ECON 104 · Discussion · Fall 2026

Granger Causality, the Linear Probability Model, and Probit

Jake Anderson

UCLA

Plan for Today

1

Granger causality

Do past values of one series improve the forecast of another?

Read two F-tests, one in each direction, and choose the model they support.

Spring 2026 Midterm (Boswell), Question 11

2

Linear probability model

A predicted probability of 1.1 and errors whose variance changes with $x$.

Two problems with the LPM and the correction for each.

Fall 2024 Midterm V1 (Boswell), Questions 15 and 17

3

Probit predictions

The same user, now with a probit model.

Read the probability $\Phi(z)$ and compare to the LPM.

Fall 2024 Midterm V1 (Boswell), Question 16

Problem 1 · Granger causality

The Question: Which Way Does Granger Causality Run?

A macroeconomist is studying whether consumer confidence (Conf) helps predict retail sales growth (Sales) in the US, using quarterly data from 2000–2023 ($T = 96$), where both series have been confirmed stationary. For the following Granger causality tests:

Granger causality test
Model 1: Sales ~ L(Sales, 1:2) + L(Conf, 1:2)
Model 2: Sales ~ L(Sales, 1:2)
  Res.Df Df      F  Pr(>F)
1     91
2     93 -2 6.8741  0.0016 **

Granger causality test
Model 1: Conf ~ L(Conf, 1:2) + L(Sales, 1:2)
Model 2: Conf ~ L(Conf, 1:2)
  Res.Df Df      F  Pr(>F)
1     91
2     93 -2 1.2033  0.3047

which of the following correctly interprets the test results and their implication for model specification?

aConsumer confidence Granger causes retail sales and retail sales Granger causes consumer confidence; a VAR model is required to capture the bidirectional relationship.
bRetail sales Granger causes consumer confidence, but consumer confidence does not Granger cause retail sales; lagged Sales should be used to forecast Conf instead.
cConsumer confidence Granger causes retail sales, but retail sales does not Granger cause consumer confidence; the ARDL model with lagged Conf as a predictor of Sales is appropriate.
dNeither series Granger causes the other; the ARDL specification is unsupported and both distributed lag terms should be dropped.
eThe test is inconclusive because the residual degrees of freedom row for Model 1 is missing from the output.

Source: Spring 2026 Midterm (Boswell), Question 11

Granger causality · The idea

Police and Crime: Which Variable Is $X$?

So far: ARDLOne outcome $Y$, and a variable $X$ that affects $Y$ but is not affected by $Y$.

Consider the size of a city’s police force and the city’s crime rate.

Police depend on past crimeHigh crime $\implies$ the city hires more officers. Hiring takes months.

Crime depends on past policeMore officers on the street $\implies$ less crime.

$\implies$ There is no clean $Y$ and $X$. Each variable belongs on both sides.

Made-up monthly data for one city, built so that each variable depends on the other’s past.

Granger causality · The idea

The Granger Test: Does Past Police Improve the Forecast of Crime?

Definition$X$ Granger causes $Y$ if past $X$ improves the forecast of $Y$, beyond what past $Y$ already gives.

$\text{Crime}_t = \delta + \theta_1 \text{Crime}_{t-1} + \theta_2 \text{Crime}_{t-2} + \beta_1 \text{Police}_{t-1} + \beta_2 \text{Police}_{t-2} + e_t$

$H_0$: $\beta_1 = \beta_2 = 0$ $\iff$ police does not Granger cause crime

Restricted model: lags of crime only$SSE_R = 441.8$

Unrestricted model: add 2 lags of police$SSE_U = 361.2$. The 2 lags lower the SSE by $80.6$.

Is a drop of 80.6 large? The F-test $F = \dfrac{(SSE_R - SSE_U)/J}{SSE_U/(T-K)} = \dfrac{80.6/2}{361.2/(98-5)} = 10.38$
$J = 2$ lags tested. $T = 98$ months, $K = 5$ coefficients.

REJECT $H_0$$10.38 > F_{(0.95,\,2,\,93)} = 3.09$ $\implies$ police Granger causes crime.

Swap the rolesPolice on 2 lags of police, then add 2 lags of crime: $F = 17.23 > 3.09$ $\implies$ crime Granger causes police. Both directions $\implies$ VAR: an equation for each variable, each with lags of both.

With real data on 56 U.S. cities and 49 states over 20+ years, Marvell and Moody (1996, Criminology) find Granger causation in both directions. The effect of police on crime is much larger.

