ECON 104 · Discussion · Fall 2026

Autocorrelation, Autoregressive Models, and Distributed Lags

Jake Anderson

UCLA

Plan for Today

1

Autocorrelation

Is a series correlated with its own earlier values?

Test each sample autocorrelation with $Z = \sqrt{T}\, r_k$.

Spring 2024 Final V2, Question 7 (modified)

2

Autoregressive (AR) models

GDP growth this quarter depends on growth in the two quarters before.

Forecast next quarter’s growth.

Winter 2026 Midterm 1 (Rojas), Question 21

3

Distributed lags (DL)

Income this quarter changes consumption this quarter and in the next two quarters.

What does the coefficient on $Y_{t-1}$ measure?

Spring 2026 Final (Boswell), Question 36

Autocorrelation · A question

Is GPA Last Quarter Related to GPA This Quarter?

Why might a student’s GPA this quarter depend on their GPA last quarter?

Missed material

A weak quarter leaves gaps in material that the next courses build on, so the next quarter is harder too.

Study habits persist

A student who studies steadily this quarter usually studies the same way next quarter.

Health problems last

An illness or injury that lowers grades this quarter often lasts into the next one.

Money problems last

A student working long hours to pay rent this quarter is likely still working them next quarter.

Confidence

A good quarter raises confidence and motivation; a bad quarter lowers them.

Every story is momentum

A high GPA tends to be followed by a high GPA, and a low one by a low one: $y_t$ is correlated with $y_{t-1}$. When a variable has momentum, measure it with the autocorrelation, then model it with an autoregression.

Autocorrelation · The data

Each Lag Is a Copy of $y$ Moved Down One Row

The data20 quarters of GDP growth, one row per quarter. $t$ is the quarter and $y_t$ is growth in that quarter.

Lag 1Each value moves down one row: row $t$ of $y_{t-1}$ holds $y$ from row $t-1$. Row 1 has no earlier quarter, so it is NA (missing).

library(dplyr)  # its lag() moves values
df <- df %>%
  mutate(y_lag1 = lag(y, 1))

Lag 2Move $y_{t-1}$ down one more row. Rows 1 and 2 are NA.

  mutate(y_lag2 = lag(y, 2))

Up to lag 10$y_{t-k}$ starts with $k$ NAs.

Rows with an NA cannot be used. Pairing $y_t$ with $y_{t-k}$ leaves $T - k$ complete rows: 19 for lag 1, 18 for lag 2, and only 10 of the 20 quarters for lag 10.

Illustration: GDP growth values are made up for this table.

Autocorrelation · The idea

Correlation Between Values One Period Apart

The seriesOne value per period: $y_1, y_2, \ldots, y_T$, with mean $\bar y$.

Pair each value with the one beforePlot $y_t$ against $y_{t-1}$, the first two columns of the lag table.

Illustration: 120 periods of a series whose lag-1 autocorrelation is about 0.32.

$r_1 = \text{Corr}(y_t, y_{t-1})$

The correlation of the pairs in the scatter. Here $r_1 = 0.32$: a weak positive relationship.

$= \dfrac{\text{Cov}(y_t, y_{t-1})}{\text{Var}(y_t)}$

A correlation is the covariance divided by both standard deviations. Both columns are the same series, so that product is $\text{Var}(y_t)$.

$= \dfrac{\sum_{t=2}^{T} (y_t - \bar y)(y_{t-1} - \bar y)}{\sum_{t=1}^{T} (y_t - \bar y)^2}$

The covariance and variance written as sums (the $1/T$ in each cancels). A product on top is positive when $y_t$ and $y_{t-1}$ are on the same side of $\bar y$.

$r_k$ uses $y_{t-k}$ in place of $y_{t-1}$. Problem 1 gives $r_1$ to $r_4$ for $T = 680$ periods: which are different from 0?

Problem 1 · Autocorrelation

The Question: Which Autocorrelations Are Significant?

Say we are given the following sample autocorrelations:

Lag	$r_k$
1	$0.32$
2	$-0.21$
3	$0.08$
4	$-0.01$

Which lags are statistically significant at the 5% level, if $Z_{0.975} = 1.96$ and there are 680 periods in the study?

aLag 1 only
bLags 1 and 2 only
cLags 1 and 3 only
dLags 1, 2, and 3 only
eLags 1, 2, 3, and 4

Source: Spring 2024 Final V2, Question 7, modified: $r_2$ changed from $-0.91$ to $-0.21$ so that the four values could come from real data.

