# Lesson 52 Autocovariance Function

## Theory

**Definition 52.1 (Autocovariance Function) **The **autocovariance function** \(C_X(s, t)\) of a random process \(\{ X(t) \}\)
is a function of *two* times \(s\) and \(t\). It is sometimes just called the
“covariance function” for short.

It specifies the covariance between the value of the process at time \(s\) and the value at time \(t\). That is, \[\begin{equation} C_X(s, t) \overset{\text{def}}{=} \text{Cov}[X(s), X(t)]. \tag{52.1} \end{equation}\]

For a discrete-time process, we notate the autocovariance function as \[ C_X[m, n] \overset{\text{def}}{=} \text{Cov}[X[m], X[n]]. \]Notice that the variance function can be obtained from the autocovariance function: \[ V(t) = \text{Var}[X(t)] = \text{Cov}[X(t), X(t)] = C(t, t). \]

Let’s calculate the autocovariance function of some random processes. Unfortunately,
the autocovariance function is difficult to visualize, since it is not just a function
of time but a function of *two* times. However, the following video gives some intuition.

**Example 52.1 (Random Amplitude Process) **Consider the random amplitude process
\[\begin{equation}
X(t) = A\cos(2\pi f t)
\tag{50.2}
\end{equation}\]
introduced in Example 48.1.

To be concrete, suppose \(A\) is a \(\text{Binomial}(n=5, p=0.5)\) random variable and \(f = 1\). To calculate the autocovariance function, observe that the only thing that is random in (50.2) is \(A\). Everything else is a constant, so they can be pulled outside the covariance. \[\begin{align*} C_X(s, t) = \text{Cov}[X(s), X(t)] &= \text{Cov}[A\cos(2\pi fs), A\cos(2\pi ft)] \\ &= \underbrace{\text{Cov}[A, A]}_{\text{Var}[A]} \cos(2\pi fs)\cos(2\pi ft) \\ &= 1.25 \cos(2\pi fs)\cos(2\pi ft), \end{align*}\] where in the last step, we used the formula for the variance of a binomial random variable, \(\text{Var}[A] = np(1-p) = 5 \cdot 0.5 \cdot (1 - 0.5) = 1.25\).

As a sanity check, we can derive the variance function from this autocovariance function: \[ V_X(t) = C_X(t, t) = 1.25 \cos(2\pi ft)\cos(2\pi ft) = 1.25 \cos^2(2\pi f t). \] This agrees with what we got in Example 51.1.**Example 52.2 (Poisson Process) **Consider the Poisson process
\(\{ N(t); t \geq 0 \}\) of rate \(\lambda\),
defined in Example 47.1.

To calculate the autocovariance function, we first calculate \(\text{Cov}[N(s), N(t)]\) assuming \(s < t\). We can do this by breaking \(N(t)\) into \(N(s)\), the number of arrivals up to time \(s\), plus \(N(t) - N(s)\), the number of arrivals between time \(s\) and time \(t\). This allows us to use the independent increments property of the Poisson process (see Example 47.1). \[\begin{align*} \text{Cov}[N(s), N(t)] &= \text{Cov}[N(s), N(s) + (N(t) - N(s))] \\ &= \text{Cov}[N(s), N(s)] + \underbrace{\text{Cov}[N(s), N(t) - N(s)]}_{\text{0 by independent increments}} \\ &= \text{Var}[N(s)] \\ &= \lambda s. \end{align*}\]

This is the covariance when \(s < t\). If we repeat the argument for \(s > t\), we will
find that the covariance is \(\lambda t\). In other words, the covariance always involves
the *smaller* of the two times \(s\) and \(t\). Therefore, we can write the autocovariance function as
\[ C_N(s, t) = \lambda \min(s, t). \]

**Example 52.3 (White Noise) **Consider the white noise process \(\{ Z[n] \}\) defined in Example 47.2,
which consists of i.i.d. random variables with variance \(\sigma^2 \overset{\text{def}}{=} \text{Var}[Z[n]]\).

By independence, the covariance between \(Z[m]\) and \(Z[n]\) is zero, unless \(m=n\), in which case \(\text{Cov}[Z[n], Z[n]] = \text{Var}[Z[n]] = \sigma^2\). That is, the autocovariance function is \[ C_Z[m, n] = \begin{cases} \sigma^2 & m = n \\ 0 & m \neq n \end{cases}. \]

We can write the autocovariance function 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 \[ C_Z[m, n] = \sigma^2 \delta[m - n]. \]

As a sanity check, we can derive the variance function from this autocovariance function: \[ V_Z[n] = C_Z[n, n] = \sigma^2 \delta[n - n] = \sigma^2 \delta[0] = \sigma^2. \] This agrees with what we got in Example 51.3.**Example 52.4 (Random Walk) **Consider the random walk process \(\{ X[n]; n\geq 0 \}\) from
Example 47.3.

To calculate the autocovariance function, we first calculate \(\text{Cov}[X[m], X[n]]\) assuming \(m < n\). Since \[ X[n] = Z[1] + Z[2] + \ldots + Z[n], \] we can write this as \[\begin{align*} \text{Cov}[X[m], X[n]] = \text{Cov}[&Z[1] + \ldots + Z[m], \\ &Z[1] + \ldots + Z[m] + \ldots + Z[n]]. \end{align*}\] I find it helpful to write the covariance in two lines, aligning like terms, since the \(Z[i]\)s are independent. When we expand the covariance, it reduces to \[ \text{Var}[Z[1]] + \ldots + \text{Var}[Z[m]] = m \text{Var}[Z[1]]. \]

This is the covariance when \(m < n\). If we repeat the argument for \(m > n\), we will
find that the covariance is \(n \text{Var}[Z[1]]\). In other words, the covariance always involves
the *smaller* of the two times \(m\) and \(n\). Therefore, we can write the autocovariance function as
\[ C_X[m, n] = \min(m, n) \text{Var}[Z[1]]. \]

## Essential Practice

Consider a grain of pollen suspended in water, whose horizontal position can be modeled by Brownian motion \(\{B(t)\}\) with parameter \(\alpha=4 \text{mm}^2/\text{s}\), as in Example 49.1. Calculate the autocovariance function of \(\{ B(t) \}\). Check that this autocovariance function agrees with the variance function you derived in Lesson 51.

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. Calculate the autocovariance function of \(\{ D(t); t \geq 3 \}\). Check that this autocovariance function agrees with the variance function you derived in Lesson 51.

(

*Hint:*Start by calculating \(\text{Cov}[D(s), D(t)]\) when \(s > t\). What happens when \(s > t + 3\)? What happens when \(t < s < t + 3\)? Once you’ve worked this out, it should be straightforward to extend this to the case \(s < t\).)Consider the moving average process \(\{ X[n] \}\) of Example 48.2, 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. Calculate the autocovariance function of \(\{ X[n] \}\). Check that this autocovariance function agrees with the variance function you derived in Lesson 51.

(

*Hint:*Consider the following cases: (1) \(m = n\), (2) \(m = n+1\), (3) \(m = n-1\), (4) \(m > n+1\), (5) \(m < n-1\).)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 \[ X(t) = \cos(2\pi f t + \Theta). \] Calculate the autocovariance function of \(\{ X(t) \}\). Check that this autocovariance function agrees with the variance function you derived in Lesson 51. (Hint: Use the shortcut formula for covariance. Calculate \(E[X(s) X(t)]\) by LOTUS.)