stationary.Rmd

# Stationary Processes {#stationary}

## Motivation {-}

What does it mean for a random process to remain the "same" over time? 
Obviously, the exact values will be different, since 
the process is random. The notion of a "stationary process" makes this notion precise.

## Theory {-}

```{definition stationary, name="Stationary Process"}
A process $\{ X(t) \}$ is called **strict-sense stationary** if the process 
is completely invariant under time shifts. That is, the distribution of 
\[ (X(t_1), X(t_2), ..., X(t_n)) \]
matches the distribution of 
\[ (X(t_1 + \tau), X(t_2 + \tau), ..., X(t_n + \tau)) \]
for any times $t_1, ..., t_n$ and any time shift $\tau$.

However, we will mostly be concerned with **wide-sense stationary** processes, which 
is less restrictive. A random process $\{ X(t) \}$ is wide-sense stationary if its mean 
and autocovariance function are invariant under time shifts. That is:

1. The mean function $\mu_X(t)$ is constant. In this case, we will write the mean function as $\mu_X(t) \equiv \mu_X$.
2. The autocovariance function $C_X(s, t)$ only depends on $s - t$, the difference between the times. 
In this case, we will write the autocovariance function as $C_X(s, t) = C_X(s - t)$.

    (For a discrete-time process, we require the autocovariance function $C_X[m, n]$ to only depend on $m - n$, 
      and we write $C_X[m, n] = C_X[m - n]$.)
```

Let's determine whether some random processes are wide-sense stationary.

```{example random-amplitude-stationary, name="Random Amplitude Process"}
Consider the random amplitude process 
\begin{equation}
X(t) = A\cos(2\pi  f t)
(\#eq:random-amplitude)
\end{equation}
introduced in Example \@ref(exm:random-amplitude). 

In Example \@ref(exm:random-amplitude-mean), we 
showed that its mean function is 
\[ \mu_X(t) = 2.5 \cos(2\pi f t). \]
This is not constant in $t$, so this process cannot be wide-sense stationary. We 
do not even need to check the autocovariance function.

If we look at a graph of the process, it is clearly not stationary. If we shift 
the process by any amount that is less than a full period, then the process looks 
different. Shown on the graph below are 20 realizations of the original process 
$\{ X(t) \}$ in blue, as well as 20 realizations of the 
time-shifted process $\{ X(t - 0.3) \}$. The peaks and troughs are in different places.
```

```{r, echo=FALSE, fig.show = "hold", fig.align = "default"}
plot(NA, NA,
     xaxt="n", yaxt="n", bty="n",
     xlim=c(-2, 3), ylim=c(-5.1, 5.1),
     xlab="Time (t)", ylab="X(t)")
axis(1, pos=0)
axis(2, pos=0)
set.seed(1)
ts <- seq(-2, 3, by=0.01)
a <- rbinom(30, 5, 0.5)
for(i in 1:20) {
  lines(ts, a[i] * cos(2 * pi * ts), col=rgb(0, 0, 1, 0.2))
  lines(ts, a[i] * cos(2 * pi * (ts - 0.3)), col=rgb(1, 0, 0, 0.2))
}
```


```{example poisson-process-stationary, name="Poisson Process"}
Consider the Poisson process 
$\{ N(t); t \geq 0 \}$ of rate $\lambda$, 
defined in Example \@ref(exm:poisson-process-as-process).

In Example \@ref(exm:poisson-process-mean), we 
showed that its mean function is 
\[ \mu_N(t) = \lambda t. \]
This is not constant in $t$, so we know that this process cannot be wide-sense stationary. We 
do not even need to check the autocovariance function.

Again, the fact that the Poisson process is not stationary is immediately apparent from a graph 
of the process. Shown below are 20 realizations of the original process $\{ N(t) \}$ in blue, as well as 
20 realizations of the time-shifted process $\{ N(t - 1) \}$ in red. The time-shifted process may not 
even start at 0 at time 0!
```

```{r, echo=FALSE, fig.show = "hold", fig.align = "default"}
plot(NA, NA,
     xaxt="n", yaxt="n", bty="n",
     xlim=c(0, 5), ylim=c(0, 8.5),
     xlab="Time (t)", ylab="Number of arrivals N(t)")
axis(1, pos=0, at=0:5)
axis(2, pos=0)
set.seed(1)
for(i in 1:20) {
  arrivals <- cumsum(rexp(10, 0.8))
  ts <- seq(0, 6, by=.01)
  N <- c()
  for(t in ts) {
    N <- c(N, sum(arrivals < t))
  }
  lines(ts, N, col=rgb(0, 0, 1, 0.3))
  lines(ts - 1, N, col=rgb(1, 0, 0, 0.3))
}
```


```{example white-noise-stationary, name="White Noise"}
Consider the white noise process $\{ Z[n] \}$ defined in Example \@ref(exm:white-noise), 
which consists of i.i.d. random variables with mean $\mu = E[Z[n]]$ and 
variance $\sigma^2 \overset{\text{def}}{=} \text{Var}[Z[n]]$. 

In Example \@ref(exm:white-noise-mean), we showed that its mean function is 
\[ \mu_Z[n] \equiv \mu, \]
which is constant, so the process could be wide-sense stationary. We need to check its 
autocovariance function.

In Example \@ref(exm:white-noise-cov), we showed that its autocovariance function is 
\[ C_Z[m, n] = \begin{cases} \sigma^2 & m = n \\ 0 & m \neq n \end{cases}. \]

To see that this is a function of $m - n$, we rewrite $m = n$ as $m - n = 0$:
  \[ C_Z[m, n] = \begin{cases} \sigma^2 & m - n = 0 \\ 0 & m - n \neq 0 \end{cases}. \]

This can be written more compactly using the discrete delta function $\delta[k]$, defined as 
\[ \delta[k] \overset{def}{=} \begin{cases} 1 & k=0 \\ 0 & k \neq 0 \end{cases}. \]
In terms of $\delta[k]$, the autocovariance function is simply 
\[ C_Z[m, n] = \sigma^2 \delta[m - n]. \]

The graph below shows white noise $\{ Z[n] \}$ in blue and time-shifted white noise $\{ Z[n-1] \}$ in 
red. Visually, it is hard to tell the difference between the two, which is intutively why the 
process is stationary.
```

```{r, echo=FALSE, fig.show = "hold", fig.align = "default"}
plot(NA, NA,
     xaxt="n", yaxt="n", bty="n",
     xlim=c(-2, 3), ylim=c(-3, 3),
     xlab="Time Sample (n)", ylab="Z[n]")
axis(1, pos=0, at=-2:3)
axis(2, pos=0)
set.seed(1)
for(i in 1:20) {
  ts <- -3:4
  Z <- rnorm(8)
  points(ts, Z, pch=19, col=rgb(0, 0, 1, 0.3))
  lines(ts, Z, col=rgb(0, 0, 1, 0.3))
  points(ts - 1, Z, pch=19, col=rgb(1, 0, 0, 0.3))
  lines(ts - 1, Z, col=rgb(1, 0, 0, 0.3))
}
points(-2:3, rep(0, 6), col="red", pch=19)
```

```{example random-walk-stationary, name="Random Walk"}
Consider the random walk process $\{ X[n]; n\geq 0 \}$ from 
Example \@ref(exm:random-walk-process). 

In Example \@ref(exm:random-walk-mean), we calculated the mean function to be
\[ \mu[n] = 0. \]
This is constant, so the process might be wide-sense stationary. But 
we also need to check the autocovariance function.

In Example \@ref(exm:random-walk-cov), we calculated the autocovariance function to be 
\[ C[m, n] = \min(m, n) \text{Var}[Z[1]]. \]
This is _not_ just a function of $m - n$, so the process is not wide-sense stationary.

You might ask, "How can we be sure that there isn't some way to manipulate $\min(m, n)$ so 
that it is a function of $m-n$?" We can prove that it is impossible by testing a few pairs of 
values. For example, $(m, n) = (3, 2)$ and $(m, n) = (6, 5)$ are two pairs of values that are both 
separated in time by $m - n = 1$ samples. If it were possible to reduce $C[m, n]$ to a function 
$C[m - n]$ of the difference only, then $C[3, 2]$ would have to equal $C[6, 5]$. 

However, they are not equal: $C[3, 2] = 2 \text{Var}[Z[1]]$, while $C[6, 5] = 5 \text{Var}[Z[1]]$.
```

```{example gaussian-process}
Let $\{ X(t) \}$ be a continuous-time random process with 

- $E[X(t)] = 2$ for all $t$, and
- $\text{Cov}[X(s), X(t)] = 5e^{-3(s - t)^2}$ for all $s$ and $t$.

The mean function $\mu_X(t) = E[X(t)]$ is constant and 
the autocovariance function $C_X(s, t) = 5e^{-3(s - t)^2}$ 
is a function of $\tau = s - t$ only, so the process 
is stationary. 

Since the process is stationary, we can 
write the mean and autocovariance functions as 
\begin{align*} 
\mu_X &= 2 & C_X(\tau) &= 5e^{-3\tau^2}.
\end{align*}
```


