Shock-to-Response Explorer

Lecture 01 lab · Four controlled experiments on what one local-projection coefficient measures

Work through the labs in order the first time. In each one, stop at the prediction before moving a control, and write the prediction in the box under it; the box travels with your settings into the handoff record at the end of the page. Use the display to test the prediction, then explain what changed and what stayed fixed. Open the explanation only after you have an answer. Every panel says whether it shows a live calculation, a stored simulation result, or a stored result estimated in Stata.

The labs follow Lecture 01 from the gap between two paths to a coefficient estimated on a century of quarterly data. Labs 2 and 3 compute on the same observations that Practicum 01 ships, so a coefficient shown here and the same regression run in Stata agree to at least six decimals. Lab 4 shows estimates only. The Ramey–Zubairy workbook is downloaded by the practicum’s script and never reaches the browser.

Lab 1: Two Paths, One Gap

The twelve-period economy of the notes follows

y_t=\rho\,y_{t-1}+\theta_0 s_t+v_t,\qquad t=1,\dots,12,\qquad y_0=0,

with interventions s_t=1 at t=1,5,10,12 and the disturbances v_t of the hand table held at their drawn values. The counterfactual path y^{c}_t removes only the intervention at t=5. The causal response h periods after it is the gap between the two paths at the same date, \theta_h=y_{5+h}-y^{c}_{5+h}, measured in units of y per unit of s. A reader of the realized path alone sees something else: the change since the date before, y_{5+h}-y_4. The defaults, \rho=0.5 and \theta_0=1, reproduce the hand table.

Predict before using the controls. If \rho rises from 0.5 to 0.9 with \theta_0 fixed, does the gap at t=5 change? Does the gap at t=7? Does the change in y since t=4?

Live calculation. Both paths are recomputed from the rule whenever a control moves. The disturbances v_t and the interventions at t=1,10,12 never change. Orange segments are the gaps from t=5 on; the dotted line holds y_4 so the change since t=4 can be read against it.

Live calculation. The same quantities read on the horizon clock, h=t-5. Orange dots are the gaps; the blue line is \theta_0\rho^h from the notes; grey rings are the changes since t=4.

Try these comparisons. Change one thing at a time.

  1. Hold \theta_0 at 1 and move \rho from 0.5 to 0.9. Read the gap at t=5, then the gap at t=7, then the change since t=4.
  2. Return \rho to 0.5 and set \theta_0 to 2. Compare each gap with its value at \theta_0=1, and do the same for the change since t=4.
  3. Return \theta_0 to 1 and switch the intervention at t=5 off. Note which of the two paths moves.

Subtract the counterfactual rule from the realized rule. The disturbances and the interventions at t=1,10,12 appear in both and cancel, so the gap g_t=y_t-y^{c}_t follows a rule of its own, g_t=\rho\,g_{t-1}+\theta_0\,(s_t-s^{c}_t). Before t=5 the gap is zero. At t=5 there is no earlier gap to carry forward, so g_5=\theta_0 whatever \rho is. After that each period keeps the share \rho of the previous gap, which gives g_{5+h}=\theta_0\rho^h, the response of the notes’ formula. Raising \rho to 0.9 leaves the gap at t=5 at 1 and raises the gap at t=7 from 0.25 to 0.81.

The change since t=4 follows no such rule. By the decomposition in the notes it is the gap plus the change the counterfactual path makes on its own, and that second piece carries every disturbance after t=4 (the draw v_7=2.1 above all) together with the echo of the intervention at t=1 in y_4, which depends on both \rho and \theta_0. Doubling \theta_0 doubles every gap exactly, but it does not double the change since t=4. Switching the intervention off moves the realized path onto the counterfactual path. The counterfactual path does not move, because it never contained s_5.

