Data.frame and different specifications linear regression models

Question

This is my data.frame:

data <- matrix(rnorm(50*5),nrow=50)
m <- data.frame(data )
m

            X1          X2           X3           X4
1  -0.47903358  1.92799699 -0.584364168 -1.475276350
2   0.05464517  1.90064618  1.721449935  1.215188405
3   2.23782855  0.16839367 -0.074212064 -0.719745849
4   0.44476924  0.37247676  0.300423948  1.722222410
5  -0.66889333  0.26094142 -0.001254037  0.617630295
6  -1.17663634 -1.93020031 -2.813433555  0.350943310
7   2.16609882 -0.06594556  0.133278642  1.693078802
8  -0.16090564 -1.14454452 -1.340407149  0.574943347
9   0.46691283  0.98394614 -0.277273498  0.030499576
10  1.19049154  1.03160103  0.626364442  0.429348535
11  0.72593476  0.89130574 -0.599052058  2.012316875
12 -0.04748923 -1.38044706  0.058364936  0.825752213
13 -0.90407306  0.14790680  0.002608154 -0.991979288
14 -1.50113350  0.78807179 -2.371160466  0.559174890
15  0.04075635 -1.14486991  1.585769771 -0.495937216
16  1.29724449  0.47729721 -0.871033202  1.524746619
17  1.12384656 -0.07589516 -0.463837527  0.206108262
18  0.65092671  0.29660588 -1.096981115 -0.326524160
19 -0.54953490 -0.06792527 -0.543922865  0.211205138
20  0.19187346 -0.65190412 -1.674653668  0.712704645
21  0.97108788  0.91087005  0.182424518 -0.296413534
22  1.64975949  0.22140448 -0.075355571 -0.472299544
23  1.45613296 -1.29101443  1.231839302  0.774475499
24 -0.15324579 -1.31596417 -0.343467527  0.087122108
25  0.20305822 -0.96867715 -0.311745324 -1.483964874
26 -2.22916775  1.46982603 -0.254705650 -0.003573105
27  0.26135095 -0.44053524  0.208284330  0.459095264
28  0.55020951 -0.37742141  0.346186077  0.825811292
29 -0.78695097 -0.21014858  0.003055816  2.142121347
30 -0.33506969  0.31572240 -0.139970552  1.344697176
31 -0.94164226 -0.88366990 -0.906886069  0.774590884
32  3.23764133 -0.47556802  0.166917302  0.959307700
33 -0.64114926  0.16089400  1.942563410  0.142073261
34  0.03848502 -0.83066179  0.934403254  1.730448144
35 -0.34621698  0.65682342  1.601365990  0.160012282
36  0.62848314  0.33126945  0.188823549  2.380319403
37  1.02807329  1.07299920  0.398083643 -0.208241920
38 -0.50616984  1.68998408  0.427342040 -0.345264684
39  1.31721397 -1.17111247  1.754036030  0.755414080
40  1.23714888 -0.12253370  0.887065669  0.001770354
41  0.59541135  0.66438057  0.880643498  1.441513269
42  0.41100680  0.52757460  1.219644246  1.450262235
43  0.08984815  1.44830794  0.113193523 -0.620546777
44 -1.51411951 -0.17346499 -0.698924584 -0.767921061
45 -1.03652530  1.27296507  1.740954529 -0.467568299
46  1.14533381 -0.67304957 -0.384012335  0.637582075
47  0.37996214  0.53566774 -1.413930797  1.222728459
48 -0.53705734  0.77591264  0.375236588 -1.430924067
49  0.64051195 -0.22562951 -1.667050192  0.345009059
50 -1.19521122  1.18436746 -0.905004199 -0.005221906

Well, I want now estimate 3 different models.

X1 = X2 + X3 + X4
X1 = X2^2 + X3(Lag1) + X4^2
ln(X1)=X2^2 + X3(Lag1) + X4^2

As im a begginer in R im not able to implement a function to do this. I'm reading the lm function and still can not implement these models. The way I could implement was very rudimentary, ie, I was riding a data.frame for each model.

Thanks

Answer 1

To use lm() you need at least two things. You need a formula and you need data to apply that formula to. In your case, here's how you would code to run a very simple, basic, linear regression on your data. Again, you need an "X3_Lag1" variable (literally a column called that which has the lag applied) in your data to regress on lagged data (make a new column with X3 pushed down a row) and then cut the data for that missing value in row one.

# Getting the first model.
formula1 <- "X1 ~ X2 + X3 + X4"
lin1 <- lm(formula1, data)
summary(lin1) # This will give you the coefficients for each value.

# Getting the second model.
formula2 <- "X1 ~ X2^2 + X3_Lag1 + X4^2"
lin2 <- lm(formula2, data)
summary(lin2)

# Getting the third model.
formula3 <- "log(X1) ~ X2^2 + X3_Lag1 + X4^2"
lin3 <- lm(formula3, data)
summary(lin3)

WARNING: I believe you will get an error on the third formula because you're taking the natural log of a negative number, which is not possible. Please take care of that being the case.