Answer:
No. See the explanation below.
Step-by-step explanation:
No. When we have the lineal model given by:
![Y_i = \beta_0 +\beta_1 X_i +\epsilon_i , i = 1,....,n](https://tex.z-dn.net/?f=%20Y_i%20%3D%20%5Cbeta_0%20%2B%5Cbeta_1%20X_i%20%2B%5Cepsilon_i%20%2C%20i%20%3D%201%2C....%2Cn)
For n observations, where y represent the dependent variable, X represent the independent variable and
are the parameters of the model, we are assuming that
is and independent and identically distrubuted variable that follows a normal distribution with the following parameters
.
So then the expected value for any error term is ![E(\epsilon_i) =0, i =1,...,n](https://tex.z-dn.net/?f=%20E%28%5Cepsilon_i%29%20%3D0%2C%20i%20%3D1%2C...%2Cn)
So then if we find the expected value for any observation we have this:
![E(Y_i) = E(\beta_0 +\beta_1 X_i +\epsilon_i) , i = 1,....,n](https://tex.z-dn.net/?f=%20E%28Y_i%29%20%3D%20E%28%5Cbeta_0%20%2B%5Cbeta_1%20X_i%20%2B%5Cepsilon_i%29%20%2C%20i%20%3D%201%2C....%2Cn)
Now we can distribute the expected value on the right by properties of the expected value like this:
![E(Y_i) = E(\beta_0) +E(\beta_1 X_i) +E(\epsilon_i), i =1,...,n](https://tex.z-dn.net/?f=%20E%28Y_i%29%20%3D%20E%28%5Cbeta_0%29%20%2BE%28%5Cbeta_1%20X_i%29%20%2BE%28%5Cepsilon_i%29%2C%20i%20%3D1%2C...%2Cn%20)
By properties of the expected value
if a is a constant and X a random variable, so then if we apply this property we got:
![E(Y_i) = \beta_0 +\beta_1 E(X_i) +0 ,i=1,...,n](https://tex.z-dn.net/?f=E%28Y_i%29%20%3D%20%5Cbeta_0%20%2B%5Cbeta_1%20E%28X_i%29%20%2B0%20%2Ci%3D1%2C...%2Cn)
![E(Y_i) =\beta_0 +\beta_1 X_i, i=1,.....,n](https://tex.z-dn.net/?f=%20E%28Y_i%29%20%3D%5Cbeta_0%20%2B%5Cbeta_1%20X_i%2C%20i%3D1%2C.....%2Cn)
And if we see that'ts not the result supported by the claim for this reason is FALSE the statement.