# How to show that $E[\hat\sigma^2_x]= \sigma^2_x$?

Suppose you want to estimate $\sigma^2_x$, the variance of a random variable $X$. We want to study the properties of a couple of estimators for the variance. You have an iid sample of $n$ observations. Let $\mu_x=E(X)$. You know that by definition, $\sigma^2_x = E[(X-\mu_x)^2]$. Note that the variance is just an expected value. How do we estimate an expected value? Using a sample mean! So, one estimator for the variance might be

$$\hat \sigma^2_x = \frac1n \sum_{i=1}^n (X_i - \mu_x)^2$$

Show that $E[\hat\sigma^2_x]= \sigma^2_x$?

How would I go about solving this?

• Hint: linearity of expectation. – Dilip Sarwate Oct 2 '15 at 19:27
• Some resources here and here? – SecretAgentMan Oct 10 '18 at 6:15

## 1 Answer

$$E[\hat \sigma^2_x]=\frac{1}{n} E \left[ \sum_{i=1}^n (X_i^2-2X_i\mu_x+\mu_x^2) \right] = E[X_1^2]-2\mu_x^2+\mu_x^2=\sigma_x^2$$

• I wonder if the value of $E[(X_i-\mu)^2]$ is called by some special name in the literature. – Dilip Sarwate Oct 3 '15 at 2:27