The Mean Of A Time Series

Introduction

A time series is one observed sequence from a process that could have produced many possible sequences. At time $t$, the process has a probability distribution across those possible outcomes. Its mean at time $t$ describes the center of that distribution; it is not, in general, the average of the values recorded at different times in one sequence.

Keeping these quantities separate matters when interpreting trends, counts, or measurements collected over time. A process can have a different mean at each time point, and a single observed series can sit above or below those means. This tutorial defines the time-specific population mean, calculates it for several simple processes, and shows how it relates to the average of a finite observed series.

Definition: A time series process is a collection of random variables $\{X_t\}$, indexed by time $t$. If $X_t$ has a finite expectation, its population mean at time $t$ is $$ \mu_t=E[X_t]. $$ The function assigning $\mu_t$ to each time point is the mean function. It may change with time. A realized series is a set of observed values $x_1,x_2,\ldots,x_T$, where each $x_t$ is one realization of $X_t$.

Population Mean and Observed-Series Average

The population mean $\mu_t$ is defined at a particular time point by considering the distribution of $X_t$. In principle, it is the average value that would be obtained across repeated realizations of the process at that same time. It does not require that repeated realizations actually be available in the data set.

For one observed series of length $T$, the observed-series average, also called its temporal average, is

\bar{x}_T=\frac{1}{T}\sum_{t=1}^{T}x_t.

These quantities answer different questions. The value $\mu_t$ concerns the distribution at one fixed time. The value $\bar{x}_T$ summarizes one recorded path across several times. If the mean changes with time, the observed-series average is not intended to represent every $\mu_t$. Even if the mean is constant, one observed average need not equal that mean.

There is also a population counterpart to the observed-series average. For the random variables $X_1,\ldots,X_T$, define their average as $T^{-1}\sum_{t=1}^{T}X_t$. Its expectation is the average of the time-specific means:

E\left[\frac{1}{T}\sum_{t=1}^{T}X_t\right] =\frac{1}{T}\sum_{t=1}^{T}E[X_t] =\frac{1}{T}\sum_{t=1}^{T}\mu_t.

This statement concerns an expectation, not a guarantee that an observed series average equals the average of its population means. The realized values can differ from their means. Importantly, the equality does not require the $X_t$ values to be independent.

Why Expectations Add

The key calculation is the linearity of expectation. This property explains why the expected average over several time points is the average of the time-specific means, without needing an independence assumption.

Result: If $A$ and $B$ are random variables with finite absolute expectations, so that $E[|A|]$ and $E[|B|]$ are finite, then $$ E[A+B]=E[A]+E[B]. $$ More generally, for a finite collection of integrable random variables and constants $c_t$, $$ E\left[\sum_{t=1}^{T}c_tX_t\right] =\sum_{t=1}^{T}c_tE[X_t]. $$

For a general probability distribution, the justification uses integration with respect to that distribution. If $A$ and $B$ are integrable, then

E[A+B] =\int (a+b)\,dP =\int a\,dP+\int b\,dP =E[A]+E[B].

The integrability condition ensures the expectations in this calculation are finite. In the special case where $A$ and $B$ are discrete, the same result can be written as a sum over possible pairs of values:

E[A+B] =\sum_a\sum_b(a+b)P(A=a,B=b) =E[A]+E[B].

That summation is a discrete-case expression; the integral argument is the general justification and also covers continuous distributions. To extend the result to a finite series, apply the two-variable result repeatedly and use the fact that a constant can be taken outside an expectation. Dividing by $T$ then gives the expected-average formula above. No step requires independence.

Worked Calculations

Worked Example: A Linear Mean Function with Centered Noise

A hypothetical monitoring process is modeled for $t=1,2,3,4$ by $X_t=12+1.5t+\varepsilon_t$. At each time, $\varepsilon_t$ equals $2$ or $-2$, each with probability $0.5$; the errors at different times are independent. Find the mean at each time, the average of those means, and the temporal average for the observed error sequence $2,-2,2,2$.

Method. First calculate the expectation of the error. Its two possible values give

E[\varepsilon_t]=(2)(0.5)+(-2)(0.5)=1-1=0.

Therefore, at each time,

\mu_t=E[X_t]=12+1.5t+E[\varepsilon_t]=12+1.5t.

Substituting each time index gives $\mu_1=12+1.5=13.5$, $\mu_2=12+3=15$, $\mu_3=12+4.5=16.5$, and $\mu_4=12+6=18$. The average of these four population means is

\frac{13.5+15+16.5+18}{4} =\frac{63}{4}=15.75.

For the specified observed errors, the recorded values are $x_1=13.5+2=15.5$, $x_2=15-2=13$, $x_3=16.5+2=18.5$, and $x_4=18+2=20$. Thus the observed-series average is

\bar{x}_4=\frac{15.5+13+18.5+20}{4} =\frac{67}{4}=16.75.

The calculation checks out: the observed average exceeds the average of the population means by $16.75-15.75=1$, which is also the average of the four realized errors, $(2-2+2+2)/4=1$. Common mistake avoided: calling $16.75$ the population mean at every time. The population means vary by time; $16.75$ summarizes this particular observed sequence.

