Variance And Autocovariance Of AR(1) Processes

Why Variance and Autocovariance Matter

An AR(1) model describes a series whose current value depends on its immediately preceding value and a new random innovation. Knowing the mean alone does not describe how variable the series is or how its values at different times move together. Those features are summarized by its stationary variance and lagged autocovariances.

For risk modeling, these quantities help distinguish a series with large one-period shocks from one that retains substantial variation over time. The autoregressive coefficient affects both: it determines how strongly past deviations persist, while the innovation variance sets the scale of new shocks.

Definition: For a weakly stationary series with mean $\mu$, define its autocovariance at lag $k$ by $\gamma_k=\operatorname{Cov}(X_t,X_{t-k})$. Weak stationarity requires a constant finite mean, a finite variance that does not depend on $t$, and covariances that depend only on the time separation $k$. In particular, $\gamma_0=\operatorname{Var}(X_t)$.

Model and Assumptions

Write an AR(1) process in centered form as

X_t-\mu=\phi(X_{t-1}-\mu)+\varepsilon_t,

where $\mu=E[X_t]$, $\phi$ is the autoregressive coefficient, and $\varepsilon_t$ is the innovation at time $t$. The innovations have mean zero and variance $\sigma_\varepsilon^2$, are uncorrelated across time, and are uncorrelated with past values of the series. For the usual causal stationary AR(1) model, the stationarity condition is $|\phi|<1$. If innovations are independent normal random variables, the process is Gaussian; normality is not required for the variance and autocovariance results below.

The intercept form $X_t=c+\phi X_{t-1}+\varepsilon_t$ has stationary mean $\mu=c/(1-\phi)$ when $\phi\ne 1$ and the stationary model conditions hold. Centering that equation gives the form used here. The results in this tutorial concern the stationary process, not an arbitrary starting value during a nonstationary or transient phase.

Deriving the Stationary Variance and Autocovariances

Let $Y_t=X_t-\mu$. Then $Y_t=\phi Y_{t-1}+\varepsilon_t$, and $E[Y_t]=0$. Because the innovation is uncorrelated with the past, $\operatorname{Cov}(Y_{t-1},\varepsilon_t)=0$. Taking the variance of both sides gives

\begin{aligned} \operatorname{Var}(Y_t) &=\operatorname{Var}(\phi Y_{t-1}+\varepsilon_t)\\ &=\phi^2\operatorname{Var}(Y_{t-1}) +\operatorname{Var}(\varepsilon_t) +2\phi\operatorname{Cov}(Y_{t-1},\varepsilon_t)\\ &=\phi^2\operatorname{Var}(Y_{t-1})+\sigma_\varepsilon^2. \end{aligned}

Under stationarity, $\operatorname{Var}(Y_t)=\operatorname{Var}(Y_{t-1})=\gamma_0$. Solving the resulting equation yields the stationary variance. For $k\geq 1$, take the covariance of the centered model with $Y_{t-k}$. The current innovation is uncorrelated with that past value, so the covariance at lag $k$ is $\phi$ times the covariance at lag $k-1$. Repeating the relation gives the autocovariance at every nonnegative lag.

Result: For a stationary AR(1) process with $|\phi|<1$, innovation variance $\sigma_\varepsilon^2$, and autocovariance $\gamma_k=\operatorname{Cov}(X_t,X_{t-k})$, $$ \gamma_0=\operatorname{Var}(X_t)=\frac{\sigma_\varepsilon^2}{1-\phi^2}, \qquad \gamma_k=\phi^k\gamma_0=\frac{\phi^k\sigma_\varepsilon^2}{1-\phi^2}\quad(k\geq 0). $$ For negative lags, $\gamma_{-k}=\gamma_k$. The formula for $\gamma_0$ requires $|\phi|<1$; it is not a finite stationary variance formula for $|\phi|\geq1$.

