Question

Let X1,X2,............Xn be a random sample with E(Xi)=m and var( Xi)= sigma^2 for all I= 1,2.....n . show that S^2=1/n-1 sum ( Xi-mean) is an unbiased estimator of sigma^2.

Answer 1

Answer on Question#65384 – Math – Statistics and Probability

Question. Let $X_{1}, X_{2}, \ldots, X_{n}$ be a random sample with $E(X_{i}) = m$ and $Var(X_{i}) = \sigma^{2}$ for all $i = 1, 2, \ldots, n$. Show that $S^{2} = \frac{1}{n - 1}\sum_{i = 1}^{n}(X_{i} - \bar{X})^{2}$ is an unbiased estimator of $\sigma^{2}$.

Proof. Let us compute $E(S^2) = \frac{1}{n - 1}\sum_{i = 1}^{n}E[(X_i - \bar{X})^2 ] = \frac{1}{n - 1}\sum_{i = 1}^{n}E\left[\left((X_i - m) - (\bar{X} -m)\right)^2\right]$

$$\begin{array}{l} = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} E [ (X _ {i} - m) ^ {2} - 2 (X _ {i} - m) (\bar {X} - m) + (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} E [ (X _ {i} - m) ^ {2} ] \\ - \frac {2}{n - 1} E \sum_ {i = 1} ^ {n} (X _ {i} - m) (\bar {X} - m) + \frac {1}{n - 1} \sum_ {i = 1} ^ {n} E [ (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} V a r (X _ {i}) \\ - \frac {2}{n - 1} E (\bar {X} - m) \sum_ {i = 1} ^ {n} (X _ {i} - m) + \frac {n}{n - 1} E [ (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} V a r (X _ {i}) \\ - \frac {2}{n - 1} E (\bar {X} - m) \left(\sum_ {i = 1} ^ {n} X _ {i} - m n\right) + \frac {n}{n - 1} E [ (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} V a r (X _ {i}) \\ - \frac {2}{n - 1} E (\bar {X} - m) (n \bar {X} - n m) + \frac {n}{n - 1} E [ (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} V a r (X _ {i}) - \frac {2 n}{n - 1} E [ (\bar {X} - m) ^ {2} ] \\ + \frac {n}{n - 1} E [ (\bar {X} - m) ^ {2} ] = \frac {1}{n - 1} \sum_ {i = 1} ^ {n} V a r (X _ {i}) - \frac {n}{n - 1} E [ (\bar {X} - m) ^ {2} ] = \frac {n \sigma^ {2}}{n - 1} - \frac {n}{n - 1} V a r (\bar {X}) \\ = \frac {n}{n - 1} \left(\sigma^ {2} - V a r \left[ \frac {1}{n} \sum_ {i = 1} ^ {n} X _ {i} \right]\right) = \frac {n}{n - 1} \left(\sigma^ {2} - \frac {1}{n ^ {2}} \sum_ {i = 1} ^ {n} V a r (X _ {i})\right) = \frac {n}{n - 1} \left(\sigma^ {2} - \frac {n \sigma^ {2}}{n ^ {2}}\right) \\ = \frac {n \sigma^ {2}}{n - 1} \left(1 - \frac {1}{n}\right) = \frac {n \sigma^ {2}}{n - 1} \cdot \frac {n - 1}{n} = \sigma^ {2}. \text{ During these computations we used the following facts: } \\ \end{array}$$ $$V a r (X) = E \left[ \left(X - E (X)\right) ^ {2} \right] \text{ by definition (see https://en.wikipedia.org/wiki/Variance)};$$ \bar {X} = \frac {1}{n} \sum_ {i = 1} ^ {n} X _ {i} \text{ by definition (see https://en.wikipedia.org/wiki/Sample_mean_and_covariance)}, $$\text{hence } \sum_ {i = 1} ^ {n} X _ {i} = n \bar {X};$$ V a r (a X) = a ^ {2} V a r (X) \text{ (see https://en.wikipedia.org/wiki/Variance#Properties)}; $$V a r \left(\sum_ {i = 1} ^ {n} X _ {i}\right) = \sum_ {i = 1} ^ {n} V a r \left(X _ {i}\right) \text{ when } X _ {1}, X _ {2}, \dots , X _ {n} \text{ are independent}$$

(see https://en.wikipedia.org/wiki/Variance#Properties); in our case $X_{1}, X_{2}, \ldots, X_{n}$ are independent by definition of data sample

(see https://en.wikipedia.org/wiki/Sample(statistics)).

Since $E(S^2) = \sigma^2$ we conclude that $S^2 = \frac{1}{n - 1}\sum_{i = 1}^{n}(X_i - \bar{X})^2$ is an unbiased estimator of $\sigma^2$ by definition of unbiased estimator (see https://en.wikipedia.org/wiki/Estimator#Bias).

Assertion is established.

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