Why Does E(X1 + X2 + ... + Xn) Equal n*E(Xi) in Statistics?

[cdux]

I'm a bit confused about a particular step in a calculation.

Given Theta_n = (2/n)(X1 + X2 + ... + Xn) being an unbiased estimator of Theta for a U(0,Theta), we have to prove it by showing E(Theta_n) = Theta.

And we go on E(Theta_n) = (2/n)E(X1+X2 + .. Xn)

Now at this point the solution is (2/n) * n * (Theta/2) (= Theta which is the sought-after result)

I understand that Theta/2 is the mean of a U() but how exactly does one go from E(X1 + X2 + .. Xn) to equaling it to n*E(Xi)? Is E(X1) = E(X2) = E(Xi)? If yes, why?

(PS. A more complex example is Var(X1+X2 + .. Xn) appearing to also result to nV(Xi) (=nσ^2) )

 

[cdux]

Hrm. Afterthought. I guess it might be simply a blatant case of "mean of the whole but it missed the 'over n' so it's n * E(Xi)".

I guess it might apply in the case of Var too..

 

[cdux]

I'm having a difficulty seeing how that could be true for Var.

Var((2/n)(X1 + X2 + ... + Xn)) = (4/n^2)nVar(Xi).

I understand 4/n^2 going out as a property of Var but how is Var(X1 + X2 + ... + Xn) = nVar(Xi)?

 

[cdux]

Nevermind. I found it. If X1, X2.. are uncorrelated then V(Σ(Xi)) = ΣV(Xi) ..after a proof involving Var's equality with E[X^2) - E^2