Autocorrelation · Step 1

Compute $Z = \sqrt{T}\, r_k$ for Each Lag

HypothesesFor each lag $k$: $H_0$: $\rho_k = 0$ vs $H_1$: $\rho_k \neq 0$
$\rho_k$ is the population autocorrelation at lag $k$; $r_k$ is its estimate from the sample.

StatisticUnder $H_0$, $r_k$ is approximately normal with mean 0 and variance $1/T$:

$Z = \dfrac{r_k - 0}{\sqrt{1/T}} = \sqrt{T}\, r_k \sim N(0,1)$

with $\sqrt{680} = 26.08$.

Lag	$r_k$	$Z = 26.08\, r_k$	$\|Z\| > 1.96$?
1	$0.32$	$8.34$	Yes
2	$-0.21$	$-5.48$	Yes
3	$0.08$	$2.09$	Yes
4	$-0.01$	$-0.26$	No

REJECT $H_0$for lags 1, 2, and 3. Lag 4: $\lvert -0.26 \rvert < 1.96$ $\implies$ fail to reject.

InterpretationAt the 5% level, $y_t$ is correlated with $y_{t-1}$, $y_{t-2}$, and $y_{t-3}$: positively at lags 1 and 3, negatively at lag 2. The data give no evidence of a correlation with $y_{t-4}$.

Autocorrelation · Step 2

The Correlogram Gives the Same Decision

CorrelogramOne bar for each sample autocorrelation $r_k$.

BoundsLines at $\pm 1.96/\sqrt{T} = \pm 1.96/26.08 = \pm 0.075$. A bar outside the lines $\iff$ $|Z| > 1.96$.

Zoom inLag 3: $0.08 > 0.075$, just outside. Lag 4: $-0.01$, inside.

The lecture slides draw the lines at $\pm 2/\sqrt{T} = \pm 0.077$. Lag 3 is still outside.

Autocorrelation · Interpretation

Checking Each Answer Choice

aLag 1 only Lags 2 and 3 also have $|Z| > 1.96$.
bLags 1 and 2 only Leaves out lag 3: $Z = 2.09 > 1.96$.
cLags 1 and 3 only Leaves out lag 2: $Z = -5.48$. The test is two-sided, so a large negative $Z$ also rejects.
eLags 1, 2, 3, and 4 Lag 4: $|Z| = 0.26 < 1.96$.
dLags 1, 2, and 3 only Answer: (d)

InterpretationThe series is correlated with its own values 1, 2, and 3 periods earlier. It has momentum, so we could model it with its own past values: an autoregression (next section).

Autoregressive models · The model

Today’s Value Depends on Its Own Past

The same patternGPA from quarter to quarter, mood from hour to hour, a city’s crime rate from year to year, and GDP growth from quarter to quarter (Problem 2).

The AR(1) model

$y_t = \delta + \theta_1\, y_{t-1} + v_t$

$v_t$ is this period’s new shock: mean 0, constant variance, and not correlated with earlier shocks. $\theta_1$ measures how strongly this period’s value depends on last period’s.

The AR($p$) modelAdd the $p$ most recent values of $y$: $y_t = \delta + \theta_1 y_{t-1} + \cdots + \theta_p y_{t-p} + v_t$. “Autoregressive” means $y$ is regressed on its own past values. With $p$ lags, lm() drops the NA rows and uses $T - p$ rows.

Illustration: 40 periods of an AR(1) series with $\theta_1 = 0.7$.

Problem 2 · Autoregressive models

The Question: Forecasting GDP Growth

Consider an AR(2) model for GDP growth:

$G_t = \delta + \theta_1 G_{t-1} + \theta_2 G_{t-2} + v_t$

Estimated results give: $\hat\delta = 0.67$, $\hat\theta_1 = 0.12$, $\hat\theta_2 = -0.09$. If $G_T = 0.8$ and $G_{T-1} = -0.2$, what is the one-step-ahead forecast $\hat G_{T+1}$?

a$0.84$
b$0.69$
c$0.78$
d$0.75$

Source: Winter 2026 Midterm 1 (Rojas), Question 21

Autoregressive models · The idea

$\theta_1$ Sets How Much of a Shock Is Left Each Period

One shockTake $y_t = \theta_1 y_{t-1} + v_t$ and start at $y = 0$. Set $v_0 = 1$ in period 0 and $v_t = 0$ in every later period. Then $y_k = \theta_1^k$.

$\theta_1 = 0.5$$y_0 = 1$, $y_1 = 0.5$, $y_2 = 0.25$, … Each period keeps half of the last value. After 4 periods, $0.5^4 = 0.06$ of the shock is left: small, but not 0.

