The expectation of variance derivation

Expectation rules

The expectation is denoted by the capital letter $E$ .

The expectation of the random variable $E(X)$ equals the mean of the random variable:

$E(X) = \mu_{X}$

Where

$\mu_{X}$ is the mean of the random variable.

Here are some basic expectation rules:

	Rule	Notes
1	$E(X) = \mu_{X} = \Sigma xp(x)$	$p(x)$ is the probability of $x$ (discrete case)
2	$E(a) = a$	$a$ is constant
3	$E(aX) = aE(X)$	$a$ is constant
4	$E(a \pm X) = a \pm E(X)$	$a$ is constant
5	$E(a \pm bX) = a \pm bE(X)$	$a$ and $b$ are constant
6	$E(X \pm Y) = E(X) \pm E(Y)$	$Y$ is another random variable
7	$E(XY) = E(X)E(Y)$	If $X$ and $Y$ are independent

All the rules are quite straightforward and don't need proof.

The expectation of variance is given by:

$V(X) = \sigma_{x}^2 = E(X^2) - \mu_{X}^2$

Where

$V(X)$ is the variance of

$X$

The proof:

	Notes
$V(X) = \sigma_{X}^2 = E((X - \mu_{X})^2)$
$= E(X^2 -2X\mu_{X} + \mu_{X}^2)$
$= E(X^2) - E(2X\mu_{X}) + E(\mu_{X}^2)$	Applied rule number 5: $E(a \pm bX) = a \pm bE(X)$
$= E(X^2) - 2\mu_{X}E(X) + E(\mu_{X}^2)$	Applied rule number 3: $E(aX) = aE(X)$
$= E(X^2) - 2\mu_{X}E(X) + \mu_{X}^2$	Applied rule number 2: $E(a) = a$
$= E(X^2) - 2\mu_{X}\mu_{X} + \mu_{X}^2$	Applied rule number 1: $E(X) = \mu_{X}$
$= E(X^2) - \mu_{X}^2$

The body position displacement variance in terms of time and velocity is given by:

$V(x) = \Delta t^{2} V(v)$ or

$\sigma_{x}^2 = \Delta t^{2} \sigma_{v}^2$

Where:

$x$	is the displacement of the body
$v$	is the velocity of the body
$\Delta(t)$	is the time interval

The proof:

	Notes
$V(x) = \sigma_{x}^2 = E(x^2) - \mu_{x}^2$
$= E((v\Delta t)^2) - (\mu_{v}\Delta t)^2$	Express the body position variance in terms of time and velocity: $x = \Delta tv$
$= E(v^{2}\Delta t^{2}) - \mu_{v}^{2}\Delta t^{2}$
$= \Delta t^{2}E(v^{2}) - \mu_{v}^{2}\Delta t^{2}$	Applied rule number 3: $E(aX) = aE(X)$
$= \Delta t^{2}(E(v^{2}) - \mu_{v}^{2})$
$= \Delta t^{2}V(v)$	Applied expectation of variance rule: $V(X) = E(X^2) - \mu_{X}^2$