What “White Noise” Requires
A time series can look irregular and still have a pattern in its dependence, mean, or variability. The term white noise is not just a visual description: it refers to a process that meets specific moment conditions at every time point. Checking all of them matters. A process with constant mean and variance is not white noise if observations at distinct times have nonzero covariance.
Let \(\{X_t\}\) be a process indexed by time, with finite second moments. Write its mean at time \(t\) as \(E[X_t]\), its variance as \(\operatorname{Var}(X_t)\), and the covariance between times \(t\) and \(s\) as \(\operatorname{Cov}(X_t,X_s)\).
The definition has three checks: the mean must not change with time, the variance must not change with time, and every covariance between distinct time points must be zero. The values of the mean and variance do not have to be zero. Some settings use “white noise” specifically for a zero-mean process; under that convention, subtracting the constant mean from a process satisfying the definition gives zero-mean white noise.
Zero covariance is a statement about linear association, not a general statement that two observations are independent. Independence is a stronger condition and is not required by this moment-based definition. In particular, matching the marginal mean and variance at each time is not enough: the cross-time covariances must also be checked.
A Reliable Check
For a process specified by a formula, first find its mean and variance at a general time \(t\). Then calculate the covariance for two distinct, otherwise arbitrary times \(t\) and \(s\). If any one condition fails, the process is not white noise. A few selected time points can reveal a failure, but a claim that a whole process qualifies requires the conditions to hold for every time point or pair in its index set.
Evaluate \(E[X_t]\). It must equal the same finite constant \(\mu\) for every \(t\).
Evaluate \(\operatorname{Var}(X_t)\). It must equal the same finite constant \(\sigma^2\) for every \(t\).
For \(t\ne s\), evaluate \(\operatorname{Cov}(X_t,X_s)\). Each such covariance must be zero.
Worked Example: A Shifted Bernoulli Process
Let \(B_t\) be independent Bernoulli random variables with \(P(B_t=1)=1/3\) and \(P(B_t=0)=2/3\), and define \(X_t=2+3B_t\) for every integer time \(t\). Determine whether \(\{X_t\}\) is white noise, and find its mean and variance.
Method. Calculate the moments at a general time and then check a covariance for any two distinct times. The stated independence applies to the \(B_t\) variables at distinct times.
Mean. Since \(E[B_t]=1/3\),
This mean does not depend on \(t\). The possible values provide a check: \(X_t=2\) with probability \(2/3\), and \(X_t=5\) with probability \(1/3\). Their mean is \(2(2/3)+5(1/3)=3\).
Variance. A Bernoulli variable with success probability \(p\) has variance \(p(1-p)\). To see this, \(B_t^2=B_t\), so
Also, adding a constant does not change variance, and multiplying a variable by 3 multiplies its variance by \(3^2\). Therefore,
This is constant over time. It can be checked directly from the two possible values: \((2-3)^2(2/3)+(5-3)^2(1/3)=2/3+4/3=2\).
Distinct-time covariance. For \(t\ne s\), the variables \(B_t\) and \(B_s\) are independent, so their covariance is zero. Constants do not affect covariance, and multiplying both variables by 3 multiplies covariance by 9:
All three conditions hold. The process is white noise with mean \(3\) and variance \(2\). No rounding is needed. The calculation also illustrates why a factor of \(3^2\) must be applied to the Bernoulli variance: the variance is \(2\), not \(1\).
Why the Covariance Check Cannot Be Skipped
The mean and variance describe each time point separately. Covariance describes how two time points vary together. A shared random component, for example, can make observations move together even when their means and variances are individually constant. The following process has the same simple marginal distribution at every time but fails the covariance condition.
Worked Example: Constant Marginal Moments but Dependence Across Time
Let \(Z\) take the values \(-1\) and \(1\), each with probability \(1/2\), and define \(X_t=(-1)^tZ\) for integer \(t\). Check the three white noise conditions.
Mean. Since \(E[Z]=(-1)(1/2)+1(1/2)=0\), at every integer \(t\),
Variance. Since \(Z^2=1\) for either possible value, \(\operatorname{Var}(Z)=E[Z^2]-(E[Z])^2=1-0=1\). Multiplication by \((-1)^t\) changes the sign at some times but not the variance:
Covariance. The process uses the same random variable \(Z\) at every time. For distinct times \(t\ne s\),
This is either \(1\) or \(-1\), never zero. For example, at times 0 and 1, \(X_0=Z\) and \(X_1=-Z\), so \(\operatorname{Cov}(X_0,X_1)=-1\). The process has a constant mean and variance but is not white noise.