Lab 2: Which Rows Enter the Horizon-h Regression

A regression row pairs the intervention s_t and the control y_{t-1} with the outcome y_{t+h}. Row t needs y_{t-1}, so row 1 never enters, and it needs y_{t+h}, so the last h rows leave. With T=12 periods and p=1 lag the sample has T_h rows, and without the control the slope is a difference in means, with \bar y_1 and \bar y_0 the mean outcomes h periods after intervention rows and after the other rows:

T_h=T-h-p,\qquad \hat\beta_h=\bar y_1-\bar y_0 .

This lab always uses the stored hand table, the realized path at \rho=0.5 and \theta_0=1 with every intervention in place, whatever Lab 1’s controls show. As in the notes, the rows start at t=2 with and without the control, so the control switch changes a regressor and never the rows.

Predict before using the controls. Which rows leave when h goes from 0 to 2, and which intervention rows leave with them? Will the slope at h=2 be closer to \theta_2=0.25 than the slope at h=0 is to \theta_0=1?

Live calculation. Each line is one regression row, dated by its intervention. Grey squares are y_{t-1}, rings are s_t (orange when the regression codes s_t=1), blue dots are the outcome y_{t+h}, and crosses mark values that lie outside the data. Faded rows are not in the horizon-h sample.

Live calculation. Group means and both slopes on the rows above, rounded to six decimals from full-precision values; compare the notes’ table of slopes.

Try these comparisons. Change one thing at a time.

  1. Keep the control off and move h from 0 to 1 to 2. Record which rows leave the sample, which intervention rows leave with them, and \hat\beta_h.
  2. At h=2, switch the control on. The rows and the group means stay where they are; the coefficient does not.
  3. Move to h=3 with the single-episode switch off and count the intervention rows. Then switch single episode on and off.

Row t enters only if both y_{t-1} and y_{t+h} exist, so the rows run from 2 to 12-h and T_h=11-h. The intervention in row 12 leaves at h=1 and the one in row 10 at h=3. From then on row 5 is the only intervention row, and the slope is the single-episode formula: row 5’s outcome minus the mean outcome of the other rows. That is why the single-episode switch changes nothing at h=3.

The slopes miss their targets because the estimate is the target plus a weighted sum of everything else in the rows, and with two or three intervention rows a few disturbances dominate that sum. At h=1 the intervention rows’ outcomes are y_6, which carries half of y_5 and so half of v_5=-1.7, and y_{11}, which receives v_{11}=-0.9. Row 4 is a comparison row, yet its outcome y_5 contains the intervention s_5. Moving from h=0 to h=2 does not bring the slope closer to its target: 0.617 against 1 becomes 2.050 against 0.25.

Adding y_{t-1} on the same rows changes the comparison, not the rows. The slope now compares outcomes after removing the part that y_{t-1} predicts, so it is no longer a difference in means; the exact weights are the subject of Lecture 3. In twelve rows the control moves the slopes by 0.04 to 0.10. Both regressions target the same \theta_h, and neither has enough rows to come close to it.

Lab 3: One Draw Against the Truth

In the smallest model the horizon-h regression

y_{t+h}=\mu_h+\beta_h s_t+\gamma_h y_{t-1}+u_{t,h},\qquad \beta_h=\theta_h=\theta_0\rho^h,

recovers the causal response in population, with the control y_{t-1} or without it (the notes). A sample of T periods is not the population. This lab runs the thirteen regressions h=0,\dots,12 on the draw behind the notes’ one-draw figure: s_t and v_t standard normal, \theta_0=1, 300 periods drawn in Stata with seed 2 and the first 100 discarded, stored once for \rho=0.5 and once for \rho=0.9. One unit of s is one standard deviation, the response is in units of y per unit of s, and horizons are periods. The draw never changes. Shortening T keeps its first T periods, and switching the control changes a regressor, not a number in the data. The defaults reproduce the notes’ figure.

Predict before using the controls. Without y_{t-1}, does the curve move toward \theta_h, away from it, or just get noisier? Decide separately for short and long horizons.

Live calculation on the stored draw: the blue curve is \hat\beta_h for the current settings, and the grey curve is the same draw with the other control choice. Stored simulation result: the shaded band holds the middle 90 percent of \hat\beta_h across R=500 samples of the same design (seed 3), estimated in Stata with the same T and control choice. It shows how the estimator behaves when the economy is known; it is not a confidence interval.