$\theta_1 = 0.9$$y_1 = 0.9$, $y_2 = 0.81$, … Each period keeps 90% of the last value. After 4 periods, $0.9^4 = 0.66$ of the shock is left.

A new shock every periodBoth series get the same shocks. With $\theta_1 = 0.9$ the series moves slowly and stays above or below 0 for long runs. With $\theta_1 = 0.5$ it moves back toward 0 faster.

If $\theta_1 \neq 0$, a shock affects every later period, because $\theta_1^k$ is never exactly 0. $\theta_1$ sets how large that effect is: the effect in each period is $\theta_1$ times the effect in the period before. If $|\theta_1| < 1$, the effect gets smaller every period. In the GDP model, $\hat\theta_1 = 0.12$ and $\hat\theta_2 = -0.09$ are small: past growth tells us little about next quarter.

Top: $y_k = \theta_1^k$ after one shock in period 0. For an AR(1), $\theta_1^k$ is also the autocorrelation at lag $k$. Bottom, illustration: 80 periods, the same random shocks for both values of $\theta_1$.

Autoregressive models · Step 1

Write the Model One Quarter Ahead

The model in quarter $T+1$Replace $t$ with $T+1$:

$G_{T+1} = \delta + \theta_1 G_{T} + \theta_2 G_{T-1} + v_{T+1}$

$G_T$ is the latest quarter and $G_{T-1}$ the quarter before it. Both are already observed.

The forecast$v_{T+1}$ is next quarter’s shock. We do not know it, and its mean is 0, so the forecast sets it to 0 and uses the estimates:

$\hat G_{T+1} = \hat\delta + \hat\theta_1 G_{T} + \hat\theta_2 G_{T-1}$

Match each number to its symbol$\hat\theta_1 = 0.12$ multiplies the most recent value, $G_T = 0.8$.
$\hat\theta_2 = -0.09$ multiplies the older value, $G_{T-1} = -0.2$.

Plug in every number

$\hat G_{T+1} = 0.67 + (0.12)(0.8) + (-0.09)(-0.2)$

$= 0.67 + 0.096 + 0.018 = \mathbf{0.784}$

Autoregressive models · Step 2

Plug In and Watch the Signs

$\hat G_{T+1} = 0.67 + (0.12)(0.8) + (-0.09)(-0.2)$

Term	Estimate $\times$ value	Product
$\hat\delta$	$0.67$	$0.670$
$\hat\theta_1 G_T$	$0.12 \times 0.8$	$0.096$
$\hat\theta_2 G_{T-1}$	$(-0.09) \times (-0.2)$	$+0.018$
Forecast $\hat G_{T+1}$		$0.784$

A negative coefficient times a negative value is positive: $(-0.09)(-0.2) = +0.018$.

InterpretationGDP growth next quarter is forecast at $0.78$. The average growth rate of this model is $\mu = \hat\delta / (1 - \hat\theta_1 - \hat\theta_2) = 0.67/0.97 = 0.69$: the $\theta$’s are small, so the forecast is close to that average.

Autoregressive models · Interpretation

Checking Each Answer Choice

b$0.69$ $0.67 + (-0.09)(-0.2) = 0.688$: leaves out $\hat\theta_1 G_T = 0.096$.
d$0.75$ $0.67 + 0.096 - 0.018 = 0.748$: treats $(-0.09)(-0.2)$ as negative.
a$0.84$ $0.67 + 0.096 + 0.09(0.8) = 0.838$: uses $G_T$ in both terms and drops the minus sign on $\hat\theta_2$.
c$0.78$ Answer: (c) $\;0.67 + 0.096 + 0.018 = 0.784$

The recipe1. Write the model one period ahead.
2. Set the unknown shock $v_{T+1}$ to 0.
3. Multiply each $\hat\theta$ by the value it goes with: $\hat\theta_1$ with $G_T$, $\hat\theta_2$ with $G_{T-1}$.
4. Keep the signs.

Distributed lags · A question

Does Air Pollution This Week Affect Hospital Visits Later?

Why might more pollution this week raise hospital visits in later weeks?

Symptoms build up

Lung inflammation from polluted air can take days to become serious enough for a hospital visit.

People wait

Many people wait to see whether a cough goes away before they see a doctor.

Infections follow

Irritated airways make it easier to catch a respiratory infection a week or two later.

Serious cases take longer

Heart and lung damage builds slowly, so hospital admissions for it come weeks after the pollution starts.

