Answer:
No. See the explanation below.
Step-by-step explanation:
No. When we have the lineal model given by:
[tex] Y_i = \beta_0 +\beta_1 X_i +\epsilon_i , i = 1,....,n[/tex]
For n observations, where y represent the dependent variable, X represent the independent variable and [tex]\beta_0, \beta_1[/tex] are the parameters of the model, we are assuming that [tex]\epsilon[/tex] is and independent and identically distrubuted variable that follows a normal distribution with the following parameters [tex] e\sim N(\mu=0,\sigma^2)[/tex].
So then the expected value for any error term is [tex] E(\epsilon_i) =0, i =1,...,n[/tex]
So then if we find the expected value for any observation we have this:
[tex] E(Y_i) = E(\beta_0 +\beta_1 X_i +\epsilon_i) , i = 1,....,n[/tex]
Now we can distribute the expected value on the right by properties of the expected value like this:
[tex] E(Y_i) = E(\beta_0) +E(\beta_1 X_i) +E(\epsilon_i), i =1,...,n [/tex]
By properties of the expected value [tex] E(aX) =aE(X)[/tex] if a is a constant and X a random variable, so then if we apply this property we got:
[tex]E(Y_i) = \beta_0 +\beta_1 E(X_i) +0 ,i=1,...,n[/tex]
[tex] E(Y_i) =\beta_0 +\beta_1 X_i, i=1,.....,n[/tex]
And if we see that'ts not the result supported by the claim for this reason is FALSE the statement.