R, AR(1) and FGLS Computations. Unable to create a lagged variable - r

Question

Problem
Data
Hello everyone. I have been tasked with trying to solve this problem for the past two weeks with little to no help from my professor. I am not really asking for the exact code, but I can't start the problem...
We are given a set of data of the variables (M,R,Y) and we are asked to run the regression shown in the image. My issue now is that I can't seem to create a lagged variable. I have been told that 1) we don't need to use time series objects 2) I should be using time series objects (based off my research). Furthermore, when we get to the part where it asks us to run the regression for t=2 and onwards to get the p_hat, the only coefficient I can get is 1.0000 and the B1 that we should be getting ends up being something like 3.14e-11, something incredibly wrong. Here is what I have currently for my variables, based off the data given. If anyone is able to help guide me in the right direction I would really appreciate it.
#Creates Time Series Objects Which Can Be Lagged using lag()
Mt2 <- ts(data=DATA$M, start=1,end=180,frequency=1)
Mt1 <- ts(data=DATA$M, start=1,end=180,frequency=1)
Rt2 <- ts(data=DATA$R, start=1,end=180,frequency=1)
Yt2 <- ts(data=DATA$Y, start=1,end=180,frequency=1)
#Dependent Variable starts at t=2 and ends at t=181
#Lag Variable starts at t=1 and ends at t=180
Model_A <- lm( Mt2 ~ lag(Mt2,1) + Rt2 + Yt2, data=DATA)
bgtest(Model_A) #Conclude there is Autocorrelation
e <- resid(Model_A)
et <- ts(e,start=2,end=180,frequency=1)
et2 <- ts(e,start=1, end=179, frequency=1)
Model_e <- lm(et ~ et2)

Answer 1

Using lag() inside a plain lm() formula does not work because lag() just shifts the time index but keeps the data vector unchanged. And as lm() (or more precisely model.frame()) ignores the time index, this leads to including the response variable as one of the regressors and hence a regression coefficient of 1. As an illustration:
lm(Nile ~ lag(Nile, -1))
## Call:
## lm(formula = Nile ~ lag(Nile, -1))
##
## Coefficients:
## (Intercept) lag(Nile, -1)
## -1.819e-13 1.000e+00
There are various solutions to this problem, including:
Using dedicated time series infrastructure such as ar() or arima(). The latter can also easily incorporate additional regressors via xreg:
arima(Nile, c(1, 0, 0))
## Call:
## arima(x = Nile, order = c(1, 0, 0))
##
## Coefficients:
## ar1 intercept
## 0.5063 919.5685
## s.e. 0.0867 29.1410
##
## sigma^2 estimated as 21125: log likelihood = -639.95, aic = 1285.9
Using convenience interfaces to lm() with additional model.frame
processing such as dyn or dynlm:
library("dynlm")
dynlm(Nile ~ L(Nile, 1))
## Time series regression with "ts" data:
## Start = 1872, End = 1970
##
## Call:
## dynlm(formula = Nile ~ L(Nile, 1))
##
## Coefficients:
## (Intercept) L(Nile, 1)
## 452.7668 0.5043
Doing the lag pre-processing "by hand" before calling lm():
nile <- ts.intersect(Nile = Nile, NileLag = lag(Nile, -1))
lm(Nile ~ NileLag, data = nile)
## Call:
## lm(formula = Nile ~ NileLag, data = nile)
##
## Coefficients:
## (Intercept) NileLag
## 452.7668 0.5043
Depending on whether you need a particular estimation method or returned model class for further computations, one or the other strategy might be more handy. In your particular case I would recommend the last solution and then go through the different steps in the exercise by using plain lm().

Answer 2

Use a data.frame and create your lagged variable "by hand" or using something like dplyr::lag. Like this:
X <- 1:10
Xlag <- c(NA, X[1:9])
The base R lag function is not very useful for anything other than ts objects and they aren't very useful except for the specific modeling functions that support them.