Day 68: Parallel Trends & Event Studies · Math for Causal Inference

Theory

The event-study regression

With one treated group adopting at date $E$ and many periods, define event time $k = t - E$ . The event-study specification is

Y_{it} = \alpha_i + \lambda_t + \sum_{k \ne -1}\beta_k\,\mathbf 1[t - E = k]\,G_i + \varepsilon_{it},

with unit fixed effects $\alpha_i$ , period fixed effects $\lambda_t$ , and the period just before treatment ( $k = -1$ ) omitted as the reference. Each $\beta_k$ is a DiD comparing period $k$ with period $-1$ :

\beta_k = (\bar Y_{G=1, k} - \bar Y_{G=1, -1}) - (\bar Y_{G=0, k} - \bar Y_{G=0, -1}).

$k \ge 0$ : dynamic treatment effects (ATT $k$ periods after adoption), under parallel trends.
$k < 0$ : placebo or pre-trend coefficients, zero if trends were parallel before treatment.

Reading pre-trends

Flat pre-period coefficients are reassuring, but:

Low power. Pre-trend coefficients are noisy; a violation large enough to matter can easily go undetected (Exercise 12).
Pre-testing distortion. Reporting only studies that pass a pre-test selects samples where noise happened to hide the trend, which can increase bias (Roth, 2022).
Parallel pre-trends do not guarantee parallel post-trends. The assumption is about the counterfactual after treatment.

Remedies and sensitivity

Group-specific linear trends: fit a line through the pre-period differences and extrapolate. Unbiased if the violation is exactly linear; can absorb part of a dynamic effect if not.
Bounded violations (Rambachan and Roth): assume the post-treatment deviation from parallel trends is at most $M$ times the largest pre-treatment deviation, and report the resulting interval for each $M$ , plus the breakdown value of $M$ at which the effect is no longer significant.
Conditional parallel trends: assume parallel trends only among units with the same pre-treatment covariates; estimate with regression, weighting or doubly robust DiD.

Anticipation

If behaviour changes one period before adoption, the coefficient at $k = -1$ is already affected. Use $k = -2$ (or earlier) as the reference.

Part 1 · Warm-up

Exercise 1 · Reading coefficients

An event study reports $\hat\beta_{-3} = 0.1$ , $\hat\beta_{-2} = -0.05$ , $\hat\beta_0 = 1.2$ , $\hat\beta_1 = 1.8$ , $\hat\beta_2 = 2.1$ (SEs around 0.2). Describe the results.

Show solution

Step 1. Pre-period coefficients are small and insignificant: no visible pre-trend.

Step 2. The effect appears on impact (1.2) and grows over two years (2.1).

Answer: a dynamic effect that builds over time, with parallel pre-trends as supporting (not conclusive) evidence.

Exercise 2 · The reference period

Why must one event-time indicator be omitted, and what does choosing $k = -1$ mean?

Show solution

Step 1. With unit and time fixed effects, the full set of event-time dummies for the treated group is collinear with the unit fixed effect.

Step 2. Omitting $k = -1$ measures every $\beta_k$ relative to the last pre-treatment period.

Answer: all coefficients are DiDs relative to period −1, which is zero by construction.

Exercise 3 · Coefficients from means

Treated means in periods −3, −2, −1, 0, 1: 10, 11, 12, 16, 17. Comparison: 8, 9, 10, 11, 12. Compute $\beta_{-3}, \beta_{-2}, \beta_0, \beta_1$ .

Show solution

Step 1. $\beta_{-3} = (10 - 12) - (8 - 10) = 0$ ; $\beta_{-2} = (11 - 12) - (9 - 10) = 0$ .

Step 2. $\beta_0 = (16 - 12) - (11 - 10) = 3$ .

Step 3. $\beta_1 = (17 - 12) - (12 - 10) = 3$ .

Answer: no pre-trend, a constant effect of 3.

Exercise 4 · Linear extrapolation

Pre-period coefficients are $\beta_{-3} = -0.6$ , $\beta_{-2} = -0.3$ (and $\beta_{-1} = 0$ ): a linear trend of +0.3 per period. If $\hat\beta_2 = 1.5$ , what is the trend-adjusted effect at $k = 2$ ?