Granger causality · Try it

Shuffle $G$ and Rerun the Granger Test

Lecture 7 testDo 4 lags of GDP growth $G$ improve the forecast of unemployment $U$, beyond 2 lags of $U$? Real data: $F = 5.98 > 2.41$, reject.

ShufflePut the 273 quarters of $G$ in a random order and rerun the test. Shuffled $G$ has no time order, so its lags carry no information about $U$.

Each dot is one shuffle. About 5% of shuffled $F$ values land above the critical value 2.41. The real data give $F = 5.98$.

Computed in the browser from usmacro (POE5): the Lecture 7 models on the same 269 quarters.

Granger causality · Step 1

Test 1: Do Lags of Conf Improve the Forecast of Sales?

Model 1: Sales ~ L(Sales, 1:2) + L(Conf, 1:2)
Model 2: Sales ~ L(Sales, 1:2)
  Res.Df Df      F  Pr(>F)
1     91
2     93 -2 6.8741  0.0016 **

Two modelsL(Conf, 1:2) means lags 1 and 2 of Conf. Model 1 (unrestricted) includes them; Model 2 (restricted) drops them. $H_0$: both coefficients on lagged Conf are 0.

Df = −2Res.Df is the residual degrees of freedom. Model 2 has 2 fewer coefficients, so its Res.Df is 2 higher; R prints the change as $-2$. That is $J = 2$ restrictions, and the F-test has $(2,\ 91)$ degrees of freedom, using Model 1’s Res.Df as printed.

Statistic$F = 6.87$. 5% critical value: $F_{(0.95,\,2,\,91)} = 3.10$.

REJECT $H_0$$6.87 > 3.10$ ($p = 0.0016$) $\implies$ Conf Granger causes Sales.

Both comparisons give the same decision: $F = 6.87 > 3.10$, or $p = 0.0016 < 0.05$. On the exam the p-value is enough.

Granger causality · Step 2

Test 2: Do Lags of Sales Improve the Forecast of Conf?

Model 1: Conf ~ L(Conf, 1:2) + L(Sales, 1:2)
Model 2: Conf ~ L(Conf, 1:2)
  Res.Df Df      F  Pr(>F)
1     91
2     93 -2 1.2033  0.3047

The other directionNow Conf is on the left. $H_0$: both coefficients on lagged Sales are 0. Same degrees of freedom, $(2,\ 91)$, so the same critical value, 3.10.

Statistic$F = 1.20$, below 3.10.

FAIL TO REJECT $H_0$$1.20 < 3.10$ ($p = 0.30$) $\implies$ no evidence that Sales Granger causes Conf.

Together: only the Conf → Sales test rejects at 5%. Conf Granger causes Sales, and we find no evidence that Sales Granger causes Conf.

Granger causality · Step 3

Choose the Model from the Two Test Results

The model for Sales

$\text{Sales}_t = \delta + \theta_1 \text{Sales}_{t-1} + \theta_2 \text{Sales}_{t-2} + \beta_1 \text{Conf}_{t-1} + \beta_2 \text{Conf}_{t-2} + e_t$

Granger causality is about forecasting. It does not show that higher confidence causes higher sales.

Granger causality · Try it

Change the Two F Statistics and Watch the Model Choice

Critical value for both tests: $F_{(0.95,\,2,\,91)} = 3.10$. Start: the exam values.

Test 1: Conf → Sales

Test 2: Sales → Conf

Test 2 rejects
(Sales → Conf)

Test 2 does not reject

Test 1 rejects
(Conf → Sales)

VAReach series on lags of both

ARDL for Saleslags of Sales and lags of Conf

Test 1 does not reject

ARDL for Conflags of Conf and lags of Sales

AR model for eachown lags only

Granger causality · Interpretation

Checking Each Answer Choice

aBoth directions; a VAR model is required. Test 2 does not reject ($p = 0.30$), so Sales does not Granger cause Conf.
bSales Granger causes Conf, but not the reverse. This reverses the direction. The test that rejects ($p = 0.0016$) has Sales on the left.
dNeither series Granger causes the other. Test 1 rejects $H_0$: $6.87 > 3.10$.
eInconclusive: the row for Model 1 is missing. Nothing is missing. R prints Df and F once, on row 2, because the test compares the two rows.
cConf Granger causes Sales, but Sales does not Granger cause Conf; the ARDL model with lagged Conf as a predictor of Sales is appropriate. Answer: (c)

InterpretationConsumer confidence in the last two quarters improves the forecast of retail sales growth, beyond what past sales growth gives. We find no evidence that past sales growth improves the forecast of confidence.