Check the model conditions.
Confirm that the series is being treated as stationary, $|\phi|<1$, and the innovations have mean zero, variance $\sigma_\varepsilon^2$, and no covariance with past values.

Find the variance.
Calculate $\gamma_0=\sigma_\varepsilon^2/(1-\phi^2)$. The denominator uses the square of $\phi$.

Find the requested lagged covariances.
For each nonnegative lag $k$, multiply $\gamma_0$ by $\phi^k$. Use symmetry for negative lags.

Check signs and scale.
The variance is positive. When $\phi$ is negative, odd-lag autocovariances are negative and even-lag autocovariances are positive.

Worked Calculations

Worked Example: Positive Autoregressive Coefficient

A hypothetical monthly risk index follows a stationary AR(1) model with $\phi=0.5$. Its innovations are independent and normally distributed as $\varepsilon_t\sim N(0,12)$, where the second parameter is the variance. Find the stationary variance and autocovariances at lags 1, 2, and 3.

Method and assumptions. The model is centered at its stationary mean, $|0.5|<1$, and the innovations have mean zero and variance $\sigma_\varepsilon^2=12$. The interest rate and mortality assumptions are not applicable to this time-series calculation.

Stationary variance. First calculate $\phi^2=0.5^2=0.25$, so $1-\phi^2=1-0.25=0.75$. Therefore,

\gamma_0=\frac{12}{0.75}=16.

Lag 1. $\gamma_1=\phi\gamma_0=0.5(16)=8$.

Lag 2. $\gamma_2=\phi^2\gamma_0=0.25(16)=4$.

Lag 3. $\gamma_3=\phi^3\gamma_0=0.125(16)=2$.

Check. Substitution into the variance relation gives $0.25(16)+12=4+12=16$. The positive coefficient gives positive autocovariances at these lags. The results are exact, so no rounding is needed. A common mistake is to treat the innovation variance 12 as the process variance; the calculation shows that the stationary process variance is 16.

Worked Example: Negative Autoregressive Coefficient

A hypothetical quarterly indicator follows a stationary AR(1) process with $\phi=-0.4$ and independent normal innovations $\varepsilon_t\sim N(0,9)$. Find the variance and autocovariances at lags 1, 2, and 3. Give final values to four decimal places.

Method and assumptions. The process is centered at its stationary mean, the innovations have mean zero and variance 9, and $|-0.4|<1$. Interest and mortality assumptions do not apply.

Variance. Square the coefficient: $(-0.4)^2=0.16$. Then $1-\phi^2=1-0.16=0.84$, giving

\gamma_0=\frac{9}{0.84}=\frac{75}{7}=10.714285\ldots

Lag 1. $\gamma_1=(-0.4)(75/7)=-30/7=-4.285714\ldots$.

Lag 2. $\gamma_2=(0.16)(75/7)=12/7=1.714285\ldots$.

Lag 3. $\gamma_3=(-0.064)(75/7)=-4.8/7=-0.685714\ldots$.

Check. Rounded to four decimals, the variance is 10.7143 and the three autocovariances are $-4.2857$, 1.7143, and $-0.6857$. Their alternating signs agree with the negative coefficient: odd powers of $-0.4$ are negative and even powers are positive. The variance relation also checks: $0.16(75/7)+9=12/7+63/7=75/7$. Do not make the common mistake of applying the negative sign to the variance; variance depends on $\phi^2$ and remains positive.

Worked Example: Recovering Parameters From Covariances

For a hypothetical stationary AR(1) risk series, the theoretical variance is $\gamma_0=25$ and the lag-one autocovariance is $\gamma_1=15$. Assume independent normal innovations with mean zero and unknown variance, and assume the AR(1) model is appropriate. Find $\phi$, the innovation variance, and the lag-two and lag-three autocovariances.

Method and assumptions. Since $\gamma_1=\phi\gamma_0$, recover the coefficient from the ratio of the given lag-one autocovariance to the variance. Then use the stationary variance equation to recover the innovation variance. The values given are theoretical quantities, not estimates from a finite sample. Interest and mortality assumptions are not applicable.

