Question

For the CLT, we try to calculate V ar(Zn) and realize that V ar(Sn) needs to be calculated first. Recall that Sn = X1 + X2 +· · ·+ Xn, where Xi ’s are all independent of each other and identically distributed. Calculate E[Sn] and V ar[Sn].

Answer 1

Answer:E[Sn] = E n Xi =

i=1

n E[Xi] = np

i=1

Var[Sn] =Var n Xi =

i=1

n Var[Xi] = np(1 − p)

i=1

Step-by-step explanation:

Retract some basic facts on expectation and variance, where Y,Y1,Y2 are random variables.

• E[Y1 + Y2] = E[Y1] + E[Y2]

• E[Y1Y2] = E[Y1]E[Y2] if random variables Y1 and Y2 are independent (Y1 ⊥ Y2)

• E[cY ] = cE[Y ], E[c + Y ] = c + E[Y ], where c is a constant

• If we denote µ = E[Y ], the variance of Y

Var[Y ] = E[(Y − µ)2] = E[Y 2 − 2µY + µ2]

= E[Y 2] − 2µE[Y ] + µ2 = E[Y 2] − E[Y ]2

• If Y1 ⊥ Y2

Var[Y1 + Y2] = E[(Y1 + Y2)2] − E[(Y1 + Y2)]2

= (E[Y 21 ] + 2 E[Y1Y2] + E[Y 22 ]) − (E[Y1]2 + 2 E[Y1]E[Y2] + E[Y2]2) = (E[Y 21 ] − E[Y1]2) + (E[Y 22 ] − E[Y2]2) =Var[Y1] +Var[Y2]

• Var[cY ] = c2Var[Y ], Var[c + Y ] =Var[Y ]

• The standard derivation σ = stddev[Y ] = Var[Y ]

For any Xi we have the expectation E[Xi] = 1·Pr[Xi = 1]+0·Pr[Xi = 0] = p, therefore E[Sn] = E n Xi =

i=1

n E[Xi] = np

i=1

Since the variance of Xi Var[Xi] = E[X2i ] − E[Xi]2 = p − p2 = p(1 − p) and Xi’s are not dependent, the variance of Sn is

Var[Sn] =Var n Xi =

i=1

n Var[Xi] = np(1 − p)