# Let `X_i` `i = 1,....,n` be independently and identically distributed (iid) stochastic variables and define `S^2 = 1/n sum_1^n (X_i-bar(X))^2` Show that `E[S^2] = ((n-1)/n)sigma^2` where ...

Let `X_i` `i = 1,....,n` be independently and identically distributed (iid) stochastic variables and define

`S^2 = 1/n sum_1^n (X_i-bar(X))^2`

Show that

`E[S^2] = ((n-1)/n)sigma^2`

where `sigma^2 = V[X_i]` and `bar(X) = sum_1^n (X_i/n)` (the mean of the `X_i`)

### 1 Answer | Add Yours

` `` ``E[S^2] = E[1/n sum_1^n (X_i- bar(X))^2]`

`= E[1/n{sum_1^n(X_i^2 - 2X_ibar(X) + bar(X)^2)}]`

`= E[1/n sum_1^n X_i^2 - 2bar(X) sum_1^n(X_i/n) + (nbar(X)^2)/n]`

`= E[1/n sum_1^n X_i^2 - 2bar(X)^2 + bar(X)^2]` since `sum_1^n(X_i/n) = bar(X)`

`= 1/n sum_1^n E[X_i^2] - E[bar(X)^2]` taking the expectation within the sum` `.

`= 1/n sum_1^n (V[X_i] + E[X_i]^2) - V[bar(X)] - E[bar(X)]^2 ` using `V[X] = E[X^2] - E[X]^2`

` = 1/n sum_1^n (sigma^2 + mu^2) - sigma^2/n - mu^2` where `sigma^2 = V[X_i]` and `mu = E[X_i] = E[bar(X)]`

`= ((n-1)/n)sigma^2 + mu^2 - mu^2`

`= ((n-1)/n)sigma^2` **as required**