Live calculation in the first five columns; stored simulation result in the last four. With the control the rows are t=2,\dots,T-h; without it they are t=1,\dots,T-h, as in the notes’ Monte Carlo table.

Try these comparisons. Change one thing at a time.

  1. Keep \rho=0.9 and T=200. Switch the control off and on. Compare the band’s width at h=0 and at h=12, then the distance between the blue curve and \theta_h.
  2. Switch the control back on and shorten T to 100, then to 50. The first 50 periods are the same numbers in all three settings.
  3. Return to T=200 and switch \rho to 0.5, which selects the other stored draw. Find the first horizon at which the band contains zero, for each \rho.

Dropping y_{t-1} does not change what the regression estimates. The omitted term \rho^{h+1}y_{t-1} moves into the residual, and s_t is unrelated to it, so the population coefficient is still \theta_h. What changes is the size of the residual. At \rho=0.9 and h=0 its variance rises from 1 to about 9.5, and across the stored samples with T=200 the standard deviation of \hat\beta_0 rises from 0.073 to 0.218. The omitted term fades as \rho^{h+1} shrinks: by h=12 the two standard deviations are 0.225 and 0.242, and at \rho=0.5 they are within one percent of each other from h=4 on. So the curve without the control mostly gets noisier, most of all at short horizons with high persistence. Samples this short also bias both regressions a little, and the one without the control visibly at impact: the stored mean of \hat\beta_0 without the control is 0.967 at T=200 and 0.855 at T=50. Lecture 2 explains that bias. The regression without the control also uses row 1, as in the notes’ Monte Carlo table; one extra row does not account for the change in spread.

Shortening T widens everything. With the control, the stored standard deviation roughly doubles when T falls from 200 to 50 (from 0.073 to 0.144 at h=0 with \rho=0.9), because a quarter as many rows vote.

With the control, the \rho=0.9 draw lies below \theta_h at every horizon. That is not thirteen independent misses. Neighbouring regressions use almost the same rows, and the residual at one horizon is \rho times the residual at the horizon before plus one new period, so an error made at h=4 is largely repeated at h=5. The seed was fixed before any estimate was computed, and the draw was kept rather than replaced by a tidier one because it shows exactly this.

Lab 4: Where the Military-News Coefficient’s Variation Comes From

First contact with data regressed y_{t+h}, real GDP divided by trend GDP, on the military news s_t, measured as a fraction of the previous quarter’s nominal trend GDP, and on y_{t-1}, for quarters 1890Q1–2015Q4. Trend GDP is Ramey and Zubairy’s rgdp_pott6, the series they call potential GDP and estimate as a sixth-degree polynomial trend in real GDP. By the slope formula, row t supplies the share

\frac{(s_t-\bar s)^2}{\sum_\tau(s_\tau-\bar s)^2}

of the variation behind the slope, so a quarter with unusually large news weighs heavily. Everything in this lab is a stored result: least-squares coefficients, row counts, and the ten largest shares, computed in StataNow/SE 19.5 from Ramey and Zubairy’s RZDAT.xlsx (February 2018 replication package, SHA-256 b2d85087…8c6120) for h=0,\dots,20, two sets of controls, and five sets of rows. “Without 1950Q3” drops the row dated 1950Q3 from every regression. Shares use s_t-\bar s over the rows of the selected regression, as in the notes’ table, even when more controls are included. Read a coefficient per 0.01 of news: news worth one percent of trend GDP, response in percent of trend GDP. No bands are shown; Lectures 5 and 6 build them.

Predict before using the controls. At h=8, will dropping Korea’s news quarter, 1950Q3, lower the coefficient, leave it unchanged, or raise it? Write a number as well as a direction. The box below is your hypothesis for Exercise 9.

Stored result. Coefficients by horizon for the selected controls: grey on the full sample, blue on the selected rows. The orange line marks the horizon h.