A note on the output: the printed Res.Df values (91 and 93) count all 96 quarters. The first 2 quarters have no lagged values, so only 94 quarters can be used, and the values should be $94 - 5 = 89$ and $94 - 3 = 91$. The 5% critical value is 3.10 either way, so the decision is the same.

Granger causality · An aside

Umbrellas Granger Cause Rain

The dataEvery day, count the umbrellas people carry across campus at 8 am, and record whether it rains at 3 pm.

The testDays with many umbrellas in the morning are more often rainy in the afternoon. Morning umbrellas improve the forecast of rain: umbrellas Granger cause rain.

The common causePeople check the weather forecast before they leave. The forecast causes both. Taking everyone’s umbrella away would not stop the rain.

Back to regressionConf may predict Sales because news about the economy appears in survey answers before it appears in sales. A causal effect needs an identification strategy: an instrument, a natural experiment, or a quasi-experiment (Lecture 7).

Problem 2 · Linear probability model

The Question: Predicted Probability from the LPM

Suppose we’re modeling whether someone will subscribe to an influencer’s YouTube channel ($Y$: 1 = subscribe, 0 = doesn’t subscribe) based on the number of Youtube video views at the time of watching the video and the number of ads that pop up during the video. Use the following table of coefficients from a sample of 5,000 users:

Model	Intercept	Ads	Video Views
Probit	$-1.0$	$0.3$	$0.05$
LPM	$0.2$	$0.1$	$0.02$

Question 15. What is the estimated predicted probability of subscribing to the YouTube channel for a user who sees 5 ads and where the video has 20 views, according to the LPM (censor your results where necessary if it falls outside the bounds of probability assumptions.)?

a0.0
b0.5
c0.6
d1.0
eNone of the above

Source: Fall 2024 Midterm V1 (Boswell), Question 15

Linear probability model · The idea

A Straight Line Leaves the Range 0 to 1

The LPM from Question 15

$\widehat{\text{Probability of subscribing}} = \hat p = 0.2 + 0.1\,\text{Ads} + 0.02\,\text{Video Views}$

Same slope everywhereEach ad adds $0.1$ to $\hat p$, and each video view adds $0.02$, however many the user has already seen.

Hold video views at 20$\hat p = 0.2 + 0.02(20) + 0.1\,\text{Ads} = 0.6 + 0.1\,\text{Ads}$

Above 1 after 4 ads$0.6 + 0.1\,\text{Ads} > 1$ $\iff$ $\text{Ads} > 4$. A straight line keeps rising, so it passes 1.

The exam user, 5 ads and 20 views:

$\hat p = 0.2 + 0.1(5) + 0.02(20) = 1.1$

Common practice: censorReport 1 for any $\hat p$ above 1 (and 0 for any below 0). The line is flat at 1 after 4 ads, so the exam user gets 1.0.

Linear probability model · Interpretation

Checking Each Answer Choice: Question 15

a0.0 Censoring moves 1.1 down to 1, not to 0.
b0.5 Only the ads term, $0.1 \times 5$.
c0.6 Leaves out the ads term: $0.2 + 0.4$.
eNone of the above Choice (d) holds after censoring.
d1.0 Answer: (d)

InterpretationThe LPM predicts 1.1 for this user. The question says to censor: replace a value above 1 with 1 (and a value below 0 with 0). The reported prediction is 1.0, the largest possible probability.

Censoring changes the reported number, not the model: the slope is still 0.1 per ad at every number of ads. The correction is a model whose predictions stay between 0 and 1 for every $x$: probit or logit (Problem 3).

Problem 2 · Linear probability model

The Question: Confidence Interval for the LPM Marginal Effect

Same setup: subscribing to a YouTube channel, 5,000 users.

Model	Intercept	Ads	Video Views
Probit	$-1.0$	$0.3$	$0.05$
LPM	$0.2$	$0.1$	$0.02$

Question 17. If the marginal effect of ads on the probability of subscribing has a standard error of 0.04 for the LPM, calculate the 95% confidence interval for the marginal effect, rounded to two decimal places. (The critical value is 1.96). Is this confidence estimate precise, and why or why not?

a$[0.02, 0.18]$; it is precise because the interval is narrow.
b$[0.1, 0.5]$; it is not precise because the interval is wide due to small sample size.
c$[0.1, 0.5]$; it is not precise because the interval is wide due to multicollinearity.
d$[0.02, 0.18]$; it is not precise because the interval is wide due to heteroscedasticity.
eNone of the above answers are correct and we need more information to speak about the precision.

