Specifying AR(p) Models

Why AR(p) Notation Needs Care

An autoregressive model describes a value by relating it to earlier values of the same series and a new innovation. For a risk analyst, a series might represent a hypothetical monthly index of claim costs or an aggregate exposure measure. The model’s order tells us how far back the equation reaches; its coefficients describe how those lagged values enter. Reading the signs and the highest lag correctly is essential before interpreting or using the equation.

The notation $AR(p)$ means an autoregressive model whose highest included lag is $p$. The intercept, lag coefficients, and innovation are distinct parts of the specification. In particular, the intercept is not automatically the process mean, and writing down an AR(p) equation by itself does not establish that the process is stationary.

Definition: An $AR(p)$ model is specified by $$ X_t=c+\phi_1X_{t-1}+\phi_2X_{t-2}+\cdots+\phi_pX_{t-p}+\varepsilon_t, $$ where $c$ is the intercept, $\phi_j$ is the coefficient of the observation at lag $j$, and $\varepsilon_t$ is a white-noise innovation. Here white noise means $E[\varepsilon_t]=0$, $\operatorname{Var}(\varepsilon_t)=\sigma_\varepsilon^2$ for a finite $\sigma_\varepsilon^2\geq 0$, and $\operatorname{Cov}(\varepsilon_t,\varepsilon_s)=0$ for $t\ne s$. White noise need not be normally distributed or independent unless those additional assumptions are stated.

Reading the Order and Coefficients

In the displayed equation, $\phi_1$ multiplies $X_{t-1}$, $\phi_2$ multiplies $X_{t-2}$, and so on. The order is the largest lag with a nonzero coefficient. Thus, if the coefficient of lag 2 is zero but the coefficient of lag 3 is nonzero, the equation is still of order 3. A zero coefficient at an intermediate lag is permitted; it means that particular lag makes no direct contribution to the equation.

The innovation $\varepsilon_t$ is the new, mean-zero disturbance at time $t$. It is not another lagged observation. When rearranging an equation, keep the innovation on the right-hand side with its sign intact. A reliable way to avoid errors is to isolate $X_t$ first, then read off the constant and the coefficient of each lag.

Worked Example: Rearranging an AR(2) Equation

A hypothetical monthly claim-cost index is specified by $X_t-0.4X_{t-1}+0.12X_{t-2}=1.5+\varepsilon_t$. Write it in standard AR form and identify the order, intercept, and lag coefficients.

Step 1: Isolate $X_t$. Add $0.4X_{t-1}$ to both sides and subtract $0.12X_{t-2}$ from both sides: $$ X_t=1.5+\varepsilon_t+0.4X_{t-1}-0.12X_{t-2}. $$

Step 2: Put the terms in standard order. $$ X_t=1.5+0.4X_{t-1}-0.12X_{t-2}+\varepsilon_t. $$

Step 3: Identify the parameters. The largest nonzero lag is 2, so this is an $AR(2)$ specification. Its intercept is $c=1.5$, $\phi_1=0.4$, and $\phi_2=-0.12$.

Check. Substituting the coefficients into the standard form and moving the lag terms back to the left gives $X_t-0.4X_{t-1}+0.12X_{t-2}=1.5+\varepsilon_t$, the original equation. The method avoids the common sign error of copying $+0.12$ as the coefficient of $X_{t-2}$ after rearrangement.

Backshift Notation

The backshift operator $B$ moves a series back by one time period: $BX_t=X_{t-1}$, and more generally $B^jX_t=X_{t-j}$. This notation compresses the lag terms but does not change their meaning. The AR(p) equation can be written $ (1-\phi_1B-\phi_2B^2-\cdots-\phi_pB^p)X_t=c+\varepsilon_t $. The minus signs in this polynomial are part of the convention; expanding and rearranging determines the signs in the standard lagged-observation form.

Key result: If $ (1-\phi_1B-\cdots-\phi_pB^p)X_t=c+\varepsilon_t $, then expansion gives $X_t=c+\phi_1X_{t-1}+\cdots+\phi_pX_{t-p}+\varepsilon_t$. Always expand the polynomial before reading coefficients if the equation is presented in backshift notation.