## Essential Practice {-}

For these practice questions, you may want to refer to the mean and autocovariance functions you 
calculated in Lessons \@ref(mean-function) and \@ref(cov-function).

1. Consider a grain of pollen suspended in water, whose horizontal position can be 
modeled by Brownian motion $\{B(t); t \geq 0\}$ with parameter 
$\alpha=4 \text{mm}^2/\text{s}$, as in Example \@ref(exm:pollen).
Is $\{ B(t); t \geq 0 \}$ wide-sense stationary?

2. Radioactive particles hit a Geiger counter according to a Poisson process 
at a rate of $\lambda=0.8$ particles per second. Let $\{ N(t); t \geq 0 \}$ represent this Poisson process.

    Define the new process $\{ D(t); t \geq 3 \}$ by 
    \[ D(t) = N(t) - N(t - 3). \]
    This process represents the number of particles that hit the Geiger counter in the last 3 seconds. 
    Is $\{ D(t); t \geq 3 \}$ wide-sense stationary?
 
3. Consider the moving average process $\{ X[n] \}$ of Example \@ref(exm:ma1), defined by 
\[ X[n] = 0.5 Z[n] + 0.5 Z[n-1], \]
where $\{ Z[n] \}$ is a sequence of i.i.d. standard normal random variables. 
Is $\{ X[n] \}$ wide-sense stationary?

    (_Hint:_ You can write $m = n + 1$ as $m - n = 1$.)
    
4. Let $\Theta$ be a $\text{Uniform}(a=-\pi, b=\pi)$ random variable, and let $f$ be a constant. 
Define the random phase process $\{ X(t) \}$ by
\[ X(t) = \cos(2\pi f t + \Theta). \]
Is $\{ X(t) \}$ wide-sense stationary?