Source: Fall 2024 Midterm V1 (Boswell), Question 17

Linear probability model · The idea

The LPM Error Variance Changes with $x$

$y$ is 0 or 1 (Bernoulli)

$\Var(y_i \mid x_i) = p_i(1 - p_i)$, $p_i = \beta_1 + \beta_2 x_i$

The variance depends on $p$$p = 0.5$: variance $0.25$. $p = 0.1$ or $0.9$: variance $0.09$.

And $p$ depends on $x$The error is $e_i = y_i - p_i$, so $\Var(e_i \mid x_i) = \Var(y_i \mid x_i)$. Because $p_i$ changes with $x_i$, the LPM errors are heteroskedastic unless $\beta_2 = 0$.

ConsequenceOLS coefficients are still unbiased (Lecture 8) but not efficient, and the usual OLS standard errors are not valid.

CorrectionUse heteroskedasticity-robust standard errors (week 2 discussion), or feasible generalized least squares (FGLS) with weights $1/\big(\hat p_i(1-\hat p_i)\big)$, which needs every $\hat p_i$ strictly between 0 and 1.

Bottom: made-up admissions data (50 applicants) and its LPM line, $\hat p = -1.75 + 0.75\,\text{GPA}$. Blue: the fitted values between 0 and 1, on both charts.

Linear probability model · Try it

Move $p$: The Variance of a 0/1 Outcome

$p = \Pr(y = 1)$

100 people with the same $x$. Blue: $y = 1$. Gray: $y = 0$.

All 1s or all 0s: no variation. Half and half: the most variation. In the LPM, $p$ moves with $x$, so the variance moves with $x$.

Linear probability model · Step 2

Build the Interval for the Ads Coefficient

Formula

$\hat\beta_{\text{Ads}} \pm 1.96 \times se(\hat\beta_{\text{Ads}})$

In the LPM the marginal effect of ads is the coefficient itself: $0.1$.

Margin$1.96 \times 0.04 = 0.078$

Interval

$0.1 \pm 0.078 = [0.022,\ 0.178] \approx [0.02,\ 0.18]$

The upper end is about 8 times the lower end: one more ad could raise the probability of subscribing by 2 percentage points or by 18.

For the interval to be valid, 0.04 must be a heteroskedasticity-robust standard error (previous slide).

Linear probability model · Interpretation

Checking Each Answer Choice: Question 17

b$[0.1, 0.5]$; wide due to small sample size. $0.1 \pm 0.078$ is not $[0.1, 0.5]$. And $n = 5{,}000$ is not small.
c$[0.1, 0.5]$; wide due to multicollinearity. Wrong interval.
a$[0.02, 0.18]$; precise because the interval is narrow. Right interval, but it runs from 0.02 to 0.18 around an estimate of 0.1.
eNone of the above; need more information. Can be argued, since the reason in (d) is weak. The posted answer is (d).
d$[0.02, 0.18]$; not precise, wide due to heteroscedasticity. Answer: (d)

On the exam, choose (d)Its reason is loosely worded. Heteroskedasticity does not by itself make an interval wider: it makes the usual OLS standard error invalid, and robust standard errors can be larger or smaller. The best case for (d): with heteroskedasticity OLS is not efficient, so FGLS would give a narrower interval.

LPM problem	Consequence	Correction
Straight line in $x$	$\hat p$ can be below 0 or above 1 (Q15: 1.1)	Probit or logit
Variance $p(1-p)$ changes with $x$	Usual standard errors not valid; OLS not efficient	Robust standard errors

Problem 3 · Probit predictions

The Question: The Same User, Now with Probit

Same setup: subscribing to a YouTube channel, 5,000 users.

Model	Intercept	Ads	Video Views
Probit	$-1.0$	$0.3$	$0.05$
LPM	$0.2$	$0.1$	$0.02$

Question 16. What is the difference between the estimate obtained in question 15, and the probability of subscribing to the YouTube channel for a user who currently sees 5 ads where the video has 20 views, using the probit model? (Note that $\Phi(1.5) = 0.9332$ (CDF) and $\phi(1.5) = 0.1295$ (PDF))

a0.0668
b0.8705
c0.1295
d0.9332
eNone of the above answers are correct

Source: Fall 2024 Midterm V1 (Boswell), Question 16

Probit predictions · Aside

Why Use Probit: GPA and the Probability of Admission

Made-up data for 50 applicants. $y = 1$ if admitted.