Worked Example: Finding the Order From Backshift Notation

A hypothetical quarterly exposure index satisfies $(1-0.3B+0.25B^3)X_t=-2+\varepsilon_t$. Write the equation using lagged observations and identify all coefficients, including any zero coefficient.

Step 1: Expand the left-hand side. By the definition of $B$, $BX_t=X_{t-1}$ and $B^3X_t=X_{t-3}$, so $$ X_t-0.3X_{t-1}+0.25X_{t-3}=-2+\varepsilon_t. $$

Step 2: Isolate $X_t$. Add $0.3X_{t-1}$ and subtract $0.25X_{t-3}$ on both sides: $$ X_t=-2+0.3X_{t-1}-0.25X_{t-3}+\varepsilon_t. $$

Step 3: State the order and coefficients. The highest nonzero lag is 3, so the specification is $AR(3)$. The coefficients are $c=-2$, $\phi_1=0.3$, $\phi_2=0$, and $\phi_3=-0.25$.

Check. Moving the two nonzero lag terms to the left recovers the given polynomial equation. The missing $B^2$ term means $\phi_2=0$, not that the model has order 2. The common mistake is to count displayed terms instead of locating the highest lag.

Intercept and Centered Representations

An AR(p) equation can also be written around a chosen center $m$: $X_t-m=\phi_1(X_{t-1}-m)+\cdots+\phi_p(X_{t-p}-m)+\varepsilon_t$. This is an algebraic representation for any chosen $m$. Expanding it gives an intercept-form equation, with the intercept determined by $m$ and the coefficients. The centered form is useful for separating deviations from a reference level from the effects of those deviations at earlier lags.

\begin{aligned} X_t-m &=\sum_{j=1}^{p}\phi_j(X_{t-j}-m)+\varepsilon_t,\\ X_t &=m\left(1-\sum_{j=1}^{p}\phi_j\right) +\sum_{j=1}^{p}\phi_jX_{t-j}+\varepsilon_t. \end{aligned}

Why this conversion holds. Distribute each $\phi_j$ across its centered term. The right side becomes $\sum_{j=1}^p\phi_jX_{t-j}-m\sum_{j=1}^p\phi_j+\varepsilon_t$. Adding $m$ to both sides produces the intercept $m-m\sum_{j=1}^p\phi_j=m(1-\sum_{j=1}^p\phi_j)$, while leaving each lag coefficient unchanged.

This calculation identifies the intercept associated with the chosen center; it does not by itself prove that $m$ is the process mean. To see when the mean interpretation follows, suppose the process is weakly stationary with finite mean $\mu=E[X_t]$, and the innovations have mean zero. Taking expectations in the intercept form gives $\mu=c+(\sum_{j=1}^p\phi_j)\mu$, hence $c=\mu(1-\sum_{j=1}^p\phi_j)$. When $1-\sum_{j=1}^p\phi_j\ne0$, this implies $\mu=c/(1-\sum_{j=1}^p\phi_j)$. Equivalently, in the centered representation, $m=\mu$ follows under that condition.

If $1-\sum_{j=1}^p\phi_j=0$, the expectation equation places no restriction on $\mu$. The centered equation alone then does not identify the mean. For example, with $p=1$, $\phi_1=1$, $\varepsilon_t=0$, and $X_t=K$ for every $t$, the process is constant and weakly stationary. It satisfies $X_t-m=X_{t-1}-m$ for any $m$, although its mean is $K$. This illustrates why the nonzero-denominator condition matters.

Worked Example: Converting a Centered AR(2) Model

A hypothetical monthly portfolio index is expressed as $X_t-10=0.5(X_{t-1}-10)-0.2(X_{t-2}-10)+\varepsilon_t$. Find its intercept form, order, and coefficients. Then state what can be concluded about its mean from the given information.

Step 1: Expand both centered lag terms. $$ X_t-10=0.5X_{t-1}-5-0.2X_{t-2}+2+\varepsilon_t. $$ The constants on the right combine to $-5+2=-3$, so $X_t-10=0.5X_{t-1}-0.2X_{t-2}-3+\varepsilon_t$.

