Question

for the u.s. economy, let gprice denote the monthly growth in the overall price level and let gwage be the monthly growth in hourly wages. [these are both obtained as differences of logarithms: gprice 5 dlog1price2 and gwage 5 dlog1wage2.4 using the monthly data in wageprc we estimate the following distributed lag model: gprice 5 2.00093 1 .119 gwage 1 .097 gwage21 1 .040 gwage22 1.000572 1.0522 1.0392 1.0392 1 .038 gwage23 1 .081 gwage24 1 .107 gwage25 1 .095 gwage26 1.0392 1.0392 1.0392 1.0392 1 .104 gwage27 1 .103 gwage28 1 .159 gwage29 1 .110 gwage210 1.0392 1.0392 1.0392 1.0392 1 .103 gwage211 1 .016 gwage212 1.0392 1.0522 n 5 273, r2 5 .317, r2 5 .283. (i) sketch the estimated lag distribution. at what lag is the effect of gwage on gprice largest? which lag has the smallest coefficient? (ii) for which lags are the t statistics less than two? (iii) what is the estimated long-run propensity? is it much different than one? explain what the lrp tells us in this example. (iv) what regression would you run to obtain the standard error of the lrp directly? (v) how would you test the joint significance of six more lags of gwage? what would be the dfs in the f distribution? (be careful here; you lose six more observations.)

Answer 1

(i) Sketch of lag distribution:

Peak effect at lag 5: .107

Smallest effect at lag 12: .016

(ii) Lags with t-stat < 2:

Lag 2: t-stat = 1.39

Lag 3: t-stat = 1.22

Lag 11: t-stat = 1.11

(iii) Long-run propensity (LRP) = sum of coefficients = 1.503

The LRP tells us the expected long-run change in gprice for a 1% change in gwage. Here the LRP is larger than 1, indicating that in the long-run a 1% change in wages leads to more than a 1% change in prices.

(iv) To get the standard error of the LRP directly, I would regress gprice on the cumulative sum of gwage lags.

(v) To test the joint significance of 6 more lags, I would run an F-test with the regression including those 6 additional lags. The dfs would be (6, n-k-6) where k is the number of lags already in the model (12). So the dfs would be (6, 272-12-6) = (6, 254).