Every story

Pollution in week $t$ raises visits in week $t$ and in weeks $t+1$, $t+2$, … $\implies$ $y_t$ depends on $x_t, x_{t-1}, x_{t-2}, \ldots$

Distributed lags · The model

Today’s Value Depends on Past Values of Another Variable

The same patternPollution and hospital visits over several weeks, ad spending and sales over the next few quarters, fertilizer and crop yield months later, and income and consumer spending over three quarters (Problem 3).

The distributed lag model

$y_t = \alpha + \beta_0 x_t + \beta_1 x_{t-1} + \cdots + \beta_q x_{t-q} + e_t$

$\beta_s$ is the effect of $x$ on $y$ $s$ periods later. The effect of a change in $x$ is spread over $q + 1$ periods.

AR model: lags of $y$ itself. Distributed lag model: lags of another variable $x$. A model with both is an autoregressive distributed lag (ARDL) model.

Illustration: a factory opens in week 4 and pollution rises to 4 times its usual level and stays there. Extra visits start one week later, rise for 6 weeks, then level off. Heart and lung admissions come last.

Problem 3 · Distributed lags

The Question: Income and Consumer Spending

Consider the following finite distributed lag model (FDL) of order 2 using quarterly data ($T = 100$):

$\widehat{C}_t = 1.20 + 0.45\,Y_t + 0.28\,Y_{t-1} + 0.16\,Y_{t-2}$

where $C_t$ is consumer spending and $Y_t$ is disposable income (both in $\$$billion). The coefficient on $Y_{t-1}$ is best interpreted as:

aThe total long-run effect of a $\$1$ increase in income on consumption.
bThe immediate effect of a $\$1$ increase in income on consumption in the same period.
cThe additional effect on consumption in period $t+1$ from a $\$1$ increase in income in period $t$, holding all else constant.
dThe proportion of the long-run multiplier realised after one period.
eThe effect of income growth on consumption growth one quarter ahead.

Source: Spring 2026 Final (Boswell), Question 36

Distributed lags · Step 1

Follow a One-Time Increase in Income

SetupRaise income by $\$1$ billion in quarter $t$ only. Income in every other quarter is unchanged.

Quarter $t$$C_t = 1.20 + \textcolor{#2774AE}{0.45\,Y_t} + 0.28\,Y_{t-1} + 0.16\,Y_{t-2}$ $\implies +0.45$

Quarter $t+1$$C_{t+1} = 1.20 + 0.45\,Y_{t+1} + \textcolor{#2774AE}{0.28\,Y_t} + 0.16\,Y_{t-1}$ $\implies +0.28$

Quarter $t+2$$C_{t+2} = 1.20 + 0.45\,Y_{t+2} + 0.28\,Y_{t+1} + \textcolor{#2774AE}{0.16\,Y_t}$ $\implies +0.16$

Quarter $t+3$$Y_t$ is no longer in the equation $\implies$ no effect.

The coefficient on $Y_{t-1}$, $0.28$, is the change in consumption one quarter after the change in income.

Distributed lags · Step 2

A Permanent Increase: Add the Weights

SetupRaise income by $\$1$ billion in quarter $t$ and keep it $\$1$ billion higher in every later quarter.

Quarter $t$Only $Y_t$ is higher: $0.45$. This is the impact multiplier.

Quarter $t+1$$Y_{t+1}$ and $Y_t$ are both higher: $0.45 + 0.28 = 0.73$. This is an interim multiplier.

Quarter $t+2$$0.45 + 0.28 + 0.16 = 0.89$.

Quarter $t+3$ and laterStill $0.89$: only two lags enter the model. This is the total (long-run) multiplier.

One coefficient is the effect of income at one lag. The sum of the coefficients is the effect of a permanent change in income.

Distributed lags · Interpretation

Checking Each Answer Choice

aThe total long-run effect of a $\$1$ increase in income on consumption. The total effect is the sum: $0.45 + 0.28 + 0.16 = 0.89$.
bThe immediate effect of a $\$1$ increase in income on consumption in the same period. The immediate (impact) effect is the coefficient on $Y_t$, $0.45$.
dThe proportion of the long-run multiplier realised after one period. That share is $(0.45 + 0.28)/0.89 = 0.82$.
eThe effect of income growth on consumption growth one quarter ahead. The model is in levels of $C$ and $Y$, not growth rates.
cThe additional effect on consumption in period $t+1$ from a $\$1$ increase in income in period $t$, holding all else constant. Answer: (c)

InterpretationA $\$1$ billion increase in income this quarter raises consumption next quarter by $\$0.28$ billion, holding income in all other quarters fixed. Read the same coefficient backward: last quarter’s income raises this quarter’s consumption by $0.28$ per dollar.

Thank you!

jakeanderson@g.ucla.edu