Step 2: Add 10 to both sides. $$ X_t=7+0.5X_{t-1}-0.2X_{t-2}+\varepsilon_t. $$ Thus $c=7$, $\phi_1=0.5$, $\phi_2=-0.2$, and the order is 2.

Step 3: Check the intercept with the conversion formula. Here $\sum_{j=1}^2\phi_j=0.5-0.2=0.3$, so $10(1-0.3)=10(0.7)=7$, agreeing with the expansion.

The equation uses 10 as its center. Whether 10 is the process mean requires additional assumptions. If the process is weakly stationary with finite mean and mean-zero innovations, then $1-\sum\phi_j=0.7\ne0$, and the mean equation gives $\mu=c/0.7=7/0.7=10$. Without those assumptions, the algebra alone establishes the centered representation, not a probabilistic claim about the mean. A common mistake is to call the intercept 10; it is 7.

Specification Checks and Common Traps

When translating a proposed model, check the equation in a fixed order: isolate the current observation, preserve the innovation’s sign, list coefficients by lag, and identify the largest nonzero lag. If converting from a centered form, calculate the constant term separately rather than treating the center as the intercept.

Sign reversal: A lag term moved across the equals sign changes sign. In $X_t-0.4X_{t-1}=\cdots$, the coefficient of $X_{t-1}$ in standard form is $+0.4$.
Wrong order: Order is the largest lag with a nonzero coefficient, not the number of nonzero terms. An $AR(3)$ equation may have $\phi_2=0$.
Intercept confused with mean: In general, $c\ne\mu$. The formula for $\mu$ requires weak stationarity, a finite mean, mean-zero innovations, and $1-\sum\phi_j\ne0$.
Extra innovation assumptions: White noise specifies zero mean, constant finite variance, and no correlation across distinct times. It does not by itself say that innovations are independent or normally distributed.
Stationarity assumed from notation: Writing an AR(p) equation specifies a relationship; it is not a stationarity test. The equation’s order and coefficients can be read without making a stationarity claim.

Key takeaway: The standard form is $X_t=c+\sum_{j=1}^p\phi_jX_{t-j}+\varepsilon_t$. Read the order from the highest nonzero lag, and treat the intercept and the process mean as different quantities unless the required conditions justify relating them.

Check Your Understanding

For each specification, identify the requested order, intercept, and lag coefficients. Keep the innovation separate from the lagged observations.

Rewrite $X_t+0.25X_{t-1}=3+\varepsilon_t$ in standard form. Give its order and coefficient. (Answer: $AR(1)$, $c=3$, $\phi_1=-0.25$.)
A centered equation is $X_t-8=0.25(X_{t-1}-8)+0.5(X_{t-2}-8)+\varepsilon_t$. Find the intercept-form equation and its order. (Answer: $AR(2)$, $c=2$, $\phi_1=0.25$, $\phi_2=0.5$.)
Expand $(1+0.2B-0.1B^3)X_t=5+\varepsilon_t$. Give the order and all lag coefficients, including any zero coefficient. (Answer: $X_t=5-0.2X_{t-1}+0X_{t-2}+0.1X_{t-3}+\varepsilon_t$; $AR(3)$, $c=5$, $\phi_1=-0.2$, $\phi_2=0$, $\phi_3=0.1$.)
A centered $AR(3)$ equation has $m=12$, $\phi_1=0.6$, $\phi_2=-0.1$, and $\phi_3=0.2$. Find its intercept. (Answer: $c=12[1-(0.6-0.1+0.2)]=12(0.3)=3.6$.)
Rewrite $X_t-0.4X_{t-1}+0.16X_{t-2}=-1+\varepsilon_t$. Give the order and coefficients, and explain why the sign of $\phi_2$ is not positive. (Answer: $AR(2)$, $X_t=-1+0.4X_{t-1}-0.16X_{t-2}+\varepsilon_t$, so $c=-1$, $\phi_1=0.4$, and $\phi_2=-0.16$; moving $+0.16X_{t-2}$ to the right changes its sign.)

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