﻿ Expectation of variance derivation

# The expectation of variance derivation

You already know what the random variable is and what the expected value (or expectation) is. If not, please read the Essential background I section.

## 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.

## Expectation of the variance

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 expectation of the body position variance

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