LPM$\widehat{\Pr}(\text{admit}) = -1.75 + 0.75\,\text{GPA}$

Probit$\widehat{\Pr}(\text{admit}) = \Phi(-21 + 7\,\text{GPA})$
Index $= 0$ at $\text{GPA} = 3.0$, so $\hat p = 0.5$ there.

Raise GPA by 0.2	LPM	Probit
2.0 → 2.2	$+0.15$	$+0.000$
2.9 → 3.1	$+0.15$	$+0.516$
3.7 → 3.9	$+0.15$	$+0.000$

The LPM says“A little bit higher GPA makes a difference, no matter where you start!”

Probit says“GPA improvements don’t really matter if you have a 2.0 or a 3.9, but it makes a big difference going from 2.9 to 3.1.”

Probit predictions · The idea

Probit Puts the Linear Index Inside the Normal CDF

Probit

$p_i = \Phi(\beta_1 + \beta_2 x_i)$

$\Phi(z) = \Pr(Z \le z)$ for a standard normal $Z$: the normal CDF.

Index and probabilityThe index $z = \beta_1 + \beta_2 x$ can be any number. $\Phi(z)$ is always between 0 and 1.

Admissions data$\hat p = \Phi(-21 + 7\,\text{GPA})$. With a GPA of 3.1, $z = -21 + 7(3.1) = 0.7$ and $\hat p = \Phi(0.7) = 0.758$.

The marginal effect is no longer constant: the curve is steepest near $p = 0.5$ and nearly flat near 0 and 1.

Made-up admissions data for 50 applicants, the same data as slide 16. Dashed: the LPM line.

Probit predictions · Try it

Move Along GPA: LPM and Probit Predictions

GPA

The LPM line goes below 0 for GPA below $2.33$ and above 1 for GPA above $3.67$. The probit curve stays between 0 and 1 for every GPA.

Made-up admissions data for 50 applicants.

Probit predictions · Try it

The Probit Marginal Effect Changes Along the Curve

GPA

The slope of the tangent line is the marginal effect: $\phi(z) \times 7$, with $z = -21 + 7\,\text{GPA}$. The LPM slope is $0.75$ at every GPA.

Made-up admissions data for 50 applicants. Slopes are per 1 point of GPA; for 0.1 points, divide by 10.

Probit predictions · Step 1

The Probability Is the Area to the Left of 1.5

The probit prediction: 5 ads, 20 views

$\begin{aligned} \hat p &= \Phi(z), \quad z = \hat\beta_0 + \hat\beta_1\,\text{Ads} + \hat\beta_2\,\text{Views} \\ z &= -1.0 + 0.3(5) + 0.05(20) \\ &= -1.0 + 1.5 + 1.0 = 1.5 \\ \hat p &= \Phi(1.5) \end{aligned}$

The question gives two numbers. Which one is the probability?

$\Phi(1.5) = 0.9332$The area under the standard normal curve to the left of 1.5. This is $\hat p$, the probit probability of subscribing.

$\phi(1.5) = 0.1295$The height of the curve at 1.5. It is used for marginal effects, $\phi(z)\,\beta$. It is not a probability.

Probit predictions · Step 2

Compare to the LPM Answer from Question 15

Question 15 (LPM)$\hat p = 1.1$, censored to $1.0$.

Question 16 (probit)$\hat p = \Phi(1.5) = 0.9332$

Difference$1.0 - 0.9332 = 0.0668$

The difference is the area to the right of 1.5: $1 - \Phi(1.5) = \Phi(-1.5) = 0.0668$, the probit probability of not subscribing. This holds only because the censored LPM answer is exactly 1.

Probit predictions · Interpretation

Checking Each Answer Choice: Question 16

c0.1295 $\phi(1.5)$: the height of the curve, not a probability.
b0.8705 $1 - \phi(1.5)$: uses the height in place of the area.
d0.9332 The probit probability itself, not the difference from the LPM.
eNone of the above answers are correct Choice (a) holds.
a0.0668 Answer: (a)

InterpretationFor a user who sees 5 ads on a video with 20 views, the probit model predicts a 93.3% chance of subscribing. The censored LPM prediction, 100%, is 6.7 percentage points higher.

Thank you!

jakeanderson@g.ucla.edu