Show solution

Step 1. Extrapolated violation at $k = 2$ : $0.3 \times 2 = 0.6$ .

Step 2. Adjusted: $1.5 - 0.6 = 0.9$ .

Exercise 5 · Joint versus individual tests

Why test pre-period coefficients jointly rather than one at a time?

Show solution

Step 1. With several pre-periods, one coefficient may be significant by chance (multiple testing), while several small same-sign coefficients may each be insignificant yet jointly informative.

Answer: a joint F (Wald) test uses all pre-period information at once with correct size.

Part 2 · Proofs

Exercise 6 · Event-study coefficients identify dynamic ATTs

With one treated and one comparison group, balanced data and the saturated event-study regression, prove $\hat\beta_k$ equals the DiD of means between periods $k$ and $-1$ , and that under parallel trends and no anticipation it is unbiased for $ATT_k$ ( $k \ge 0$ ) and for 0 ( $k < 0$ ).

Show solution

Step 1. The regression has one parameter per group-period cell (unit effects absorb group levels, period effects the comparison path, $\beta_k$ the treated deviations); it is saturated in group × period means.

Step 2. So fitted cell means equal sample cell means, and $\hat\beta_k$ is the treated-minus-comparison change from period $-1$ to $k$ .

Step 3. By Day 67, Exercise 9, its expectation is $ATT_k + \Delta_k$ , where $\Delta_k$ is the difference in untreated trends between $-1$ and $k$ ; under parallel trends $\Delta_k = 0$ .

Step 4. For $k < 0$ the ATT is 0 (no anticipation), so $E[\hat\beta_k] = 0$ . $\blacksquare$

Exercise 7 · Linear trend adjustment

Suppose the violation is linear: $\Delta_k = s\,(k + 1)$ for all $k$ . Show that $\hat\beta_k - \hat s(k + 1)$ , with $\hat s$ fitted on pre-periods, is unbiased for $ATT_k$ .

Show solution

Step 1. Pre-period coefficients have expectation $s(k + 1)$ , so a least-squares slope through them (and the origin at $k = -1$ ) is unbiased for $s$ .

Step 2. $E[\hat\beta_k] = ATT_k + s(k + 1)$ for $k \ge 0$ .

Step 3. Subtract $E[\hat s](k + 1) = s(k + 1)$ . $\blacksquare$ If the violation is not linear, the adjustment is biased.

Exercise 8 · Power of a pre-trend test

A pre-period coefficient has expectation $\mu$ and standard error $\sigma$ (normal). Show the power of a two-sided 5% test is $\Phi(-1.96 - \mu/\sigma) + 1 - \Phi(1.96 - \mu/\sigma)$ .

Show solution

Step 1. The t-statistic is $N(\mu/\sigma, 1)$ .

Step 2. Rejection: $|t| > 1.96$ .

Step 3. $P(t > 1.96) = 1 - \Phi(1.96 - \mu/\sigma)$ and $P(t < -1.96) = \Phi(-1.96 - \mu/\sigma)$ . Add. $\blacksquare$

Exercise 9 · Bounded violations

Assume the post-period violation satisfies $|\Delta_{post}| \le M\bar\Delta_{pre}$ , where $\bar\Delta_{pre}$ is the largest absolute pre-period deviation (treated as known). Show that $ATT \in [\hat\beta - M\bar\Delta_{pre},\; \hat\beta + M\bar\Delta_{pre}]$ in population, and give a 95% version.

Show solution

Step 1. $\beta = ATT + \Delta_{post}$ , so $ATT = \beta - \Delta_{post}$ .

Step 2. With $\Delta_{post} \in [-M\bar\Delta_{pre}, M\bar\Delta_{pre}]$ , $ATT \in [\beta - M\bar\Delta_{pre}, \beta + M\bar\Delta_{pre}]$ .

Step 3. Accounting for sampling error in $\hat\beta$ : $[\hat\beta - M\bar\Delta_{pre} - 1.96\,\operatorname{SE},\; \hat\beta + M\bar\Delta_{pre} + 1.96\,\operatorname{SE}]$ covers the ATT with probability at least 95%. $\blacksquare$

Exercise 10 · Anticipation and the reference period

Suppose units react one period early, so $E[\hat\beta_{-1}]$ would be $a \ne 0$ if it were estimated. Show that normalising $\beta_{-1} = 0$ shifts every other coefficient by $-a$ , and that using $k = -2$ as the reference removes the bias.