Worked Example: Time-Varying Bernoulli Means

In a hypothetical three-period study, $X_t$ records whether a specified event occurs, with $X_t=1$ for occurrence and $X_t=0$ otherwise. The variables are independent, with $X_1\sim\operatorname{Bernoulli}(0.2)$, $X_2\sim\operatorname{Bernoulli}(0.5)$, and $X_3\sim\operatorname{Bernoulli}(0.8)$. Find each population mean and the expected average across the three periods. If the observed values are $0,1,1$, calculate their temporal average and compare.

Method. For a Bernoulli random variable with probability $p$ of value $1$, its expectation is $1p+0(1-p)=p$. Hence $\mu_1=0.2$, $\mu_2=0.5$, and $\mu_3=0.8$. The expected average of the three random variables is

E\left[\frac{X_1+X_2+X_3}{3}\right] =\frac{0.2+0.5+0.8}{3} =\frac{1.5}{3}=0.5.

For the observed sequence, the temporal average is $(0+1+1)/3=2/3=0.6667$, rounded to four decimal places. This differs from the expected average $0.5$; the expected average describes the process across possible outcomes, while $0.6667$ describes this one observed sequence. In this example the means are time-varying, so there is no single time-specific mean that equals all three values. Common mistake avoided: interpreting the observed fraction $2/3$ as the mean at each of the three times, despite the different Bernoulli parameters.

Worked Example: A Shared Continuous Random Quantity

Suppose a hypothetical process is $X_t=2+t\Theta$ for $t=1,2,3$, where the same random quantity $\Theta$ is used at all three times and has a uniform distribution on the interval from $0$ to $4$. Find the time-specific means and their average. Then calculate the observed-series average if the realized value of $\Theta$ is $1.5$.

Method. A uniform random variable on $[0,4]$ has expectation equal to the midpoint of the interval:

E[\Theta]=\frac{0+4}{2}=2.

Using linearity, $\mu_t=E[2+t\Theta]=2+tE[\Theta]=2+2t$. Thus $\mu_1=4$, $\mu_2=6$, and $\mu_3=8$. The average of the three means is $(4+6+8)/3=18/3=6$.

For the realized value $\Theta=1.5$, the observed values are $x_1=2+1.5=3.5$, $x_2=2+3=5$, and $x_3=2+4.5=6.5$. Their average is $(3.5+5+6.5)/3=15/3=5$. The observed average is therefore $1$ below the average of the population means. This difference also follows from the realized value of $\Theta$: its departure from its expectation is $1.5-2=-0.5$, and the average coefficient of $\Theta$ over the three times is $(1+2+3)/3=2$, giving $2(-0.5)=-1$. Common mistake avoided: requiring independence to calculate the expected average. Here the same $\Theta$ enters every time point, but linearity still applies.

Interpreting the Mean Function

The examples illustrate three useful points. First, a mean function can change with time: the formula $\mu_t=E[X_t]$ does not assume a constant level. Second, the distributional description at each time determines the corresponding expectation. Third, a single observed sequence provides values $x_t$, not direct access to every population mean $\mu_t$.

If all time-specific means are equal to a common value $\mu$, then the expected average of $X_1,\ldots,X_T$ is also $\mu$:

\mu_t=\mu\text{ for every }t \quad\Longrightarrow\quad E\left[\frac{1}{T}\sum_{t=1}^{T}X_t\right] =\frac{1}{T}\sum_{t=1}^{T}\mu =\mu.

This conclusion is about the expectation of the random average. It does not say that every realized average equals $\mu$. Nor does a constant mean by itself say that a particular observed average will be close to $\mu$; that question involves additional properties of the process. Conversely, when the means vary, the expected average over the chosen time points is their arithmetic average, not the mean at an arbitrary single time.

Key takeaway: At a fixed time $t$, the population mean is $\mu_t=E[X_t]$. For a finite set of times, the expected average is the average of the $\mu_t$ values. An observed-series average is calculated from one realized path and need not equal either a particular $\mu_t$ or the average of the population means.

Check Your Understanding

For each problem, distinguish the time-specific population means from any average calculated from an observed series.

A process is $X_t=7+2t+\varepsilon_t$ for $t=1,2,3$, where the independent errors each equal $-1$ or $1$ with probability $0.5$. Find the three time-specific means and their average. (Answer: $\mu_1=9,\ \mu_2=11,\ \mu_3=13$; average of means $=11$.)
For three periods, $X_t$ is Bernoulli with respective success probabilities $0.1,\ 0.4,\ 0.7$. Find the expected average of the three variables. (Answer: $(0.1+0.4+0.7)/3=0.4000$.)
A four-period observed series is $3,6,4,7$, and the process has $\mu_t=4+t$ for $t=1,2,3,4$. Calculate the observed-series average and the average of the time-specific means. (Answer: observed average $=5$; mean values $5,6,7,8$, average of means $=6.5$.)
Let $X_t=1+tU$ for $t=1,2$, where $U$ is uniform on $[2,6]$ and is shared across both times. Find $\mu_1,\mu_2$, and the expected average of $X_1,X_2$. (Answer: $E[U]=4$, so $\mu_1=5,\ \mu_2=9$, and expected average $=7$.)
Three integrable variables have time-specific means $2,-1,5$. No independence assumption is given. Find the expected average of the variables. (Answer: $(2-1+5)/3=2$; independence is not required.)

What You'll Learn