Coefficient. $\phi=\gamma_1/\gamma_0=15/25=0.6$, which satisfies $|\phi|<1$.

Innovation variance. Rearranging $\gamma_0=\sigma_\varepsilon^2/(1-\phi^2)$ gives $\sigma_\varepsilon^2=\gamma_0(1-\phi^2)$. Here, $\phi^2=0.36$, so

\sigma_\varepsilon^2=25(1-0.36)=25(0.64)=16.

Lag 2. $\gamma_2=\phi^2\gamma_0=0.36(25)=9$.

Lag 3. $\gamma_3=\phi^3\gamma_0=0.216(25)=5.4$.

Check. The recovered parameters return the stated variance: $16/(1-0.36)=16/0.64=25$. They also return the stated lag-one covariance: $0.6(25)=15$. All results are exact. The key trap is reversing the ratio: $\gamma_0/\gamma_1$ is not the AR coefficient.

Interpretation and Common Errors

The innovation variance measures the variability of the new shocks, not the total stationary variability of the series. When $\phi$ is nonzero, the effects of earlier shocks contribute to current variation as well, which is why $\gamma_0$ generally differs from $\sigma_\varepsilon^2$. A larger absolute coefficient, while still below one, increases the stationary variance for a fixed innovation variance because $1-\phi^2$ becomes smaller.

Autocovariance has squared units: if $X_t$ is measured in index points, then $\gamma_k$ is in squared index points. Its sign describes the direction of linear co-movement at the specified lag. For negative $\phi$, odd lags have negative autocovariance and even lags have positive autocovariance. For negative $k$, use $\gamma_{-k}=\gamma_k$, since covariance is symmetric.

Using $1-\phi$ instead of $1-\phi^2$. The variance equation contains $\phi^2$, because multiplying a random variable by $\phi$ multiplies its variance by $\phi^2$.
Using the innovation variance as $\gamma_0$. The stationary variance includes variation carried forward from earlier periods.
Dropping signs at odd lags. If $\phi<0$, preserve the sign of each power $\phi^k$.
Ignoring stationarity. The formulas require $|\phi|<1$ and the stated innovation assumptions; they do not provide a finite stationary variance when that condition fails.

The central derivation is the variance balance $\gamma_0=\phi^2\gamma_0+\sigma_\varepsilon^2$: current variation is the carried-forward variance plus new innovation variance. The autocovariance recursion then tracks how that dependence changes with lag. These calculations are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each hypothetical stationary AR(1) process, assume independent normal innovations with mean zero and the stated variance. Give exact values where convenient; otherwise round numerical answers to four decimal places.

A process has $\phi=0.25$ and innovation variance 15. Find $\gamma_0$ and $\gamma_1$. (Answer: $\gamma_0=16,\ \gamma_1=4$.)
A process has $\phi=-0.5$ and innovation variance 12. Find $\gamma_0$ and $\gamma_3$. (Answer: $\gamma_0=16,\ \gamma_3=-2$.)
A process has $\phi=0.7$ and stationary variance 20. Find the innovation variance and $\gamma_2$. (Answer: $\sigma_\varepsilon^2=10.2,\ \gamma_2=9.8$.)
A stationary AR(1) process has theoretical variance 36 and lag-one autocovariance 27. Find $\phi$, the innovation variance, and $\gamma_2$. (Answer: $\phi=0.75,\ \sigma_\varepsilon^2=15.75,\ \gamma_2=20.25$.)
A process has $\phi=0.9$ and innovation variance 1.9. Find $\gamma_0,\gamma_1,\gamma_2,$ and $\gamma_4$. (Answer: $\gamma_0=10,\ \gamma_1=9,\ \gamma_2=8.1,\ \gamma_4=6.561$.)

What You'll Learn