Show solution

Step 1. Each $\hat\beta_k$ is a DiD relative to period −1: $E[\hat\beta_k] = (ATT_k + \Delta_k) - (a + \Delta_{-1})$ ; with parallel trends $\Delta = 0$ , it is $ATT_k - a$ .

Step 2. With $k = -2$ as reference (no anticipation there), $E[\hat\beta_k] = ATT_k$ , and the coefficient at $k = -1$ estimates $a$ . $\blacksquare$

Part 3 · Applied

Exercise 11 · State minimum wage and teen employment

Teen employment rates (%) in a state that raised its minimum wage in period 0, periods −3 to 2: 30.0, 30.5, 31.0, 30.0, 29.5, 29.3. In comparison states: 32.0, 32.4, 32.9, 33.2, 33.5, 33.7. Compute the event-study coefficients (reference −1).

Show solution

Step 1. $\beta_{-3} = (30.0 - 31.0) - (32.0 - 32.9) = -1.0 + 0.9 = -0.1$ .

Step 2. $\beta_{-2} = (30.5 - 31.0) - (32.4 - 32.9) = -0.5 + 0.5 = 0$ .

Step 3. $\beta_0 = (30.0 - 31.0) - (33.2 - 32.9) = -1.3$ .

Step 4. $\beta_1 = (29.5 - 31.0) - (33.5 - 32.9) = -2.1$ ; $\beta_2 = (29.3 - 31.0) - (33.7 - 32.9) = -2.5$ .

Answer: flat pre-trends and a growing employment decline of 1.3 to 2.5 points.

Exercise 12 · How likely is a pre-test to catch a trend?

In a study of hospital mergers on prices, a linear violation would make $E[\hat\beta_{-3}] = -0.9$ (percent), with SE 0.5. What is the probability the pre-trend test at $k = -3$ rejects?

Show solution

Step 1. $\mu/\sigma = -1.8$ .

Step 2. Power $= \Phi(-1.96 + 1.8) + 1 - \Phi(1.96 + 1.8) = \Phi(-0.16) + 1 - \Phi(3.76) \approx 0.436 + 0.000 = 0.44$ .

Answer: about 44%: more often than not, this sizeable trend goes undetected.

Exercise 13 · Sensitivity for a paid-leave mandate

A paid-family-leave mandate raises mothers' employment by $\hat\beta = 2.0$ points (SE 0.5) one year after; the largest pre-period deviation is 0.4. Compute the 95% sensitivity intervals for $M = 1$ and $M = 2$ , and the breakdown value of $M$ .

Show solution

Step 1. $M = 1$ : $[2.0 - 0.4 - 0.98,\; 2.0 + 0.4 + 0.98] = [0.62, 3.38]$ .

Step 2. $M = 2$ : $[2.0 - 0.8 - 0.98,\; 2.0 + 0.8 + 0.98] = [0.22, 3.78]$ .

Step 3. Breakdown: lower bound $2.0 - 0.4M - 0.98 = 0 \Rightarrow M = 1.02/0.4 = 2.55$ .

Answer: the effect stays positive unless post-period violations are more than 2.5 times the largest pre-period one.

Exercise 14 · Anticipation in hospital mergers

Hospitals announce a merger a year before it closes and start renegotiating contracts. An event study normalised at $k = -1$ shows $\hat\beta_0 = 3\%$ and $\hat\beta_{-2} = -2\%$ . Reinterpret with $k = -2$ as the reference.

Show solution

Step 1. Re-normalising to $k = -2$ adds 2 points to every coefficient: $\beta_{-1} = 2\%$ (anticipation effect), $\beta_0 = 5\%$ .

Answer: prices rose 2% already after the announcement and 5% by closing; the original normalisation understated the effect.

Exercise 15 · Conditional parallel trends

States that expanded Medicaid had lower baseline uninsured rates, and uninsured rates predict mortality trends. How can covariates restore parallel trends, and what is assumed?

