Question

Suppose you have a box with values [-1 0 1] each with equal probability. You want to find the sum of 100 draws, but instead of the usual sampling with replacement, you envision a different procedure.
You decide to sample the box 50 times with replacement to get the initial 50 draws.
X1,X2,X3,··· ,X50
Then, randomly re-order the first 50 to get the next 50 draws (i.e. a total of 100). That is, the values remain the same, just the order has changed. For instance, a possible re-ordering could be: X48, X25, X33, · · · , X8. This would give you the final sequence of 100 draws:
X1,X2,X3,··· ,X50, X48,X25,X33,··· ,X8

Calculate the variance for the sum.

Answer 1

E(Xi) = 1/3*(-1) + 1/3*0 +1/3*1 = 0     ( given, P(Xi=-1) = P(Xi=0) = P(Xi=1) )

Variance of each Xi = E(X²) -E(X)² = 1/3 * (1²+0²+(-1)²) - 0² = 2/3

Variance of the sum of first 50 Xi's (say V1) in the 100-long sequence is the same as that of the next 50 Xi's (say V2), since the same set of variables have been reordered.

Now, the variance of sum of first 50 Xi's or (X1 + X2 + ....X50),

V1 = 50*(2/3)² (recall that V(X+Y) = Var(X) + Var(Y) if X and Y are independent)

=> Variance of sum of next 50 Xi's, V2 = V1 = 50*(2/3)² (as it's basically the same sum re-ordered)

Variance for the sum (X1+X2+.....X50 + X48+X25+......+X8) = V1+V2 = 100*(2/3)² = 44.444