Mean and variance

E(Y)

= E[(X1 + X2 +...+ Xn)/n]

= (1/n)E(X1 + X2 +...+ Xn)

= (1/n)(u + u +... + u)

= (1/n)(nu)

= u

Var(Y)

= Var [(X1 + X2 +...+ Xn)/n]

= (1/n^2)Var (X1 + X2 +...+ Xn)

= (1/n^2)(nv^2)

= v^2/n

As n tends to infinity, Var(Y) tends to 0 and so the expected deviation of sample mean E(Y) and the population mean u will tends to 0.

E(Y) = E(( X1+X2+...+Xn) / n)
=1/n*E(X1+X2+...+Xn)
=nu/n
=u

Var(Y) = Var(( X1+X2+...+Xn) / n)
=1/n^2*Var( X1+X2+...+Xn)
=nv^2/n^2
=v^2/n

2013-05-13 12:02:24 補充：
E(A+B) = E(A)+E(B):
If A follows a(x) and b follows b(y),
E(A+B) = sum(x)sum(y)(x+y)a(x)b(y)
= sum(x)sum(y)(xa(x)b(y)+ya(x)b(y))
= sum(x)xa(x)+sum(y)yb(y)
= E(A) + E(B)

2013-05-13 12:05:27 補充：
Var(A+B) = Var(A)+Var(B) if A,B independent:
Var(A+B) = E((A+B)^2) - E(A+B)^2
= E(A^2+2AB+B^2)-(E(A)+E(B))^2
= E(A^2)+2E(AB)+E(B^2)-(E(A)^2+2E(A)E(B)+E(B)^2)
= E(A^2)-E(A)^2+E(B^2)-E(B)^2 - since E(AB)=E(A)E(B) for A,B independent
= Var(A)+Var(B)

2013-05-13 12:07:46 補充：
For A,B independent, assume A follows a(x) and b follows b(y).
E(AB) = sum(x)sum(y)(xy)a(x)b(y)
= sum(x)xa(x)sum(y)yb(y)
= E(A)E(B)

2013-05-13 21:19:56 補充：
When n tends to infinity, E(Y) -> u (independent of n) and Var(Y) -> 0 (proportional to 1/n)