# 2.9 Second-order description of stochastic processes

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## Second-order description

Practical and incomplete statistics

Mean
The mean function of a random process ${X}_{t}$ is defined as the expected value of ${X}_{t}$ for all $t$ 's.
${\mu }_{{X}_{t}}()=({X}_{t})=\begin{cases}\int_{()} \,d x & \text{if }\end{cases}$ x f X t x continuous k x k p X t x k discrete
Autocorrelation
The autocorrelation function of the random process ${X}_{t}$ is defined as
${R}_{X}({t}_{2}, {t}_{1})=({X}_{{t}_{2}}\overline{{X}_{{t}_{1}}})=\begin{cases}\int_{()} \,d {x}_{2} & \text{if }\end{cases}$ x 1 x 2 x 1 f X t 2 X t 1 x 2 x 1 continuous k l x l x k p X t 2 X t 1 x l x k discrete

Fact

If ${X}_{t}$ is second-order stationary, then ${R}_{X}({t}_{2}, {t}_{1})$ only depends on ${t}_{2}-{t}_{1}$ .

${R}_{X}({t}_{2}, {t}_{1})=({X}_{{t}_{2}}\overline{{X}_{{t}_{1}}})=\int_{()} \,d {x}_{1}$ x 2 x 2 x 1 f X t 2 X t 1 x 2 x 1
${R}_{X}({t}_{2}, {t}_{1})=\int_{()} \,d {x}_{1}$ x 2 x 2 x 1 f X t 2 - t 1 X 0 x 2 x 1 R X t 2 t 1 0

If ${R}_{X}({t}_{2}, {t}_{1})$ depends on ${t}_{2}-{t}_{1}$ only, then we will represent the autocorrelation with only one variable $\tau ={t}_{2}-{t}_{1}$

${R}_{X}(\tau )={R}_{X}({t}_{2}-{t}_{1})={R}_{X}({t}_{2}, {t}_{1})$

## Properties

1. ${R}_{X}(0)\ge 0$
2. ${R}_{X}(\tau )=\overline{{R}_{X}(-\tau )}$
3. $\left|{R}_{X}(\tau )\right|\le {R}_{X}(0)$

${X}_{t}=\cos (2\pi {f}_{0}t+\Theta (\omega ))$ and $\Theta$ is uniformly distributed between $0$ and $2\pi$ . The mean function

${\mu }_{X}(t)=({X}_{t})=(\cos (2\pi {f}_{0}t+\Theta ))=\int_{0}^{2\pi } \cos (2\pi {f}_{0}t+\theta )\frac{1}{2\pi }\,d \theta =0$

The autocorrelation function

${R}_{X}(t+\tau , t)=({X}_{t+\tau }\overline{{X}_{t}})=(\cos (2\pi {f}_{0}(t+\tau )+\Theta )\cos (2\pi {f}_{0}t+\Theta ))=1/2(\cos (2\pi {f}_{0}\tau ))+1/2(\cos (2\pi {f}_{0}(2t+\tau )+2\Theta ))=1/2\cos (2\pi {f}_{0}\tau )+1/2\int_{0}^{2\pi } \cos (2\pi {f}_{0}(2t+\tau )+2\theta )\frac{1}{2\pi }\,d \theta =1/2\cos (2\pi {f}_{0}\tau )$
Not a function of $t$ since the second term in the right hand side of the equality in [link] is zero.

Toss a fair coin every $T$ seconds. Since ${X}_{t}$ is a discrete valued random process, the statistical characteristics can be captured by the pmf and the mean function is written as

${\mu }_{X}(t)=({X}_{t})=1/2\times -1+1/2\times 1=0$
${R}_{X}({t}_{2}, {t}_{1})=\sum \sum {x}_{k}{x}_{l}p({X}_{{t}_{2}}, , {X}_{{t}_{1}}, {x}_{k}, {x}_{l})=1\times 1\times 1/2-1\times -1\times 1/2=1$
when $nT\le {t}_{1}< (n+1)T$ and $nT\le {t}_{2}< (n+1)T$
${R}_{X}({t}_{2}, {t}_{1})=1\times 1\times 1/4-1\times -1\times 1/4-1\times 1\times 1/4+1\times -1\times 1/4=0$
when $nT\le {t}_{1}< (n+1)T$ and $mT\le {t}_{2}< (m+1)T$ with $n\neq m$
${R}_{X}({t}_{2}, {t}_{1})=\begin{cases}1 & \text{if (nT\le {t}_{1}< (n+1)T)\land (nT\le {t}_{2}< (n+1)T)}\\ 0 & \text{otherwise}\end{cases}$
A function of ${t}_{1}$ and ${t}_{2}$ .

Wide Sense Stationary
A process is said to be wide sense stationary if ${\mu }_{X}$ is constant and ${R}_{X}({t}_{2}, {t}_{1})$ is only a function of ${t}_{2}-{t}_{1}$ .
Fact

If ${X}_{t}$ is strictly stationary, then it is wide sense stationary. The converse is not necessarily true.

Autocovariance
Autocovariance of a random process is defined as
${C}_{X}({t}_{2}, {t}_{1})=(({X}_{{t}_{2}}-{\mu }_{X}({t}_{2}))\overline{{X}_{{t}_{1}}-{\mu }_{X}({t}_{1})})={R}_{X}({t}_{2}, {t}_{1})-{\mu }_{X}({t}_{2})\overline{{\mu }_{X}({t}_{1})}$

The variance of ${X}_{t}$ is $\mathrm{Var}({X}_{t})={C}_{X}(t, t)$

Two processes defined on one experiment ( [link] ).

Crosscorrelation
The crosscorrelation function of a pair of random processes is defined as
${R}_{XY}({t}_{2}, {t}_{1})=({X}_{{t}_{2}}\overline{{Y}_{{t}_{1}}})=\int_{()} \,d y$ x x y f X t 2 Y t 1 x y
${C}_{XY}({t}_{2}, {t}_{1})={R}_{XY}({t}_{2}, {t}_{1})-{\mu }_{X}({t}_{2})\overline{{\mu }_{Y}({t}_{1})}$
Jointly Wide Sense Stationary
The random processes ${X}_{t}$ and ${Y}_{t}$ are said to be jointly wide sense stationary if ${R}_{XY}({t}_{2}, {t}_{1})$ is a function of ${t}_{2}-{t}_{1}$ only and ${\mu }_{X}(t)$ and ${\mu }_{Y}(t)$ are constant.

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