Stored result. The ten quarters with the largest shares of \sum_t(s_t-\bar s)^2 among the rows of the selected regression at horizon h. Orange bars are 1941Q4 and 1950Q3 whenever they are in the rows.

Try these comparisons. Change one thing at a time.

  1. Keep one lag of y and h=8. Choose “Without 1950Q3”, then “Without 1941Q4”, then both. Record the coefficient, the number of rows, and the largest remaining share each time.
  2. Keep “Without 1950Q3” and switch to the Ramey–Zubairy controls. Does dropping the quarter move the coefficient by as much as it did before?
  3. Choose “From 1947Q1 on”. Read the share chart before the coefficient.

The slope is a weighted sum of outcomes, with each quarter’s weight proportional to how far its news was from the average (the notes’ slope formula). 1941Q4 and 1950Q3 carry the two largest weights, so the coefficient leans on what followed them. At h=8 the outcome after the Korean news was smaller, measured against the regression’s line, than the outcomes after the other large episodes, which 1941Q4 leads. Removing a heavily weighted quarter that sat below the line raises the slope from 0.362 to 0.434; removing 1941Q4, which sat above it, lowers the slope to 0.344. This is the one-episode formula at a larger scale. When a few quarters dominate the variation, the coefficient behaves like an average over a handful of episodes, not over 496 rows.

From 1947Q1 on, every world-war quarter leaves the rows. Korea’s two news quarters, 1950Q3 and 1950Q4, then supply 93 percent of the variation at h=8, and the coefficient falls to 0.059. That describes which rows carry the estimate. It does not say whether wartime or postwar news is the better experiment, which is the identification question of Lecture 3; Lecture 13 turns drop-one comparisons into a systematic sensitivity analysis.

With Ramey and Zubairy’s controls the same exclusion moves the h=8 coefficient from 0.229 to 0.278, a smaller move in the same direction. Dropping the row dated 1950Q3 removes it as an intervention row, but its news still enters the next four rows as a lagged control. The shares shown use s_t-\bar s; with controls in the regression the exact weights use s_t after the controls have been partialled out, which Lecture 3 develops.

From One Gap to One Coefficient

The four labs form one chain. Before reading the next paragraph, explain that chain aloud without formulas, one sentence per lab, and then compare.

A causal response is the gap between two paths at the same date, and in the smallest model that gap is fixed by the impact response and persistence, whatever else happens. A regression recovers it by comparing the rows that follow larger interventions with the rows that follow smaller ones, and each horizon has its own rows, fewer as h grows. One sample of those rows can land far from the truth, with errors that persist across neighbouring horizons. In the military-news data, a few quarters carry most of the comparison.

If a step of your version was missing or different, return to its lab and change only the input responsible for that step. The derivations are in the Lecture 01 notes, and the exercises take each step by hand and in Stata.

Transfer to Stata

The record below collects your predictions and the current settings of all four labs, and it changes as you move the controls. Set each lab the way you want it recorded, then copy the record or download it. Paste it at the top of your problem-set do-file as a comment block and run the checks it lists. The whole record, predictions included, is the lab record in the submission package of Practicum 01. Three of its parts feed specific tasks:

From Object Used in
Lab 2 hand_table.csv (t s v y, written as Stata’s default export writes it) and the chosen h, control, and single-episode setting Exercise 4; STA01 Tasks 1–2
Lab 3 onedraw_rho09_seed2.csv or onedraw_rho05_seed2.csv (t s v y), the chosen T and control, and the thirteen browser estimates Exercises 6 and 7; STA01 Tasks 3–4
Lab 4 Your hypothesis about the h=8 coefficient without 1950Q3 Exercise 9; STA01 Task 5

Run in Stata on the downloaded files, the lines below reproduce the browser’s coefficients to at least six decimals. hand_table.csv is written exactly as Stata’s default export delimited writes the notes’ table, so its v column shows float storage (0.89999998 for 0.9); the draw file follows Lab 3’s \rho setting.

The checks the record describes, written out for the current settings. Run them after hand_table.csv and the draw file are in your working directory.