Common mistake and correction. The incorrect argument is: “Every \(X_t\) has mean zero and variance one, so the process is white noise.” That checks only the marginal conditions. The correction is to calculate a covariance at distinct times; the adjacent-time covariance is \(-1\), which violates the definition.
Constant Mean and Constant Variance Are Separate Conditions
A process may have zero covariance between distinct times and still fail because its mean or variance changes over time. These checks are logically separate. The next example keeps the mean constant and the distinct-time covariances at zero, but varies the variance.
Worked Example: Independent Observations with Changing Variance
Let \(Z_t\) be independent random variables, each taking \(-1\) and \(1\) with probability \(1/2\). For positive integer times \(t\), define \(X_t=\sqrt{t}\,Z_t\). Determine which white noise conditions hold.
Mean. Since \(E[Z_t]=0\),
for every \(t\). The mean is constant. Also, \(Z_t^2=1\), so \(E[Z_t^2]=1\) and \(\operatorname{Var}(Z_t)=1\).
Variance. Scaling by \(\sqrt{t}\) multiplies variance by \(t\):
Thus \(\operatorname{Var}(X_1)=1\) and \(\operatorname{Var}(X_2)=2\); the variance is not constant. This condition fails.
Distinct-time covariance. For \(t\ne s\), independence of \(Z_t\) and \(Z_s\) gives
The mean condition and the zero-covariance condition hold, but the variance condition does not. Therefore, the process is not white noise. Independence alone does not compensate for changing variance.
Variance Zero and the Autocorrelation Qualification
The definition allows \(\sigma^2=0\). For example, if \(X_t=7\) with probability one at every time, its mean is 7, its variance is 0, and its covariance between any distinct times is 0. This is white noise under the stated moment-based definition, although it is a degenerate, constant process.
When \(\sigma^2>0\), the autocorrelation at lag \(h\) is the autocovariance divided by the variance at a time point. The white noise conditions then imply autocorrelation 1 at lag zero and 0 at nonzero lags. If \(\sigma^2=0\), that division is undefined; it is not correct to assign an autocorrelation pattern by dividing by zero.
This distinction matters when interpreting an ACF: the stated zero-at-nonzero-lags pattern applies to the autocorrelation only when the variance is positive. The defining covariance condition itself remains meaningful when the variance is zero.
Common Checks Before Classifying a Series
- Do not stop at the mean. A constant mean says nothing by itself about variance or cross-time covariance.
- Do not stop at the marginal distribution. Identical distributions at each time would not establish zero covariance between distinct times.
- Do not infer zero covariance from a plot that looks irregular. Use the process specification or the relevant covariance calculation.
- Keep the distinct-time condition explicit. The covariance requirement applies when \(t\ne s\); it does not require the variance at a time point to be zero.
- Handle degeneracy carefully. Zero variance is permitted by the definition, but autocorrelation is then undefined.
Check Your Understanding
For each process, check the stated moments and covariances before deciding whether it is white noise.
- Let \(B_t\) be independent Bernoulli variables with success probability \(1/4\), and set \(X_t=4+2B_t\). Find the mean and variance, and classify the process. (Answer: Mean \(4.5\), variance \(0.75\), and white noise.)
- Let \(U\) be equally likely to be \(-2\) or \(2\), and set \(X_t=(-1)^tU\). What is the covariance between times 0 and 1, and is the process white noise? (Answer: The covariance is \(-4\); the process is not white noise.)
- At times 1 and 2, let \(X_1\) and \(X_2\) be independent, each with mean zero, with variances 1 and 4, respectively. Does zero covariance suffice to make this two-time process white noise? (Answer: No; the variance is not constant across the two times.)
- Let \(Z\) be equally likely to be \(-1\) or \(1\), and let \(\varepsilon_t\) be independent equally likely \(-1\) or \(1\) variables, independent of \(Z\). Define \(X_t=Z+\varepsilon_t\). Find the mean, variance, and covariance between distinct times, and classify the process. (Answer: Mean \(0\), variance \(2\), distinct-time covariance \(1\); not white noise.)
- Suppose \(X_t=7\) with probability one for every \(t\). Does it satisfy the stated definition, and is its autocorrelation defined? (Answer: Yes, it satisfies the definition with mean \(7\) and variance \(0\); its autocorrelation is undefined.)