Show solution

Step 1. Assume parallel trends conditional on baseline covariates $X$ (uninsured rate, income, age structure): $E[\Delta Y(0) \mid G = 1, X] = E[\Delta Y(0) \mid G = 0, X]$ .

Step 2. Estimate by comparing changes within $X$ cells, by outcome regression of $\Delta Y$ on $X$ in comparison states, or by doubly robust DiD (propensity weighting plus outcome regression).

Step 3. Use only pre-treatment covariates: post-treatment ones may be affected by the expansion (Day 53).

Answer: conditional parallel trends plus overlap in $X$ .

Verify in Python

import numpy as np
from scipy.stats import norm

def es(T, C, ref):
    """Event-study coefficients from treated and comparison means; ref = index of period -1."""
    return np.round([(T[k] - T[ref]) - (C[k] - C[ref]) for k in range(len(T))], 3)
print("Ex 3:", es([10, 11, 12, 16, 17], [8, 9, 10, 11, 12], 2))
print("Ex 11:", es([30, 30.5, 31, 30, 29.5, 29.3], [32, 32.4, 32.9, 33.2, 33.5, 33.7], 2))
m = -.9 / .5
print("pre-test power:", round(norm.cdf(-1.96 - m) + 1 - norm.cdf(1.96 - m), 3))
for M in (1, 2):
    print(f"M={M}: [{2 - .4 * M - .98:.2f}, {2 + .4 * M + .98:.2f}]", end="  ")
print("breakdown M:", round((2 - .98) / .4, 2))

# Event-study regression with unit and time fixed effects on a simulated panel (linear pre-trend 0.3)
rng = np.random.default_rng(68)
N, Tn, E = 200, 8, 5
G = (np.arange(N) < N // 2).astype(float)
t = np.arange(Tn); k = t - E
Y = (rng.normal(0, 2, N)[:, None] + 0.5 * t[None, :] + 0.3 * k[None, :] * G[:, None]
     + G[:, None] * np.where(k >= 0, 1 + 0.5 * k, 0)[None, :] + rng.normal(0, 1, (N, Tn)))
rows = []
for i in range(N):
    for s in range(Tn):
        r = [1.0] + [float(i == j) for j in range(1, N)] + [float(s == q) for q in range(1, Tn)]
        r += [G[i] * float(k[s] == kk) for kk in range(-E, Tn - E) if kk != -1]
        rows.append(r)
b = np.linalg.lstsq(np.array(rows), Y.ravel(), rcond=None)[0][-(Tn - 1):]
labels = [kk for kk in range(-E, Tn - E) if kk != -1]
print("event-study betas:", {kk: round(float(v), 2) for kk, v in zip(labels, b)})
print("truth (effect + 0.3*(k+1) trend):", {kk: round((1 + .5 * kk if kk >= 0 else 0) + .3 * (kk + 1), 2) for kk in labels})

Key takeaways

Event-study coefficients are DiDs relative to a reference period; pre-period coefficients test for diverging trends before treatment.
Pre-trend tests have limited power, and passing them does not establish parallel trends after treatment.
Report sensitivity to bounded violations (and the breakdown value), re-normalise if there is anticipation, and use conditional parallel trends when observable differences predict trends.

Day 68. Parallel Trends & Event Studies

Theory

The event-study regression

Reading pre-trends

Remedies and sensitivity

Anticipation

Part 1 · Warm-up

Exercise 1 · Reading coefficients

Exercise 2 · The reference period

Exercise 3 · Coefficients from means

Exercise 4 · Linear extrapolation

Exercise 5 · Joint versus individual tests

Part 2 · Proofs

Exercise 6 · Event-study coefficients identify dynamic ATTs

Exercise 7 · Linear trend adjustment

Exercise 8 · Power of a pre-trend test

Exercise 9 · Bounded violations

Exercise 10 · Anticipation and the reference period

Part 3 · Applied

Exercise 11 · State minimum wage and teen employment

Exercise 12 · How likely is a pre-test to catch a trend?

Exercise 13 · Sensitivity for a paid-leave mandate

Exercise 14 · Anticipation in hospital mergers

Exercise 15 · Conditional parallel trends

Verify in Python

Key takeaways

Simulation Lab: Event Studies & Pre-Trends