$\displaystyle=$	$\displaystyle\int_{F}\left[Y\left(\omega\right)-X\left(\omega\right)\right]dP% \geq 0.$

∎

Title	properties of expected value
Canonical name	PropertiesOfExpectedValue
Date of creation	2013-03-22 16:16:05
Last modified on	2013-03-22 16:16:05
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	6
Author	Andrea Ambrosio (7332)
Entry type	Theorem
Classification	msc 60-00