Submitted to: Dr. Tanweer ul Islam Submitted by: Kaneez Fatima MS ECONOMICS 2K20 NUST SCHOOL OF SOCIAL SCIENCES & HUMANITIES Applied Econometrics Assignment # 2 Q. Proof ππππ(ππ , ππ−π ) = πππ ππππ(ππ , ππ−π ) = ππ Solution: Before giving the proof, we have to check the data for stationarity If it is stationary, then the mean value will remain constant over time i.e. π = 0. If ππ is constant, then its mean value is also constant; the absolute value of ππ is less than 1. Using autoregressive model AR (1): Let ππ =ππ ππ−π + πΊπ i.e., put ππ = π → π = π πππ(ππ , ππ−π ) = π¬[(ππ ππ−π + πΊπ )ππ−π ] πππ(ππ , ππ−π ) = ππ π¬(πππ−π ) πππ(ππ , ππ−π ) = ππ πππ ππππ(ππ , ππ−π ) = πππ(ππ , ππ−π ) πππ(ππ , ππ−π ) Due to stationarity, the variance of ππ = the variance of ππ−π ππππ(ππ , ππ−π ) = ππ πππ(ππ , ππ−π ) = π¬[(ππ ππ−π + πΊπ )ππ−π ] πππ(ππ , ππ−π ) = π¬[(ππ (ππ ππ−π + πΊπ−π ) + πΊπ )ππ−π ] πππ(ππ , ππ−π ) = π¬(πππ πππ−π + ππ πΊπ−π ππ−π + πΊπ ππ−π ) πππ(ππ , ππ−π ) = πππ πππ Due to stationarity, the variance of ππ = the variance of ππ−π ππππ(ππ , ππ−π ) = πππ(ππ , ππ−π ) πππ(ππ , ππ−π ) ππππ(ππ , ππ−π ) = πππ